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Computing a trigonometric integral
Help computing integral of quarter contourA trigonometric integral identityComputing an integral arising in potential theoryDefinite integral which does not evaluateEvaluate $int_0^piln(cos(x)+1)cos(nx),dx$solving exponential of trigonometric function inside an integral.How to convert this integral to an elliptic integral?Why does this integral not depend on the parameter?Difficult trigonometric integral. [Solved]Solving the Integral without Cauchy Integral formula.
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I am studying the integral
$$I=int_-pi/2^pi/2
frac28cos^2(theta)+10cos(theta)sin(theta)-28sin^2(theta)2cos^4(theta)+3cos^2(theta)sin^2(theta)+msin^4(theta)dtheta,$$
where $m>0$. In the problem I am working, it is very important to know what value of $m$ makes this integral positive, negative or equal to zero. I found that if $m=2$, then the integral is zero (introducing it in Wolfram Alpha). However, I don't know how to prove it formally.
On the other hand, I tried some values of $m$ in Wolfram and it seems that if $m<2$, then the integral is negative, and if $m>2$, the integral is positive. But again, I have no proof of this.
Any ideas of how to approach this problem?
Just in case, I found this alternative representation of the integral
$$I=int_-pi/2^pi/2
frac8[5sin(2theta)+28cos(2theta)]cos(4theta)+15+8(m-2)sin^4(theta)dtheta.$$
Any help would be appreciated.
integration definite-integrals
asked 8 hours ago
user326159user326159
1,4771 gold badge9 silver badges22 bronze badges
$endgroup$
add a comment |
$begingroup$
I am studying the integral
$$I=int_-pi/2^pi/2
frac28cos^2(theta)+10cos(theta)sin(theta)-28sin^2(theta)2cos^4(theta)+3cos^2(theta)sin^2(theta)+msin^4(theta)dtheta,$$
where $m>0$. In the problem I am working, it is very important to know what value of $m$ makes this integral positive, negative or equal to zero. I found that if $m=2$, then the integral is zero (introducing it in Wolfram Alpha). However, I don't know how to prove it formally.
On the other hand, I tried some values of $m$ in Wolfram and it seems that if $m<2$, then the integral is negative, and if $m>2$, the integral is positive. But again, I have no proof of this.
Any ideas of how to approach this problem?
Just in case, I found this alternative representation of the integral
$$I=int_-pi/2^pi/2
frac8[5sin(2theta)+28cos(2theta)]cos(4theta)+15+8(m-2)sin^4(theta)dtheta.$$
Any help would be appreciated.
integration definite-integrals
asked 8 hours ago
user326159user326159
1,4771 gold badge9 silver badges22 bronze badges
$endgroup$
$begingroup$
This looks like a messy integral. Do you really need a formal proof? Why not just plot or table the integral numerically (in Mathematica, Matlab etc) for some range of $m$ and see if your guess is right
$endgroup$
– Yuriy S
8 hours ago
$begingroup$
Surely it is $8(m-2)sin^4theta$ not $(m-2)sin^4theta$ in the denominator of the last equation?
$endgroup$
– user10354138
7 hours ago
$begingroup$
Yes. I've corrected it
$endgroup$
– user326159
7 hours ago
1
$begingroup$
Writing $alpha^2 = m/2$, we get $$ I(alpha) = 14 pi left( 1-frac1alpharight)sqrtfrac24alpha+3. $$ From this, we easily deduce the sign of $I$ as function of $alpha$, i.e., $operatornamesign(I(alpha)) = operatornamesign(alpha - 1)$. Now if you ask me how I arrived this expression, it is basically from an ugly residue computation applied to $$I(alpha) = int_-infty^infty frac28(1-t^2)2+3t^2+2alpha^2 t^4, mathrmdt, $$ but I have not enough energy to write up all the intermediate steps...
$endgroup$
– Sangchul Lee
5 hours ago
add a comment |
$begingroup$
I am studying the integral
$$I=int_-pi/2^pi/2
frac28cos^2(theta)+10cos(theta)sin(theta)-28sin^2(theta)2cos^4(theta)+3cos^2(theta)sin^2(theta)+msin^4(theta)dtheta,$$
where $m>0$. In the problem I am working, it is very important to know what value of $m$ makes this integral positive, negative or equal to zero. I found that if $m=2$, then the integral is zero (introducing it in Wolfram Alpha). However, I don't know how to prove it formally.
On the other hand, I tried some values of $m$ in Wolfram and it seems that if $m<2$, then the integral is negative, and if $m>2$, the integral is positive. But again, I have no proof of this.
Any ideas of how to approach this problem?
Just in case, I found this alternative representation of the integral
$$I=int_-pi/2^pi/2
frac8[5sin(2theta)+28cos(2theta)]cos(4theta)+15+8(m-2)sin^4(theta)dtheta.$$
Any help would be appreciated.
integration definite-integrals
asked 8 hours ago
user326159user326159
1,4771 gold badge9 silver badges22 bronze badges
$endgroup$
I am studying the integral
$$I=int_-pi/2^pi/2
frac28cos^2(theta)+10cos(theta)sin(theta)-28sin^2(theta)2cos^4(theta)+3cos^2(theta)sin^2(theta)+msin^4(theta)dtheta,$$
where $m>0$. In the problem I am working, it is very important to know what value of $m$ makes this integral positive, negative or equal to zero. I found that if $m=2$, then the integral is zero (introducing it in Wolfram Alpha). However, I don't know how to prove it formally.
On the other hand, I tried some values of $m$ in Wolfram and it seems that if $m<2$, then the integral is negative, and if $m>2$, the integral is positive. But again, I have no proof of this.
Any ideas of how to approach this problem?
Just in case, I found this alternative representation of the integral
$$I=int_-pi/2^pi/2
frac8[5sin(2theta)+28cos(2theta)]cos(4theta)+15+8(m-2)sin^4(theta)dtheta.$$
Any help would be appreciated.
integration definite-integrals
integration definite-integrals
asked 8 hours ago
user326159user326159
1,4771 gold badge9 silver badges22 bronze badges
asked 8 hours ago
user326159user326159
1,4771 gold badge9 silver badges22 bronze badges
edited 7 hours ago
user326159
asked 8 hours ago
user326159user326159
1,4771 gold badge9 silver badges22 bronze badges
asked 8 hours ago
user326159user326159
1,4771 gold badge9 silver badges22 bronze badges
asked 8 hours ago
user326159user326159
1,4771 gold badge9 silver badges22 bronze badges
1,4771 gold badge9 silver badges22 bronze badges
$begingroup$
This looks like a messy integral. Do you really need a formal proof? Why not just plot or table the integral numerically (in Mathematica, Matlab etc) for some range of $m$ and see if your guess is right
$endgroup$
– Yuriy S
8 hours ago
$begingroup$
Surely it is $8(m-2)sin^4theta$ not $(m-2)sin^4theta$ in the denominator of the last equation?
$endgroup$
– user10354138
7 hours ago
$begingroup$
Yes. I've corrected it
$endgroup$
– user326159
7 hours ago
1
$begingroup$
Writing $alpha^2 = m/2$, we get $$ I(alpha) = 14 pi left( 1-frac1alpharight)sqrtfrac24alpha+3. $$ From this, we easily deduce the sign of $I$ as function of $alpha$, i.e., $operatornamesign(I(alpha)) = operatornamesign(alpha - 1)$. Now if you ask me how I arrived this expression, it is basically from an ugly residue computation applied to $$I(alpha) = int_-infty^infty frac28(1-t^2)2+3t^2+2alpha^2 t^4, mathrmdt, $$ but I have not enough energy to write up all the intermediate steps...
$endgroup$
– Sangchul Lee
5 hours ago
add a comment |
$begingroup$
This looks like a messy integral. Do you really need a formal proof? Why not just plot or table the integral numerically (in Mathematica, Matlab etc) for some range of $m$ and see if your guess is right
$endgroup$
– Yuriy S
8 hours ago
$begingroup$
Surely it is $8(m-2)sin^4theta$ not $(m-2)sin^4theta$ in the denominator of the last equation?
$endgroup$
– user10354138
7 hours ago
$begingroup$
Yes. I've corrected it
$endgroup$
– user326159
7 hours ago
1
$begingroup$
Writing $alpha^2 = m/2$, we get $$ I(alpha) = 14 pi left( 1-frac1alpharight)sqrtfrac24alpha+3. $$ From this, we easily deduce the sign of $I$ as function of $alpha$, i.e., $operatornamesign(I(alpha)) = operatornamesign(alpha - 1)$. Now if you ask me how I arrived this expression, it is basically from an ugly residue computation applied to $$I(alpha) = int_-infty^infty frac28(1-t^2)2+3t^2+2alpha^2 t^4, mathrmdt, $$ but I have not enough energy to write up all the intermediate steps...
$endgroup$
– Sangchul Lee
5 hours ago
$begingroup$
This looks like a messy integral. Do you really need a formal proof? Why not just plot or table the integral numerically (in Mathematica, Matlab etc) for some range of $m$ and see if your guess is right
$endgroup$
– Yuriy S
8 hours ago
$begingroup$
This looks like a messy integral. Do you really need a formal proof? Why not just plot or table the integral numerically (in Mathematica, Matlab etc) for some range of $m$ and see if your guess is right
$endgroup$
– Yuriy S
8 hours ago
$begingroup$
Surely it is $8(m-2)sin^4theta$ not $(m-2)sin^4theta$ in the denominator of the last equation?
$endgroup$
– user10354138
7 hours ago
$begingroup$
Surely it is $8(m-2)sin^4theta$ not $(m-2)sin^4theta$ in the denominator of the last equation?
$endgroup$
– user10354138
7 hours ago
$begingroup$
Yes. I've corrected it
$endgroup$
– user326159
7 hours ago
$begingroup$
Yes. I've corrected it
$endgroup$
– user326159
7 hours ago
1
1
$begingroup$
Writing $alpha^2 = m/2$, we get $$ I(alpha) = 14 pi left( 1-frac1alpharight)sqrtfrac24alpha+3. $$ From this, we easily deduce the sign of $I$ as function of $alpha$, i.e., $operatornamesign(I(alpha)) = operatornamesign(alpha - 1)$. Now if you ask me how I arrived this expression, it is basically from an ugly residue computation applied to $$I(alpha) = int_-infty^infty frac28(1-t^2)2+3t^2+2alpha^2 t^4, mathrmdt, $$ but I have not enough energy to write up all the intermediate steps...
$endgroup$
– Sangchul Lee
5 hours ago
$begingroup$
Writing $alpha^2 = m/2$, we get $$ I(alpha) = 14 pi left( 1-frac1alpharight)sqrtfrac24alpha+3. $$ From this, we easily deduce the sign of $I$ as function of $alpha$, i.e., $operatornamesign(I(alpha)) = operatornamesign(alpha - 1)$. Now if you ask me how I arrived this expression, it is basically from an ugly residue computation applied to $$I(alpha) = int_-infty^infty frac28(1-t^2)2+3t^2+2alpha^2 t^4, mathrmdt, $$ but I have not enough energy to write up all the intermediate steps...
$endgroup$
– Sangchul Lee
5 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
note that since the function part of the function is odd i.e:
$$f(x)=frac28cos^2x+10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4x$$
$$f(-x)=frac28cos^2x-10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4x$$
you could notice that the integral can be simplified to:
$$int_-pi/2^pi/2frac28cos^2x+10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=int_-pi/2^pi/2frac28cos^2x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=2int_0^pi/2frac28cos^2x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=56int_0^pi/2fraccos^2x-sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
One route you could try to take is Tangent half-angle substitution, which yields:
$$112int_0^1(1+t^2)frac(1-t^2)^2-(2t)^22(1-t^2)^4+3(1-t^2)^2(2t)^2+m(2t)^4dt$$
the bottom of this fraction can be expanded to:
$$2t^8+4t^6+16t^4m-12t^4+4t^2+2$$
this may be factorisable for certain values of $m$
answered 7 hours ago
Henry LeeHenry Lee
2,6891 gold badge4 silver badges19 bronze badges
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add a comment |
$begingroup$
This problem is "nice" in the sense that the integrand is really trig function of $2theta$
$$
I=int_-pi/2^pi/2frac28cos 2theta+5sin2thetafrac12(1+cos2theta)^2+frac34sin^2 2theta+fracm4(cos2theta-1)^2,mathrmdtheta
$$
So
$$
I=frac12int_-pi^pifrac28cosphi+5sinphifrac12(1+cosphi)^2+frac34sin^2 phi+fracm4(cosphi-1)^2,mathrmdphi
$$
which we can rewrite as a contour integral of a rational function over the unit circle $z=e^iphi$, $-pileqphileqpi$ in $mathbbC$, hence it is just a matter of computing residues.
So
$$
I=frac12int_mathbbTfrac28cdotfrac12(z+z^-1)+5cdotfrac12i(z-z^-1)frac12(1+frac12(z+z^-1))^2+frac34(-frac14(z-z^-1)^2)+fracm4(frac12(z+z^-1)-1)^2,fracmathrmdziz
$$
which simplifies to
$$
I=frac12iint_mathbbT
frac8 ((28 + 5 i) + (28 - 5 i) z^2),mathrmdz(m-1)(z^4+1)-4(m-2)(z^3+z)+6(m-3)z^2
$$
The poles are at, if $mneq 1$,
$$
z+z^-1=frac2(m-2pmsqrtm)m-1.
$$
so, since $m>0$
$$labeleq:poles
z=
begincases
0,frac12(3pmsqrt5)& m=1\
frac2pmsqrtmpmsqrt3pm 2sqrtm1pmsqrtm&mneq 1.
endcasestag$star$
$$
So all poles are at worst simple unless $m=frac49$ (in which case we get a double pole at $z=4$ which we can ignore), and
$$
I=pisum_substacklvert z_irvert<1\z_iineqrefeq:polesoperatornameres_z_i(dots)+(textcorrection if lvert z_irvert=1).
$$
answered 7 hours ago
user10354138user10354138
17.3k2 gold badges12 silver badges32 bronze badges
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$begingroup$
Can you give more details of this method?
$endgroup$
– user326159
7 hours ago
add a comment |
$begingroup$
Define the function $mathcalI:mathbbR_>0rightarrowmathbbR$ via the trigonometric integral
$$beginalign
mathcalIleft(muright)
&:=int_-fracpi2^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright).\
endalign$$
Let $muinmathbbR_>0$. Since the integral $mathcalI$ has a symmetric interval of integration, the integral of the odd component of the integrand vanishes identically:
$$beginalign
mathcalIleft(muright)
&=int_-fracpi2^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_-fracpi2^0mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&~~~~~+int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)-10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright);~~~smallleft[thetamapsto-thetaright]\
&~~~~~+int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_0^fracpi2mathrmdtheta,frac56cos^2left(thetaright)-56sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,fraccos^2left(thetaright)-sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright).\
endalign$$
Using the double-angle formulas for sine and cosine,
$$beginalign
sinleft(2thetaright)
&=2sinleft(thetaright)cosleft(thetaright),\
cosleft(2thetaright)
&=cos^2left(thetaright)-sin^2left(thetaright)\
&=2cos^2left(thetaright)-1\
&=1-2sin^2left(thetaright),\
endalign$$
we can rewrite the integral as
$$beginalign
mathcalIleft(muright)
&=56int_0^fracpi2mathrmdtheta,fraccos^2left(thetaright)-sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,fraccosleft(2thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,frac4cosleft(2thetaright)8cos^4left(thetaright)+12cos^2left(thetaright)sin^2left(thetaright)+4musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,frac4cosleft(2thetaright)2left[1+cosleft(2thetaright)right]^2+3sin^2left(2thetaright)+muleft[1-cosleft(2thetaright)right]^2\
&=56int_0^pimathrmdtheta,frac2cosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2;~~~smallleft[thetamapstofrac12thetaright]\
&=112int_0^pimathrmdtheta,fraccosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2.\
endalign$$
Using the tangent half-angle substitution, the trigonometric integral transforms as
$$beginalign
mathcalIleft(muright)
&=112int_0^pimathrmdtheta,fraccosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2\
&=112int_0^inftymathrmdt,frac21+t^2cdotfracleft(frac1-t^21+t^2right)2left(1+frac1-t^21+t^2right)^2+3left(frac2t1+t^2right)^2+muleft(1-frac1-t^21+t^2right)^2;~~~smallleft[theta=2arctanleft(tright)right]\
&=int_0^inftymathrmdt,frac56left(1-t^2right)2+3t^2+mu,t^4.\
endalign$$
Setting $sqrtfrac2mu=:ainmathbbR_>0$ and $frac34a=:binmathbbR_>0$, we then have
$$beginalign
mathcalIleft(muright)
&=mathcalIleft(frac2a^2right)\
&=int_0^inftymathrmdt,frac56left(1-t^2right)2+3t^2+2a^-2t^4\
&=sqrtaint_0^inftymathrmdu,frac56left(1-au^2right)2+3au^2+2u^4;~~~smallleft[t=usqrtaright]\
&=frac13sqrtaint_0^inftymathrmdu,frac28left(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^inftymathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&~~~~~+frac283sqrt[4]frac2muint_1^inftymathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&~~~~~+frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3u^2-4bright)1+2bu^2+u^4;~~~smallleft[umapstofrac1uright]\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bright)left(1+u^2right)1+2bu^2+u^4\
&=28left(1-frac43bright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=28left(1-aright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=28left(1-sqrtfrac2muright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=frac28left(mu-2right)left(sqrtmu+sqrt2right)sqrtmusqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4.\
endalign$$
Since the integral factor in the last line above is a positive quantity, we don't need to actually solve the integral to determine the sign of $mathcalI$. We have
$$operatornamesgnleft(mathcalIleft(muright)right)=operatornamesgnleft(mu-2right),$$
as you originally conjectured.
answered 3 hours ago
David HDavid H
22.2k2 gold badges49 silver badges96 bronze badges
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3 Answers
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3 Answers
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votes
$begingroup$
note that since the function part of the function is odd i.e:
$$f(x)=frac28cos^2x+10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4x$$
$$f(-x)=frac28cos^2x-10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4x$$
you could notice that the integral can be simplified to:
$$int_-pi/2^pi/2frac28cos^2x+10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=int_-pi/2^pi/2frac28cos^2x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=2int_0^pi/2frac28cos^2x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=56int_0^pi/2fraccos^2x-sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
One route you could try to take is Tangent half-angle substitution, which yields:
$$112int_0^1(1+t^2)frac(1-t^2)^2-(2t)^22(1-t^2)^4+3(1-t^2)^2(2t)^2+m(2t)^4dt$$
the bottom of this fraction can be expanded to:
$$2t^8+4t^6+16t^4m-12t^4+4t^2+2$$
this may be factorisable for certain values of $m$
answered 7 hours ago
Henry LeeHenry Lee
2,6891 gold badge4 silver badges19 bronze badges
$endgroup$
add a comment |
$begingroup$
note that since the function part of the function is odd i.e:
$$f(x)=frac28cos^2x+10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4x$$
$$f(-x)=frac28cos^2x-10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4x$$
you could notice that the integral can be simplified to:
$$int_-pi/2^pi/2frac28cos^2x+10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=int_-pi/2^pi/2frac28cos^2x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=2int_0^pi/2frac28cos^2x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=56int_0^pi/2fraccos^2x-sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
One route you could try to take is Tangent half-angle substitution, which yields:
$$112int_0^1(1+t^2)frac(1-t^2)^2-(2t)^22(1-t^2)^4+3(1-t^2)^2(2t)^2+m(2t)^4dt$$
the bottom of this fraction can be expanded to:
$$2t^8+4t^6+16t^4m-12t^4+4t^2+2$$
this may be factorisable for certain values of $m$
answered 7 hours ago
Henry LeeHenry Lee
2,6891 gold badge4 silver badges19 bronze badges
$endgroup$
add a comment |
$begingroup$
note that since the function part of the function is odd i.e:
$$f(x)=frac28cos^2x+10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4x$$
$$f(-x)=frac28cos^2x-10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4x$$
you could notice that the integral can be simplified to:
$$int_-pi/2^pi/2frac28cos^2x+10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=int_-pi/2^pi/2frac28cos^2x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=2int_0^pi/2frac28cos^2x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=56int_0^pi/2fraccos^2x-sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
One route you could try to take is Tangent half-angle substitution, which yields:
$$112int_0^1(1+t^2)frac(1-t^2)^2-(2t)^22(1-t^2)^4+3(1-t^2)^2(2t)^2+m(2t)^4dt$$
the bottom of this fraction can be expanded to:
$$2t^8+4t^6+16t^4m-12t^4+4t^2+2$$
this may be factorisable for certain values of $m$
answered 7 hours ago
Henry LeeHenry Lee
2,6891 gold badge4 silver badges19 bronze badges
$endgroup$
note that since the function part of the function is odd i.e:
$$f(x)=frac28cos^2x+10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4x$$
$$f(-x)=frac28cos^2x-10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4x$$
you could notice that the integral can be simplified to:
$$int_-pi/2^pi/2frac28cos^2x+10cos xsin x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=int_-pi/2^pi/2frac28cos^2x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=2int_0^pi/2frac28cos^2x-28sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
$$=56int_0^pi/2fraccos^2x-sin^2x2cos^4x+3cos^2xsin^2x+msin^4xdx$$
One route you could try to take is Tangent half-angle substitution, which yields:
$$112int_0^1(1+t^2)frac(1-t^2)^2-(2t)^22(1-t^2)^4+3(1-t^2)^2(2t)^2+m(2t)^4dt$$
the bottom of this fraction can be expanded to:
$$2t^8+4t^6+16t^4m-12t^4+4t^2+2$$
this may be factorisable for certain values of $m$
answered 7 hours ago
Henry LeeHenry Lee
2,6891 gold badge4 silver badges19 bronze badges
edited 7 hours ago
answered 7 hours ago
Henry LeeHenry Lee
2,6891 gold badge4 silver badges19 bronze badges
answered 7 hours ago
Henry LeeHenry Lee
2,6891 gold badge4 silver badges19 bronze badges
answered 7 hours ago
Henry LeeHenry Lee
2,6891 gold badge4 silver badges19 bronze badges
2,6891 gold badge4 silver badges19 bronze badges
add a comment |
add a comment |
$begingroup$
This problem is "nice" in the sense that the integrand is really trig function of $2theta$
$$
I=int_-pi/2^pi/2frac28cos 2theta+5sin2thetafrac12(1+cos2theta)^2+frac34sin^2 2theta+fracm4(cos2theta-1)^2,mathrmdtheta
$$
So
$$
I=frac12int_-pi^pifrac28cosphi+5sinphifrac12(1+cosphi)^2+frac34sin^2 phi+fracm4(cosphi-1)^2,mathrmdphi
$$
which we can rewrite as a contour integral of a rational function over the unit circle $z=e^iphi$, $-pileqphileqpi$ in $mathbbC$, hence it is just a matter of computing residues.
So
$$
I=frac12int_mathbbTfrac28cdotfrac12(z+z^-1)+5cdotfrac12i(z-z^-1)frac12(1+frac12(z+z^-1))^2+frac34(-frac14(z-z^-1)^2)+fracm4(frac12(z+z^-1)-1)^2,fracmathrmdziz
$$
which simplifies to
$$
I=frac12iint_mathbbT
frac8 ((28 + 5 i) + (28 - 5 i) z^2),mathrmdz(m-1)(z^4+1)-4(m-2)(z^3+z)+6(m-3)z^2
$$
The poles are at, if $mneq 1$,
$$
z+z^-1=frac2(m-2pmsqrtm)m-1.
$$
so, since $m>0$
$$labeleq:poles
z=
begincases
0,frac12(3pmsqrt5)& m=1\
frac2pmsqrtmpmsqrt3pm 2sqrtm1pmsqrtm&mneq 1.
endcasestag$star$
$$
So all poles are at worst simple unless $m=frac49$ (in which case we get a double pole at $z=4$ which we can ignore), and
$$
I=pisum_substacklvert z_irvert<1\z_iineqrefeq:polesoperatornameres_z_i(dots)+(textcorrection if lvert z_irvert=1).
$$
answered 7 hours ago
user10354138user10354138
17.3k2 gold badges12 silver badges32 bronze badges
$endgroup$
$begingroup$
Can you give more details of this method?
$endgroup$
– user326159
7 hours ago
add a comment |
$begingroup$
This problem is "nice" in the sense that the integrand is really trig function of $2theta$
$$
I=int_-pi/2^pi/2frac28cos 2theta+5sin2thetafrac12(1+cos2theta)^2+frac34sin^2 2theta+fracm4(cos2theta-1)^2,mathrmdtheta
$$
So
$$
I=frac12int_-pi^pifrac28cosphi+5sinphifrac12(1+cosphi)^2+frac34sin^2 phi+fracm4(cosphi-1)^2,mathrmdphi
$$
which we can rewrite as a contour integral of a rational function over the unit circle $z=e^iphi$, $-pileqphileqpi$ in $mathbbC$, hence it is just a matter of computing residues.
So
$$
I=frac12int_mathbbTfrac28cdotfrac12(z+z^-1)+5cdotfrac12i(z-z^-1)frac12(1+frac12(z+z^-1))^2+frac34(-frac14(z-z^-1)^2)+fracm4(frac12(z+z^-1)-1)^2,fracmathrmdziz
$$
which simplifies to
$$
I=frac12iint_mathbbT
frac8 ((28 + 5 i) + (28 - 5 i) z^2),mathrmdz(m-1)(z^4+1)-4(m-2)(z^3+z)+6(m-3)z^2
$$
The poles are at, if $mneq 1$,
$$
z+z^-1=frac2(m-2pmsqrtm)m-1.
$$
so, since $m>0$
$$labeleq:poles
z=
begincases
0,frac12(3pmsqrt5)& m=1\
frac2pmsqrtmpmsqrt3pm 2sqrtm1pmsqrtm&mneq 1.
endcasestag$star$
$$
So all poles are at worst simple unless $m=frac49$ (in which case we get a double pole at $z=4$ which we can ignore), and
$$
I=pisum_substacklvert z_irvert<1\z_iineqrefeq:polesoperatornameres_z_i(dots)+(textcorrection if lvert z_irvert=1).
$$
answered 7 hours ago
user10354138user10354138
17.3k2 gold badges12 silver badges32 bronze badges
$endgroup$
$begingroup$
Can you give more details of this method?
$endgroup$
– user326159
7 hours ago
add a comment |
$begingroup$
This problem is "nice" in the sense that the integrand is really trig function of $2theta$
$$
I=int_-pi/2^pi/2frac28cos 2theta+5sin2thetafrac12(1+cos2theta)^2+frac34sin^2 2theta+fracm4(cos2theta-1)^2,mathrmdtheta
$$
So
$$
I=frac12int_-pi^pifrac28cosphi+5sinphifrac12(1+cosphi)^2+frac34sin^2 phi+fracm4(cosphi-1)^2,mathrmdphi
$$
which we can rewrite as a contour integral of a rational function over the unit circle $z=e^iphi$, $-pileqphileqpi$ in $mathbbC$, hence it is just a matter of computing residues.
So
$$
I=frac12int_mathbbTfrac28cdotfrac12(z+z^-1)+5cdotfrac12i(z-z^-1)frac12(1+frac12(z+z^-1))^2+frac34(-frac14(z-z^-1)^2)+fracm4(frac12(z+z^-1)-1)^2,fracmathrmdziz
$$
which simplifies to
$$
I=frac12iint_mathbbT
frac8 ((28 + 5 i) + (28 - 5 i) z^2),mathrmdz(m-1)(z^4+1)-4(m-2)(z^3+z)+6(m-3)z^2
$$
The poles are at, if $mneq 1$,
$$
z+z^-1=frac2(m-2pmsqrtm)m-1.
$$
so, since $m>0$
$$labeleq:poles
z=
begincases
0,frac12(3pmsqrt5)& m=1\
frac2pmsqrtmpmsqrt3pm 2sqrtm1pmsqrtm&mneq 1.
endcasestag$star$
$$
So all poles are at worst simple unless $m=frac49$ (in which case we get a double pole at $z=4$ which we can ignore), and
$$
I=pisum_substacklvert z_irvert<1\z_iineqrefeq:polesoperatornameres_z_i(dots)+(textcorrection if lvert z_irvert=1).
$$
answered 7 hours ago
user10354138user10354138
17.3k2 gold badges12 silver badges32 bronze badges
$endgroup$
This problem is "nice" in the sense that the integrand is really trig function of $2theta$
$$
I=int_-pi/2^pi/2frac28cos 2theta+5sin2thetafrac12(1+cos2theta)^2+frac34sin^2 2theta+fracm4(cos2theta-1)^2,mathrmdtheta
$$
So
$$
I=frac12int_-pi^pifrac28cosphi+5sinphifrac12(1+cosphi)^2+frac34sin^2 phi+fracm4(cosphi-1)^2,mathrmdphi
$$
which we can rewrite as a contour integral of a rational function over the unit circle $z=e^iphi$, $-pileqphileqpi$ in $mathbbC$, hence it is just a matter of computing residues.
So
$$
I=frac12int_mathbbTfrac28cdotfrac12(z+z^-1)+5cdotfrac12i(z-z^-1)frac12(1+frac12(z+z^-1))^2+frac34(-frac14(z-z^-1)^2)+fracm4(frac12(z+z^-1)-1)^2,fracmathrmdziz
$$
which simplifies to
$$
I=frac12iint_mathbbT
frac8 ((28 + 5 i) + (28 - 5 i) z^2),mathrmdz(m-1)(z^4+1)-4(m-2)(z^3+z)+6(m-3)z^2
$$
The poles are at, if $mneq 1$,
$$
z+z^-1=frac2(m-2pmsqrtm)m-1.
$$
so, since $m>0$
$$labeleq:poles
z=
begincases
0,frac12(3pmsqrt5)& m=1\
frac2pmsqrtmpmsqrt3pm 2sqrtm1pmsqrtm&mneq 1.
endcasestag$star$
$$
So all poles are at worst simple unless $m=frac49$ (in which case we get a double pole at $z=4$ which we can ignore), and
$$
I=pisum_substacklvert z_irvert<1\z_iineqrefeq:polesoperatornameres_z_i(dots)+(textcorrection if lvert z_irvert=1).
$$
answered 7 hours ago
user10354138user10354138
17.3k2 gold badges12 silver badges32 bronze badges
edited 7 hours ago
answered 7 hours ago
user10354138user10354138
17.3k2 gold badges12 silver badges32 bronze badges
answered 7 hours ago
user10354138user10354138
17.3k2 gold badges12 silver badges32 bronze badges
answered 7 hours ago
user10354138user10354138
17.3k2 gold badges12 silver badges32 bronze badges
17.3k2 gold badges12 silver badges32 bronze badges
$begingroup$
Can you give more details of this method?
$endgroup$
– user326159
7 hours ago
add a comment |
$begingroup$
Can you give more details of this method?
$endgroup$
– user326159
7 hours ago
$begingroup$
Can you give more details of this method?
$endgroup$
– user326159
7 hours ago
$begingroup$
Can you give more details of this method?
$endgroup$
– user326159
7 hours ago
add a comment |
$begingroup$
Define the function $mathcalI:mathbbR_>0rightarrowmathbbR$ via the trigonometric integral
$$beginalign
mathcalIleft(muright)
&:=int_-fracpi2^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright).\
endalign$$
Let $muinmathbbR_>0$. Since the integral $mathcalI$ has a symmetric interval of integration, the integral of the odd component of the integrand vanishes identically:
$$beginalign
mathcalIleft(muright)
&=int_-fracpi2^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_-fracpi2^0mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&~~~~~+int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)-10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright);~~~smallleft[thetamapsto-thetaright]\
&~~~~~+int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_0^fracpi2mathrmdtheta,frac56cos^2left(thetaright)-56sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,fraccos^2left(thetaright)-sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright).\
endalign$$
Using the double-angle formulas for sine and cosine,
$$beginalign
sinleft(2thetaright)
&=2sinleft(thetaright)cosleft(thetaright),\
cosleft(2thetaright)
&=cos^2left(thetaright)-sin^2left(thetaright)\
&=2cos^2left(thetaright)-1\
&=1-2sin^2left(thetaright),\
endalign$$
we can rewrite the integral as
$$beginalign
mathcalIleft(muright)
&=56int_0^fracpi2mathrmdtheta,fraccos^2left(thetaright)-sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,fraccosleft(2thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,frac4cosleft(2thetaright)8cos^4left(thetaright)+12cos^2left(thetaright)sin^2left(thetaright)+4musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,frac4cosleft(2thetaright)2left[1+cosleft(2thetaright)right]^2+3sin^2left(2thetaright)+muleft[1-cosleft(2thetaright)right]^2\
&=56int_0^pimathrmdtheta,frac2cosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2;~~~smallleft[thetamapstofrac12thetaright]\
&=112int_0^pimathrmdtheta,fraccosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2.\
endalign$$
Using the tangent half-angle substitution, the trigonometric integral transforms as
$$beginalign
mathcalIleft(muright)
&=112int_0^pimathrmdtheta,fraccosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2\
&=112int_0^inftymathrmdt,frac21+t^2cdotfracleft(frac1-t^21+t^2right)2left(1+frac1-t^21+t^2right)^2+3left(frac2t1+t^2right)^2+muleft(1-frac1-t^21+t^2right)^2;~~~smallleft[theta=2arctanleft(tright)right]\
&=int_0^inftymathrmdt,frac56left(1-t^2right)2+3t^2+mu,t^4.\
endalign$$
Setting $sqrtfrac2mu=:ainmathbbR_>0$ and $frac34a=:binmathbbR_>0$, we then have
$$beginalign
mathcalIleft(muright)
&=mathcalIleft(frac2a^2right)\
&=int_0^inftymathrmdt,frac56left(1-t^2right)2+3t^2+2a^-2t^4\
&=sqrtaint_0^inftymathrmdu,frac56left(1-au^2right)2+3au^2+2u^4;~~~smallleft[t=usqrtaright]\
&=frac13sqrtaint_0^inftymathrmdu,frac28left(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^inftymathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&~~~~~+frac283sqrt[4]frac2muint_1^inftymathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&~~~~~+frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3u^2-4bright)1+2bu^2+u^4;~~~smallleft[umapstofrac1uright]\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bright)left(1+u^2right)1+2bu^2+u^4\
&=28left(1-frac43bright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=28left(1-aright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=28left(1-sqrtfrac2muright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=frac28left(mu-2right)left(sqrtmu+sqrt2right)sqrtmusqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4.\
endalign$$
Since the integral factor in the last line above is a positive quantity, we don't need to actually solve the integral to determine the sign of $mathcalI$. We have
$$operatornamesgnleft(mathcalIleft(muright)right)=operatornamesgnleft(mu-2right),$$
as you originally conjectured.
answered 3 hours ago
David HDavid H
22.2k2 gold badges49 silver badges96 bronze badges
$endgroup$
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$begingroup$
Define the function $mathcalI:mathbbR_>0rightarrowmathbbR$ via the trigonometric integral
$$beginalign
mathcalIleft(muright)
&:=int_-fracpi2^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright).\
endalign$$
Let $muinmathbbR_>0$. Since the integral $mathcalI$ has a symmetric interval of integration, the integral of the odd component of the integrand vanishes identically:
$$beginalign
mathcalIleft(muright)
&=int_-fracpi2^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_-fracpi2^0mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&~~~~~+int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)-10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright);~~~smallleft[thetamapsto-thetaright]\
&~~~~~+int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_0^fracpi2mathrmdtheta,frac56cos^2left(thetaright)-56sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,fraccos^2left(thetaright)-sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright).\
endalign$$
Using the double-angle formulas for sine and cosine,
$$beginalign
sinleft(2thetaright)
&=2sinleft(thetaright)cosleft(thetaright),\
cosleft(2thetaright)
&=cos^2left(thetaright)-sin^2left(thetaright)\
&=2cos^2left(thetaright)-1\
&=1-2sin^2left(thetaright),\
endalign$$
we can rewrite the integral as
$$beginalign
mathcalIleft(muright)
&=56int_0^fracpi2mathrmdtheta,fraccos^2left(thetaright)-sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,fraccosleft(2thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,frac4cosleft(2thetaright)8cos^4left(thetaright)+12cos^2left(thetaright)sin^2left(thetaright)+4musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,frac4cosleft(2thetaright)2left[1+cosleft(2thetaright)right]^2+3sin^2left(2thetaright)+muleft[1-cosleft(2thetaright)right]^2\
&=56int_0^pimathrmdtheta,frac2cosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2;~~~smallleft[thetamapstofrac12thetaright]\
&=112int_0^pimathrmdtheta,fraccosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2.\
endalign$$
Using the tangent half-angle substitution, the trigonometric integral transforms as
$$beginalign
mathcalIleft(muright)
&=112int_0^pimathrmdtheta,fraccosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2\
&=112int_0^inftymathrmdt,frac21+t^2cdotfracleft(frac1-t^21+t^2right)2left(1+frac1-t^21+t^2right)^2+3left(frac2t1+t^2right)^2+muleft(1-frac1-t^21+t^2right)^2;~~~smallleft[theta=2arctanleft(tright)right]\
&=int_0^inftymathrmdt,frac56left(1-t^2right)2+3t^2+mu,t^4.\
endalign$$
Setting $sqrtfrac2mu=:ainmathbbR_>0$ and $frac34a=:binmathbbR_>0$, we then have
$$beginalign
mathcalIleft(muright)
&=mathcalIleft(frac2a^2right)\
&=int_0^inftymathrmdt,frac56left(1-t^2right)2+3t^2+2a^-2t^4\
&=sqrtaint_0^inftymathrmdu,frac56left(1-au^2right)2+3au^2+2u^4;~~~smallleft[t=usqrtaright]\
&=frac13sqrtaint_0^inftymathrmdu,frac28left(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^inftymathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&~~~~~+frac283sqrt[4]frac2muint_1^inftymathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&~~~~~+frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3u^2-4bright)1+2bu^2+u^4;~~~smallleft[umapstofrac1uright]\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bright)left(1+u^2right)1+2bu^2+u^4\
&=28left(1-frac43bright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=28left(1-aright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=28left(1-sqrtfrac2muright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=frac28left(mu-2right)left(sqrtmu+sqrt2right)sqrtmusqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4.\
endalign$$
Since the integral factor in the last line above is a positive quantity, we don't need to actually solve the integral to determine the sign of $mathcalI$. We have
$$operatornamesgnleft(mathcalIleft(muright)right)=operatornamesgnleft(mu-2right),$$
as you originally conjectured.
answered 3 hours ago
David HDavid H
22.2k2 gold badges49 silver badges96 bronze badges
$endgroup$
add a comment |
$begingroup$
Define the function $mathcalI:mathbbR_>0rightarrowmathbbR$ via the trigonometric integral
$$beginalign
mathcalIleft(muright)
&:=int_-fracpi2^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright).\
endalign$$
Let $muinmathbbR_>0$. Since the integral $mathcalI$ has a symmetric interval of integration, the integral of the odd component of the integrand vanishes identically:
$$beginalign
mathcalIleft(muright)
&=int_-fracpi2^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_-fracpi2^0mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&~~~~~+int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)-10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright);~~~smallleft[thetamapsto-thetaright]\
&~~~~~+int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_0^fracpi2mathrmdtheta,frac56cos^2left(thetaright)-56sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,fraccos^2left(thetaright)-sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright).\
endalign$$
Using the double-angle formulas for sine and cosine,
$$beginalign
sinleft(2thetaright)
&=2sinleft(thetaright)cosleft(thetaright),\
cosleft(2thetaright)
&=cos^2left(thetaright)-sin^2left(thetaright)\
&=2cos^2left(thetaright)-1\
&=1-2sin^2left(thetaright),\
endalign$$
we can rewrite the integral as
$$beginalign
mathcalIleft(muright)
&=56int_0^fracpi2mathrmdtheta,fraccos^2left(thetaright)-sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,fraccosleft(2thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,frac4cosleft(2thetaright)8cos^4left(thetaright)+12cos^2left(thetaright)sin^2left(thetaright)+4musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,frac4cosleft(2thetaright)2left[1+cosleft(2thetaright)right]^2+3sin^2left(2thetaright)+muleft[1-cosleft(2thetaright)right]^2\
&=56int_0^pimathrmdtheta,frac2cosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2;~~~smallleft[thetamapstofrac12thetaright]\
&=112int_0^pimathrmdtheta,fraccosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2.\
endalign$$
Using the tangent half-angle substitution, the trigonometric integral transforms as
$$beginalign
mathcalIleft(muright)
&=112int_0^pimathrmdtheta,fraccosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2\
&=112int_0^inftymathrmdt,frac21+t^2cdotfracleft(frac1-t^21+t^2right)2left(1+frac1-t^21+t^2right)^2+3left(frac2t1+t^2right)^2+muleft(1-frac1-t^21+t^2right)^2;~~~smallleft[theta=2arctanleft(tright)right]\
&=int_0^inftymathrmdt,frac56left(1-t^2right)2+3t^2+mu,t^4.\
endalign$$
Setting $sqrtfrac2mu=:ainmathbbR_>0$ and $frac34a=:binmathbbR_>0$, we then have
$$beginalign
mathcalIleft(muright)
&=mathcalIleft(frac2a^2right)\
&=int_0^inftymathrmdt,frac56left(1-t^2right)2+3t^2+2a^-2t^4\
&=sqrtaint_0^inftymathrmdu,frac56left(1-au^2right)2+3au^2+2u^4;~~~smallleft[t=usqrtaright]\
&=frac13sqrtaint_0^inftymathrmdu,frac28left(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^inftymathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&~~~~~+frac283sqrt[4]frac2muint_1^inftymathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&~~~~~+frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3u^2-4bright)1+2bu^2+u^4;~~~smallleft[umapstofrac1uright]\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bright)left(1+u^2right)1+2bu^2+u^4\
&=28left(1-frac43bright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=28left(1-aright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=28left(1-sqrtfrac2muright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=frac28left(mu-2right)left(sqrtmu+sqrt2right)sqrtmusqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4.\
endalign$$
Since the integral factor in the last line above is a positive quantity, we don't need to actually solve the integral to determine the sign of $mathcalI$. We have
$$operatornamesgnleft(mathcalIleft(muright)right)=operatornamesgnleft(mu-2right),$$
as you originally conjectured.
answered 3 hours ago
David HDavid H
22.2k2 gold badges49 silver badges96 bronze badges
$endgroup$
Define the function $mathcalI:mathbbR_>0rightarrowmathbbR$ via the trigonometric integral
$$beginalign
mathcalIleft(muright)
&:=int_-fracpi2^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright).\
endalign$$
Let $muinmathbbR_>0$. Since the integral $mathcalI$ has a symmetric interval of integration, the integral of the odd component of the integrand vanishes identically:
$$beginalign
mathcalIleft(muright)
&=int_-fracpi2^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_-fracpi2^0mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&~~~~~+int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)-10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright);~~~smallleft[thetamapsto-thetaright]\
&~~~~~+int_0^fracpi2mathrmdtheta,frac28cos^2left(thetaright)+10cosleft(thetaright)sinleft(thetaright)-28sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=int_0^fracpi2mathrmdtheta,frac56cos^2left(thetaright)-56sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,fraccos^2left(thetaright)-sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright).\
endalign$$
Using the double-angle formulas for sine and cosine,
$$beginalign
sinleft(2thetaright)
&=2sinleft(thetaright)cosleft(thetaright),\
cosleft(2thetaright)
&=cos^2left(thetaright)-sin^2left(thetaright)\
&=2cos^2left(thetaright)-1\
&=1-2sin^2left(thetaright),\
endalign$$
we can rewrite the integral as
$$beginalign
mathcalIleft(muright)
&=56int_0^fracpi2mathrmdtheta,fraccos^2left(thetaright)-sin^2left(thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,fraccosleft(2thetaright)2cos^4left(thetaright)+3cos^2left(thetaright)sin^2left(thetaright)+musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,frac4cosleft(2thetaright)8cos^4left(thetaright)+12cos^2left(thetaright)sin^2left(thetaright)+4musin^4left(thetaright)\
&=56int_0^fracpi2mathrmdtheta,frac4cosleft(2thetaright)2left[1+cosleft(2thetaright)right]^2+3sin^2left(2thetaright)+muleft[1-cosleft(2thetaright)right]^2\
&=56int_0^pimathrmdtheta,frac2cosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2;~~~smallleft[thetamapstofrac12thetaright]\
&=112int_0^pimathrmdtheta,fraccosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2.\
endalign$$
Using the tangent half-angle substitution, the trigonometric integral transforms as
$$beginalign
mathcalIleft(muright)
&=112int_0^pimathrmdtheta,fraccosleft(thetaright)2left[1+cosleft(thetaright)right]^2+3sin^2left(thetaright)+muleft[1-cosleft(thetaright)right]^2\
&=112int_0^inftymathrmdt,frac21+t^2cdotfracleft(frac1-t^21+t^2right)2left(1+frac1-t^21+t^2right)^2+3left(frac2t1+t^2right)^2+muleft(1-frac1-t^21+t^2right)^2;~~~smallleft[theta=2arctanleft(tright)right]\
&=int_0^inftymathrmdt,frac56left(1-t^2right)2+3t^2+mu,t^4.\
endalign$$
Setting $sqrtfrac2mu=:ainmathbbR_>0$ and $frac34a=:binmathbbR_>0$, we then have
$$beginalign
mathcalIleft(muright)
&=mathcalIleft(frac2a^2right)\
&=int_0^inftymathrmdt,frac56left(1-t^2right)2+3t^2+2a^-2t^4\
&=sqrtaint_0^inftymathrmdu,frac56left(1-au^2right)2+3au^2+2u^4;~~~smallleft[t=usqrtaright]\
&=frac13sqrtaint_0^inftymathrmdu,frac28left(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^inftymathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&~~~~~+frac283sqrt[4]frac2muint_1^inftymathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bu^2right)1+2bu^2+u^4\
&~~~~~+frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3u^2-4bright)1+2bu^2+u^4;~~~smallleft[umapstofrac1uright]\
&=frac283sqrt[4]frac2muint_0^1mathrmdu,fracleft(3-4bright)left(1+u^2right)1+2bu^2+u^4\
&=28left(1-frac43bright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=28left(1-aright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=28left(1-sqrtfrac2muright)sqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4\
&=frac28left(mu-2right)left(sqrtmu+sqrt2right)sqrtmusqrt[4]frac2muint_0^1mathrmdu,frac1+u^21+2bu^2+u^4.\
endalign$$
Since the integral factor in the last line above is a positive quantity, we don't need to actually solve the integral to determine the sign of $mathcalI$. We have
$$operatornamesgnleft(mathcalIleft(muright)right)=operatornamesgnleft(mu-2right),$$
as you originally conjectured.
answered 3 hours ago
David HDavid H
22.2k2 gold badges49 silver badges96 bronze badges
answered 3 hours ago
David HDavid H
22.2k2 gold badges49 silver badges96 bronze badges
answered 3 hours ago
David HDavid H
22.2k2 gold badges49 silver badges96 bronze badges
answered 3 hours ago
David HDavid H
22.2k2 gold badges49 silver badges96 bronze badges
22.2k2 gold badges49 silver badges96 bronze badges
add a comment |
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$begingroup$
This looks like a messy integral. Do you really need a formal proof? Why not just plot or table the integral numerically (in Mathematica, Matlab etc) for some range of $m$ and see if your guess is right
$endgroup$
– Yuriy S
8 hours ago
$begingroup$
Surely it is $8(m-2)sin^4theta$ not $(m-2)sin^4theta$ in the denominator of the last equation?
$endgroup$
– user10354138
7 hours ago
$begingroup$
Yes. I've corrected it
$endgroup$
– user326159
7 hours ago
1
$begingroup$
Writing $alpha^2 = m/2$, we get $$ I(alpha) = 14 pi left( 1-frac1alpharight)sqrtfrac24alpha+3. $$ From this, we easily deduce the sign of $I$ as function of $alpha$, i.e., $operatornamesign(I(alpha)) = operatornamesign(alpha - 1)$. Now if you ask me how I arrived this expression, it is basically from an ugly residue computation applied to $$I(alpha) = int_-infty^infty frac28(1-t^2)2+3t^2+2alpha^2 t^4, mathrmdt, $$ but I have not enough energy to write up all the intermediate steps...
$endgroup$
– Sangchul Lee
5 hours ago