10
$\begingroup$

I would like to determine the general term of the following sequence defined by an infinite integral: $$ I_n = \int_0^\infty \left| \frac{\sin t}{t} \right|^n \, \mathrm{d}t \, , $$ wherein $n =3, 5, 7, \dots$ is an odd integer.

It can be checked that the integral is convergent for all values of $n$ in the prescribed range. The case of even $n$ is solved in A sine integral $\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x$. Also, $I_1 = \infty$. I have tried to use the method of multiple integrations by parts but in vein. I was wondering whether there exists a suitable approach to address this problem more effectively.

$\endgroup$
  • 1
    $\begingroup$ Can you write $|\sin^n x|$ as a Fourier series first ? $\endgroup$ – Empy2 Aug 28 at 12:36
  • 1
    $\begingroup$ $|\sin^n x|$ is an even function with period $\pi$ so it is a sum of cosines $\sum_ka_{nk}\cos 2kx$. Find $a_{nk}$ and $\int_0^\infty \cos(2kx)/x^n$ $\endgroup$ – Empy2 Aug 28 at 12:43
  • 1
    $\begingroup$ Hmm, they are singular at zero. $\endgroup$ – Empy2 Aug 28 at 12:49
  • 2
    $\begingroup$ @CalvinKhor (and Daddy) I am almost sure that Empy2 referred to the $\int_0^{\infty} \cos(2kx)/x^n\,dx$ integrals mentioned in the previous comment. Those integrands indeed have a non-integrable singularity at $0$. $\endgroup$ – Daniel Fischer Aug 28 at 13:28
  • 1
    $\begingroup$ @DanielFischer thank you very much. I believe you're right. Sorry to Empy2 $\endgroup$ – Calvin Khor Aug 28 at 13:29
12
+250
$\begingroup$

Here is a partial answer: If $n \geq 2$ is an integer, then

$$ I_n = \frac{1}{(n-1)!2^{n-1}} \sum_{l=0}^{\lfloor n/2 \rfloor} (-1)^l \binom{n}{l} (n-2l)^{n-1} J_{n-2l}, \tag{1} $$

where $J_p$ is defined by

$$ J_{p} = \begin{cases} \displaystyle \frac{4p}{\pi} \sum_{j=1}^{\infty} \frac{\log(2\pi j)}{4j^2-p^2}, & \text{if $p$ is odd}, \\ \displaystyle \frac{\pi}{2}, & \text{if $p$ is even}. \end{cases} $$


Proof of $\text{(1)}$. The case of even $n$ has already been discussed in other postings, so we focus on odd $n$. We first note that, if $n \geq 1$ is an odd integer, then

\begin{align*} \frac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}} \sin^n x &= \frac{1}{(2i)^n} \sum_{l=0}^{\frac{n-1}{2}} (-1)^l \binom{n}{l} \frac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}} (e^{(n-2l)ix} - e^{-(n-2l)ix}) \\ &= \frac{1}{2^{n-1}} \sum_{l=0}^{\frac{n-1}{2}} (-1)^l \binom{n}{l} (n-2l)^{n-1} \sin((n-2l)x). \end{align*}

So by applying integration by parts $(n-1)$-times, we get

\begin{align*} I_n &= \sum_{k=0}^{\infty} \int_{0}^{\pi} \frac{\sin^n x}{(x+k\pi)^n} \, \mathrm{d}x \\ &= \frac{1}{(n-1)!} \sum_{k=0}^{\infty} \int_{0}^{\pi} \biggl( \frac{1}{x+k\pi} - \frac{1}{(k+1)\pi} \biggr) \biggl( \frac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}} \sin^n x \biggr) \, \mathrm{d}x \\ &= \frac{1}{(n-1)!2^{n-1}} \sum_{l=0}^{\frac{n-1}{2}} (-1)^l \binom{n}{l} (n-2l)^{n-1} J_{n-2l}, \end{align*}

where $J_{p}$ is defined by

\begin{align*} J_{p} &= \sum_{k=0}^{\infty} \int_{0}^{\pi} \biggl( \frac{1}{x+k\pi} - \frac{1}{(k+1)\pi} \biggr) \sin(px) \, \mathrm{d}x. \end{align*}

If $p$ is odd, then the above definition is recast as

\begin{align*} J_{p} &= \sum_{k=0}^{\infty} \biggl( \int_{0}^{\pi} \frac{1}{x+k\pi} \sin(px) \, \mathrm{d}x - \frac{2}{p\pi(k+1)} \biggr) \\ &= \lim_{N \to \infty} \biggl( \int_{0}^{N \pi} \frac{\sin(p(x \text{ mod } \pi))}{x} \, \mathrm{d}x - \frac{2}{p\pi} H_N \biggr), \end{align*}

where $H_N = 1 + \frac{1}{2} + \dots + \frac{1}{N}$ is the $N$-th harmonic number. Still assuming that $p$ is an odd integer, Fourier series computation shows that

\begin{align*} \sin(p(x \text{ mod } \pi)) &= \frac{2}{p\pi} - \frac{4p}{\pi} \sum_{n=1}^{\infty} \frac{\cos(2\pi n x)}{4n^2-p^2} \\ &= \frac{4p}{\pi} \sum_{j=1}^{\infty} \frac{1 - \cos(2\pi j x)}{4j^2-p^2}, \end{align*}

and so,

\begin{align*} \int_{0}^{N \pi} \frac{\sin(p(x \mathrm{ mod } \pi))}{x} \, \mathrm{d}x &= \frac{4p}{\pi} \sum_{j=1}^{\infty} \frac{1}{4j^2-p^2} \int_{0}^{N \pi} \frac{1 - \cos(2 j x)}{x} \, \mathrm{d}x \\ &= \frac{4p}{\pi} \sum_{j=1}^{\infty} \frac{1}{4j^2-p^2} (\gamma + \log(2\pi j N) - \operatorname{Ci}(2\pi j N) ). \end{align*}

Plugging this back and using the identity $\frac{4p}{\pi} \sum_{j=1}^{\infty} \frac{1}{4j^2-p^2} = \frac{2}{p\pi}$, which itself follows from the Fourier series of $\sin(p(x \text{ mod } \pi))$, we finally obtain

\begin{align*} J_{p} &= \frac{4p}{\pi} \lim_{N \to \infty} \sum_{j=1}^{\infty} \frac{1}{4j^2-p^2} (\gamma + \log(2\pi j N) - \operatorname{Ci}(2\pi j N) - H_N ) \biggr) \\ &= \frac{4p}{\pi} \sum_{j=1}^{\infty} \frac{\log(2\pi j)}{4j^2-p^2} \end{align*}

as desired. $\square$


Addendum. Here is a Mathematica code for numerical verification of $\text{(1)}$:

n = 5; (* Choose your favorite odd integer >= 3*)
NIntegrate[Evaluate[Sum[1/(x + k Pi)^n, {k, 0, Infinity}] Sin[x]^n], {x, 0, Pi}, WorkingPrecision -> 20]
TermJ[p_] := (4 p)/Pi NSum[Log[2 Pi j]/(4 j^2 - p^2), {j, 1, Infinity}, WorkingPrecision -> 20];
1/((n - 1)! 2^(n - 1)) Sum[Binomial[n, l] (-1)^l (n - 2 l)^(n - 1) TermJ[n - 2 l], {l, 0, (n - 1)/2}]
Clear[n, TermJ];

Numerical verification

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.