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Contour integration with infinite poles
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Contour integration with infinite poles
contour integration of logarithm functionBest way to evaluate integral with contour integration?How many poles have to be inside the contour?Integral with contour integrationCan Cauchy principal values of functions with nonsimple poles be evaluated using complex contour integration methods?Are there real definite integrals which can only be evaluated by contour integration?Contour integral with two polesContour integration - complex analysis
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$begingroup$
I am trying to find the integral of the following expression:-
$$int_-infty^infty fracx^2e^k+e^x^2dx$$
I think this can be solved using contour integration. The poles will be the points with $z^2 = k+i(2n+1)pi$ according to me. But this results in infinite poles if you choose a contour covering the entire real axis. Does anybody have any insights to solve this integral?
integration complex-analysis contour-integration
asked 8 hours ago
Rishabh JainRishabh Jain
1958 bronze badges
$endgroup$
add a comment
|
$begingroup$
I am trying to find the integral of the following expression:-
$$int_-infty^infty fracx^2e^k+e^x^2dx$$
I think this can be solved using contour integration. The poles will be the points with $z^2 = k+i(2n+1)pi$ according to me. But this results in infinite poles if you choose a contour covering the entire real axis. Does anybody have any insights to solve this integral?
integration complex-analysis contour-integration
asked 8 hours ago
Rishabh JainRishabh Jain
1958 bronze badges
$endgroup$
1
$begingroup$
It is not a problem of infinitely poles. You can't use contour integration because the function doesn't $to 0$ for $|x|$ large on an half-plane.
$endgroup$
– reuns
3 hours ago
add a comment
|
$begingroup$
I am trying to find the integral of the following expression:-
$$int_-infty^infty fracx^2e^k+e^x^2dx$$
I think this can be solved using contour integration. The poles will be the points with $z^2 = k+i(2n+1)pi$ according to me. But this results in infinite poles if you choose a contour covering the entire real axis. Does anybody have any insights to solve this integral?
integration complex-analysis contour-integration
asked 8 hours ago
Rishabh JainRishabh Jain
1958 bronze badges
$endgroup$
I am trying to find the integral of the following expression:-
$$int_-infty^infty fracx^2e^k+e^x^2dx$$
I think this can be solved using contour integration. The poles will be the points with $z^2 = k+i(2n+1)pi$ according to me. But this results in infinite poles if you choose a contour covering the entire real axis. Does anybody have any insights to solve this integral?
integration complex-analysis contour-integration
integration complex-analysis contour-integration
asked 8 hours ago
Rishabh JainRishabh Jain
1958 bronze badges
asked 8 hours ago
Rishabh JainRishabh Jain
1958 bronze badges
asked 8 hours ago
Rishabh JainRishabh Jain
1958 bronze badges
asked 8 hours ago
Rishabh JainRishabh Jain
1958 bronze badges
asked 8 hours ago
Rishabh JainRishabh Jain
1958 bronze badges
1958 bronze badges
1
$begingroup$
It is not a problem of infinitely poles. You can't use contour integration because the function doesn't $to 0$ for $|x|$ large on an half-plane.
$endgroup$
– reuns
3 hours ago
add a comment
|
1
$begingroup$
It is not a problem of infinitely poles. You can't use contour integration because the function doesn't $to 0$ for $|x|$ large on an half-plane.
$endgroup$
– reuns
3 hours ago
1
1
$begingroup$
It is not a problem of infinitely poles. You can't use contour integration because the function doesn't $to 0$ for $|x|$ large on an half-plane.
$endgroup$
– reuns
3 hours ago
$begingroup$
It is not a problem of infinitely poles. You can't use contour integration because the function doesn't $to 0$ for $|x|$ large on an half-plane.
$endgroup$
– reuns
3 hours ago
add a comment
|
2 Answers
2
active
oldest
votes
$begingroup$
Let's assume, for the moment, that $kleq 0$. Then we can rewrite our integral in the following way:
$$int_-infty^infty fracx^2e^k + e^x^2dx = e^-kint_-infty^infty fracx^2e^ke^-x^2e^ke^-x^2+1dx = e^-ksum_n=1^infty (-1)^n+1e^nkint_-infty^infty x^2e^-nx^2dx$$
Then, by Feynman's trick
$$= e^-ksum_n=1^infty (-1)^n+1 e^nkBiggr(-fracddnint_-infty^infty e^-nx^2dxBiggr) = e^-ksum_n=1^infty (-1)^n+1 e^nk left(-fracddnsqrtfracpinright)$$
$$ = -fracsqrtpi2e^-k sum_n=1^infty frac(-e^k)^nn^frac32 equiv -fracsqrtpi2e^-k textLi_frac32(-e^k)$$
While these specific steps don't apply when $k>0$, we can use analytic continuation to extend the polylogarithm (and taking the antilimit).
answered 7 hours ago
Ninad MunshiNinad Munshi
2,6573 silver badges12 bronze badges
$endgroup$
1
$begingroup$
Nice use of Feynman's trick. $to +1$
$endgroup$
– Claude Leibovici
7 hours ago
$begingroup$
Alternatively I think you could use the identity $$fracGamma(s/2)n^spi^s/2=int_0^infty x^s/2-1e^-n^2pi xdx,$$ instead of Feynman's trick since your integral is symmetric about $x=0$. You'd need to apply a change of variable and substitute a value for $s$ first.
$endgroup$
– Pixel
6 hours ago
add a comment
|
$begingroup$
Using a CAS, I did not find any antiderivative. However, for the integral
$$int_-infty^infty fracx^2K+e^x^2,dx=-fracsqrtpi 2 K textLi_frac32(-K)$$ where appears the polylogarithm function.
$$int_-infty^infty fracx^2e^k+e^x^2,dx=-fracsqrtpi 2 e^-k ,textLi_frac32left(-e^kright)$$
answered 7 hours ago
Claude LeiboviciClaude Leibovici
136k11 gold badges64 silver badges146 bronze badges
$endgroup$
$begingroup$
Hmm, interesting. Originally, I came across the above integral while manipulating a function related to fermi-Dirac statistics. I think if we do go along the road of expanding the above function in terms of the infinite poles, we may end up in the polylogarithmic function only.
$endgroup$
– Rishabh Jain
7 hours ago
add a comment
|
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's assume, for the moment, that $kleq 0$. Then we can rewrite our integral in the following way:
$$int_-infty^infty fracx^2e^k + e^x^2dx = e^-kint_-infty^infty fracx^2e^ke^-x^2e^ke^-x^2+1dx = e^-ksum_n=1^infty (-1)^n+1e^nkint_-infty^infty x^2e^-nx^2dx$$
Then, by Feynman's trick
$$= e^-ksum_n=1^infty (-1)^n+1 e^nkBiggr(-fracddnint_-infty^infty e^-nx^2dxBiggr) = e^-ksum_n=1^infty (-1)^n+1 e^nk left(-fracddnsqrtfracpinright)$$
$$ = -fracsqrtpi2e^-k sum_n=1^infty frac(-e^k)^nn^frac32 equiv -fracsqrtpi2e^-k textLi_frac32(-e^k)$$
While these specific steps don't apply when $k>0$, we can use analytic continuation to extend the polylogarithm (and taking the antilimit).
answered 7 hours ago
Ninad MunshiNinad Munshi
2,6573 silver badges12 bronze badges
$endgroup$
1
$begingroup$
Nice use of Feynman's trick. $to +1$
$endgroup$
– Claude Leibovici
7 hours ago
$begingroup$
Alternatively I think you could use the identity $$fracGamma(s/2)n^spi^s/2=int_0^infty x^s/2-1e^-n^2pi xdx,$$ instead of Feynman's trick since your integral is symmetric about $x=0$. You'd need to apply a change of variable and substitute a value for $s$ first.
$endgroup$
– Pixel
6 hours ago
add a comment
|
$begingroup$
Let's assume, for the moment, that $kleq 0$. Then we can rewrite our integral in the following way:
$$int_-infty^infty fracx^2e^k + e^x^2dx = e^-kint_-infty^infty fracx^2e^ke^-x^2e^ke^-x^2+1dx = e^-ksum_n=1^infty (-1)^n+1e^nkint_-infty^infty x^2e^-nx^2dx$$
Then, by Feynman's trick
$$= e^-ksum_n=1^infty (-1)^n+1 e^nkBiggr(-fracddnint_-infty^infty e^-nx^2dxBiggr) = e^-ksum_n=1^infty (-1)^n+1 e^nk left(-fracddnsqrtfracpinright)$$
$$ = -fracsqrtpi2e^-k sum_n=1^infty frac(-e^k)^nn^frac32 equiv -fracsqrtpi2e^-k textLi_frac32(-e^k)$$
While these specific steps don't apply when $k>0$, we can use analytic continuation to extend the polylogarithm (and taking the antilimit).
answered 7 hours ago
Ninad MunshiNinad Munshi
2,6573 silver badges12 bronze badges
$endgroup$
1
$begingroup$
Nice use of Feynman's trick. $to +1$
$endgroup$
– Claude Leibovici
7 hours ago
$begingroup$
Alternatively I think you could use the identity $$fracGamma(s/2)n^spi^s/2=int_0^infty x^s/2-1e^-n^2pi xdx,$$ instead of Feynman's trick since your integral is symmetric about $x=0$. You'd need to apply a change of variable and substitute a value for $s$ first.
$endgroup$
– Pixel
6 hours ago
add a comment
|
$begingroup$
Let's assume, for the moment, that $kleq 0$. Then we can rewrite our integral in the following way:
$$int_-infty^infty fracx^2e^k + e^x^2dx = e^-kint_-infty^infty fracx^2e^ke^-x^2e^ke^-x^2+1dx = e^-ksum_n=1^infty (-1)^n+1e^nkint_-infty^infty x^2e^-nx^2dx$$
Then, by Feynman's trick
$$= e^-ksum_n=1^infty (-1)^n+1 e^nkBiggr(-fracddnint_-infty^infty e^-nx^2dxBiggr) = e^-ksum_n=1^infty (-1)^n+1 e^nk left(-fracddnsqrtfracpinright)$$
$$ = -fracsqrtpi2e^-k sum_n=1^infty frac(-e^k)^nn^frac32 equiv -fracsqrtpi2e^-k textLi_frac32(-e^k)$$
While these specific steps don't apply when $k>0$, we can use analytic continuation to extend the polylogarithm (and taking the antilimit).
answered 7 hours ago
Ninad MunshiNinad Munshi
2,6573 silver badges12 bronze badges
$endgroup$
Let's assume, for the moment, that $kleq 0$. Then we can rewrite our integral in the following way:
$$int_-infty^infty fracx^2e^k + e^x^2dx = e^-kint_-infty^infty fracx^2e^ke^-x^2e^ke^-x^2+1dx = e^-ksum_n=1^infty (-1)^n+1e^nkint_-infty^infty x^2e^-nx^2dx$$
Then, by Feynman's trick
$$= e^-ksum_n=1^infty (-1)^n+1 e^nkBiggr(-fracddnint_-infty^infty e^-nx^2dxBiggr) = e^-ksum_n=1^infty (-1)^n+1 e^nk left(-fracddnsqrtfracpinright)$$
$$ = -fracsqrtpi2e^-k sum_n=1^infty frac(-e^k)^nn^frac32 equiv -fracsqrtpi2e^-k textLi_frac32(-e^k)$$
While these specific steps don't apply when $k>0$, we can use analytic continuation to extend the polylogarithm (and taking the antilimit).
answered 7 hours ago
Ninad MunshiNinad Munshi
2,6573 silver badges12 bronze badges
edited 7 hours ago
answered 7 hours ago
Ninad MunshiNinad Munshi
2,6573 silver badges12 bronze badges
answered 7 hours ago
Ninad MunshiNinad Munshi
2,6573 silver badges12 bronze badges
answered 7 hours ago
Ninad MunshiNinad Munshi
2,6573 silver badges12 bronze badges
2,6573 silver badges12 bronze badges
1
$begingroup$
Nice use of Feynman's trick. $to +1$
$endgroup$
– Claude Leibovici
7 hours ago
$begingroup$
Alternatively I think you could use the identity $$fracGamma(s/2)n^spi^s/2=int_0^infty x^s/2-1e^-n^2pi xdx,$$ instead of Feynman's trick since your integral is symmetric about $x=0$. You'd need to apply a change of variable and substitute a value for $s$ first.
$endgroup$
– Pixel
6 hours ago
add a comment
|
1
$begingroup$
Nice use of Feynman's trick. $to +1$
$endgroup$
– Claude Leibovici
7 hours ago
$begingroup$
Alternatively I think you could use the identity $$fracGamma(s/2)n^spi^s/2=int_0^infty x^s/2-1e^-n^2pi xdx,$$ instead of Feynman's trick since your integral is symmetric about $x=0$. You'd need to apply a change of variable and substitute a value for $s$ first.
$endgroup$
– Pixel
6 hours ago
1
1
$begingroup$
Nice use of Feynman's trick. $to +1$
$endgroup$
– Claude Leibovici
7 hours ago
$begingroup$
Nice use of Feynman's trick. $to +1$
$endgroup$
– Claude Leibovici
7 hours ago
$begingroup$
Alternatively I think you could use the identity $$fracGamma(s/2)n^spi^s/2=int_0^infty x^s/2-1e^-n^2pi xdx,$$ instead of Feynman's trick since your integral is symmetric about $x=0$. You'd need to apply a change of variable and substitute a value for $s$ first.
$endgroup$
– Pixel
6 hours ago
$begingroup$
Alternatively I think you could use the identity $$fracGamma(s/2)n^spi^s/2=int_0^infty x^s/2-1e^-n^2pi xdx,$$ instead of Feynman's trick since your integral is symmetric about $x=0$. You'd need to apply a change of variable and substitute a value for $s$ first.
$endgroup$
– Pixel
6 hours ago
add a comment
|
$begingroup$
Using a CAS, I did not find any antiderivative. However, for the integral
$$int_-infty^infty fracx^2K+e^x^2,dx=-fracsqrtpi 2 K textLi_frac32(-K)$$ where appears the polylogarithm function.
$$int_-infty^infty fracx^2e^k+e^x^2,dx=-fracsqrtpi 2 e^-k ,textLi_frac32left(-e^kright)$$
answered 7 hours ago
Claude LeiboviciClaude Leibovici
136k11 gold badges64 silver badges146 bronze badges
$endgroup$
$begingroup$
Hmm, interesting. Originally, I came across the above integral while manipulating a function related to fermi-Dirac statistics. I think if we do go along the road of expanding the above function in terms of the infinite poles, we may end up in the polylogarithmic function only.
$endgroup$
– Rishabh Jain
7 hours ago
add a comment
|
$begingroup$
Using a CAS, I did not find any antiderivative. However, for the integral
$$int_-infty^infty fracx^2K+e^x^2,dx=-fracsqrtpi 2 K textLi_frac32(-K)$$ where appears the polylogarithm function.
$$int_-infty^infty fracx^2e^k+e^x^2,dx=-fracsqrtpi 2 e^-k ,textLi_frac32left(-e^kright)$$
answered 7 hours ago
Claude LeiboviciClaude Leibovici
136k11 gold badges64 silver badges146 bronze badges
$endgroup$
$begingroup$
Hmm, interesting. Originally, I came across the above integral while manipulating a function related to fermi-Dirac statistics. I think if we do go along the road of expanding the above function in terms of the infinite poles, we may end up in the polylogarithmic function only.
$endgroup$
– Rishabh Jain
7 hours ago
add a comment
|
$begingroup$
Using a CAS, I did not find any antiderivative. However, for the integral
$$int_-infty^infty fracx^2K+e^x^2,dx=-fracsqrtpi 2 K textLi_frac32(-K)$$ where appears the polylogarithm function.
$$int_-infty^infty fracx^2e^k+e^x^2,dx=-fracsqrtpi 2 e^-k ,textLi_frac32left(-e^kright)$$
answered 7 hours ago
Claude LeiboviciClaude Leibovici
136k11 gold badges64 silver badges146 bronze badges
$endgroup$
Using a CAS, I did not find any antiderivative. However, for the integral
$$int_-infty^infty fracx^2K+e^x^2,dx=-fracsqrtpi 2 K textLi_frac32(-K)$$ where appears the polylogarithm function.
$$int_-infty^infty fracx^2e^k+e^x^2,dx=-fracsqrtpi 2 e^-k ,textLi_frac32left(-e^kright)$$
answered 7 hours ago
Claude LeiboviciClaude Leibovici
136k11 gold badges64 silver badges146 bronze badges
answered 7 hours ago
Claude LeiboviciClaude Leibovici
136k11 gold badges64 silver badges146 bronze badges
answered 7 hours ago
Claude LeiboviciClaude Leibovici
136k11 gold badges64 silver badges146 bronze badges
answered 7 hours ago
Claude LeiboviciClaude Leibovici
136k11 gold badges64 silver badges146 bronze badges
136k11 gold badges64 silver badges146 bronze badges
$begingroup$
Hmm, interesting. Originally, I came across the above integral while manipulating a function related to fermi-Dirac statistics. I think if we do go along the road of expanding the above function in terms of the infinite poles, we may end up in the polylogarithmic function only.
$endgroup$
– Rishabh Jain
7 hours ago
add a comment
|
$begingroup$
Hmm, interesting. Originally, I came across the above integral while manipulating a function related to fermi-Dirac statistics. I think if we do go along the road of expanding the above function in terms of the infinite poles, we may end up in the polylogarithmic function only.
$endgroup$
– Rishabh Jain
7 hours ago
$begingroup$
Hmm, interesting. Originally, I came across the above integral while manipulating a function related to fermi-Dirac statistics. I think if we do go along the road of expanding the above function in terms of the infinite poles, we may end up in the polylogarithmic function only.
$endgroup$
– Rishabh Jain
7 hours ago
$begingroup$
Hmm, interesting. Originally, I came across the above integral while manipulating a function related to fermi-Dirac statistics. I think if we do go along the road of expanding the above function in terms of the infinite poles, we may end up in the polylogarithmic function only.
$endgroup$
– Rishabh Jain
7 hours ago
add a comment
|
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It is not a problem of infinitely poles. You can't use contour integration because the function doesn't $to 0$ for $|x|$ large on an half-plane.
$endgroup$
– reuns
3 hours ago