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3

$begingroup$

I've been studying infinite sums and came across the Möbius function, and I was wondering if there exists a definite integral that represents it, or equivalently if there exists some $f(x,n)$ and $a, b$ such that

$$
int_a^b f(x,n)dx = mu(n).
$$

I have no idea how to begin to go about solving this problem, but through my basic workings, I have the suspicion that there may be no closed form that satisfies this equation.

Thanks for your time!

complex-analysis number-theory

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asked 8 hours ago

trytryagaintrytryagain

2744 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

I've been studying infinite sums and came across the Möbius function, and I was wondering if there exists a definite integral that represents it, or equivalently if there exists some $f(x,n)$ and $a, b$ such that

$$
int_a^b f(x,n)dx = mu(n).
$$

I have no idea how to begin to go about solving this problem, but through my basic workings, I have the suspicion that there may be no closed form that satisfies this equation.

Thanks for your time!

complex-analysis number-theory

share|cite|improve this question

asked 8 hours ago

trytryagaintrytryagain

2744 bronze badges

$endgroup$

add a comment | 

$begingroup$

I've been studying infinite sums and came across the Möbius function, and I was wondering if there exists a definite integral that represents it, or equivalently if there exists some $f(x,n)$ and $a, b$ such that

$$
int_a^b f(x,n)dx = mu(n).
$$

I have no idea how to begin to go about solving this problem, but through my basic workings, I have the suspicion that there may be no closed form that satisfies this equation.

Thanks for your time!

complex-analysis number-theory

share|cite|improve this question

asked 8 hours ago

trytryagaintrytryagain

2744 bronze badges

$endgroup$

share|cite|improve this question

asked 8 hours ago

trytryagaintrytryagain

2744 bronze badges

share|cite|improve this question

asked 8 hours ago

trytryagaintrytryagain

2744 bronze badges

asked 8 hours ago

trytryagaintrytryagain

2744 bronze badges

asked 8 hours ago

trytryagaintrytryagain

2744 bronze badges

3 Answers
3

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5

$begingroup$

Sure. Fix any $a<b$, and put $f(x,n)=tfrac1b-amu(n)$ (independent of $x$).

share|cite|improve this answer

answered 8 hours ago

MPWMPW

31.9k11 gold badge2222 silver badges5959 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  well that was trivial!
  $endgroup$
  – vidyarthi
  8 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

As you stated it you can define $f(x,n)$ is many ways to generate any sequence.

A more interesting answer is that what we want is a generating function of the sequence $mu(n)$, and that the natural ones are $$frac1zeta(s) = sum_n=1^infty mu(n) n^-s$$ The partial sums admitting explicit formulas in term of the non-trivial zeros $$f(s)=sum_n le x mu(n) n^-s$$ The inverse Mellin transform of $fracGamma(s)zeta(s)$ (again admitting an explicit formula) $$sum_n=1^infty mu(n) e^-nx$$
Its periodic versions
$$F(x)=sum_n=1^infty fracmu(n)n^s e^inx$$
and the weird Riesz function $$sum_n=1^infty fracmu(n)n^2 e^-x/n^2 = sum_k=0^infty frac(-x)^kk! frac1zeta(2+2k)=sum_k=0^infty frac(-x)^kk! frac1 frac(-1)^k+1(2pi)^2k B_2k(2k)!$$
All the other useful generating functions of $mu(n)$ are more or less directly related to one of them.

share|cite|improve this answer

answered 7 hours ago

reunsreuns

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add a comment | 

1

$begingroup$

If we allow bending the rules a bit to study the Mertens function
$$M(x)=sum_n=1^lfloor x rfloormu(x)$$
and improper integral representations, then there's the Perron formula
$$M(x)=lim_sigmato 1^+int_-infty^inftyfracmathrmdt2pifracx^sigma+mathrmit(sigma+mathrmit)zeta(sigma+mathrmit)text.$$

share|cite|improve this answer

answered 7 hours ago

K B DaveK B Dave

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5

$begingroup$

Sure. Fix any $a<b$, and put $f(x,n)=tfrac1b-amu(n)$ (independent of $x$).

share|cite|improve this answer

answered 8 hours ago

MPWMPW

31.9k11 gold badge2222 silver badges5959 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  well that was trivial!
  $endgroup$
  – vidyarthi
  8 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

Sure. Fix any $a<b$, and put $f(x,n)=tfrac1b-amu(n)$ (independent of $x$).

share|cite|improve this answer

answered 8 hours ago

MPWMPW

31.9k11 gold badge2222 silver badges5959 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  well that was trivial!
  $endgroup$
  – vidyarthi
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Sure. Fix any $a<b$, and put $f(x,n)=tfrac1b-amu(n)$ (independent of $x$).

share|cite|improve this answer

answered 8 hours ago

MPWMPW

31.9k11 gold badge2222 silver badges5959 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

MPWMPW

31.9k11 gold badge2222 silver badges5959 bronze badges

answered 8 hours ago

MPWMPW

31.9k11 gold badge2222 silver badges5959 bronze badges

answered 8 hours ago

MPWMPW

31.9k11 gold badge2222 silver badges5959 bronze badges

5

$begingroup$

As you stated it you can define $f(x,n)$ is many ways to generate any sequence.

A more interesting answer is that what we want is a generating function of the sequence $mu(n)$, and that the natural ones are $$frac1zeta(s) = sum_n=1^infty mu(n) n^-s$$ The partial sums admitting explicit formulas in term of the non-trivial zeros $$f(s)=sum_n le x mu(n) n^-s$$ The inverse Mellin transform of $fracGamma(s)zeta(s)$ (again admitting an explicit formula) $$sum_n=1^infty mu(n) e^-nx$$
Its periodic versions
$$F(x)=sum_n=1^infty fracmu(n)n^s e^inx$$
and the weird Riesz function $$sum_n=1^infty fracmu(n)n^2 e^-x/n^2 = sum_k=0^infty frac(-x)^kk! frac1zeta(2+2k)=sum_k=0^infty frac(-x)^kk! frac1 frac(-1)^k+1(2pi)^2k B_2k(2k)!$$
All the other useful generating functions of $mu(n)$ are more or less directly related to one of them.

share|cite|improve this answer

answered 7 hours ago

reunsreuns

24k22 gold badges1414 silver badges6262 bronze badges

$endgroup$

add a comment | 

5

$begingroup$

As you stated it you can define $f(x,n)$ is many ways to generate any sequence.

A more interesting answer is that what we want is a generating function of the sequence $mu(n)$, and that the natural ones are $$frac1zeta(s) = sum_n=1^infty mu(n) n^-s$$ The partial sums admitting explicit formulas in term of the non-trivial zeros $$f(s)=sum_n le x mu(n) n^-s$$ The inverse Mellin transform of $fracGamma(s)zeta(s)$ (again admitting an explicit formula) $$sum_n=1^infty mu(n) e^-nx$$
Its periodic versions
$$F(x)=sum_n=1^infty fracmu(n)n^s e^inx$$
and the weird Riesz function $$sum_n=1^infty fracmu(n)n^2 e^-x/n^2 = sum_k=0^infty frac(-x)^kk! frac1zeta(2+2k)=sum_k=0^infty frac(-x)^kk! frac1 frac(-1)^k+1(2pi)^2k B_2k(2k)!$$
All the other useful generating functions of $mu(n)$ are more or less directly related to one of them.

share|cite|improve this answer

answered 7 hours ago

reunsreuns

24k22 gold badges1414 silver badges6262 bronze badges

$endgroup$

add a comment | 

$begingroup$

As you stated it you can define $f(x,n)$ is many ways to generate any sequence.

A more interesting answer is that what we want is a generating function of the sequence $mu(n)$, and that the natural ones are $$frac1zeta(s) = sum_n=1^infty mu(n) n^-s$$ The partial sums admitting explicit formulas in term of the non-trivial zeros $$f(s)=sum_n le x mu(n) n^-s$$ The inverse Mellin transform of $fracGamma(s)zeta(s)$ (again admitting an explicit formula) $$sum_n=1^infty mu(n) e^-nx$$
Its periodic versions
$$F(x)=sum_n=1^infty fracmu(n)n^s e^inx$$
and the weird Riesz function $$sum_n=1^infty fracmu(n)n^2 e^-x/n^2 = sum_k=0^infty frac(-x)^kk! frac1zeta(2+2k)=sum_k=0^infty frac(-x)^kk! frac1 frac(-1)^k+1(2pi)^2k B_2k(2k)!$$
All the other useful generating functions of $mu(n)$ are more or less directly related to one of them.

share|cite|improve this answer

answered 7 hours ago

reunsreuns

24k22 gold badges1414 silver badges6262 bronze badges

$endgroup$

share|cite|improve this answer

answered 7 hours ago

reunsreuns

24k22 gold badges1414 silver badges6262 bronze badges

answered 7 hours ago

reunsreuns

24k22 gold badges1414 silver badges6262 bronze badges

answered 7 hours ago

reunsreuns

24k22 gold badges1414 silver badges6262 bronze badges

1

$begingroup$

If we allow bending the rules a bit to study the Mertens function
$$M(x)=sum_n=1^lfloor x rfloormu(x)$$
and improper integral representations, then there's the Perron formula
$$M(x)=lim_sigmato 1^+int_-infty^inftyfracmathrmdt2pifracx^sigma+mathrmit(sigma+mathrmit)zeta(sigma+mathrmit)text.$$

share|cite|improve this answer

answered 7 hours ago

K B DaveK B Dave

4,14644 silver badges1717 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

If we allow bending the rules a bit to study the Mertens function
$$M(x)=sum_n=1^lfloor x rfloormu(x)$$
and improper integral representations, then there's the Perron formula
$$M(x)=lim_sigmato 1^+int_-infty^inftyfracmathrmdt2pifracx^sigma+mathrmit(sigma+mathrmit)zeta(sigma+mathrmit)text.$$

share|cite|improve this answer

answered 7 hours ago

K B DaveK B Dave

4,14644 silver badges1717 bronze badges

$endgroup$

add a comment | 

$begingroup$

If we allow bending the rules a bit to study the Mertens function
$$M(x)=sum_n=1^lfloor x rfloormu(x)$$
and improper integral representations, then there's the Perron formula
$$M(x)=lim_sigmato 1^+int_-infty^inftyfracmathrmdt2pifracx^sigma+mathrmit(sigma+mathrmit)zeta(sigma+mathrmit)text.$$

share|cite|improve this answer

answered 7 hours ago

K B DaveK B Dave

4,14644 silver badges1717 bronze badges

$endgroup$

share|cite|improve this answer

answered 7 hours ago

K B DaveK B Dave

4,14644 silver badges1717 bronze badges

answered 7 hours ago

K B DaveK B Dave

4,14644 silver badges1717 bronze badges

answered 7 hours ago

K B DaveK B Dave

4,14644 silver badges1717 bronze badges