# Evaluating $\int_0^\infty \left| \frac{\sin t}{t} \right|^n \, \mathrm{d}t$ for $n = 3, 5, 7, \dots$

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.

• Can you write $|\sin^n x|$ as a Fourier series first ? – Empy2 Aug 28 at 12:36
• $|\sin^n x|$ is an even function with period $\pi$ so it is a sum of cosines $\sum_ka_{nk}\cos 2kx$. Find $a_{nk}$ and $\int_0^\infty \cos(2kx)/x^n$ – Empy2 Aug 28 at 12:43
• Hmm, they are singular at zero. – Empy2 Aug 28 at 12:49
• @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 $0$. – Daniel Fischer Aug 28 at 13:28
• @DanielFischer thank you very much. I believe you're right. Sorry to Empy2 – Calvin Khor Aug 28 at 13:29

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];