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I have done one with $\displaystyle\int_0^{\infty}\frac{x-\sin x}{x^3}\ \text{d}x$, but I have no ideas with these: $$\begin{align*} I&=\int_{0}^{\infty }{\frac{\sin x}{\cosh x+\cos x}\cdot \frac{{{x}^{n}}}{n!}\ \text{d}x}\tag1 \\ J&= \int_{0}^{\infty }{\frac{x-\sin x}{\left( {{\pi }^{2}}+{{x}^{2}} \right){{x}^{3}}}\ \text{d}x}\tag2 \\ \end{align*}$$

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Are you looking for the convergent or divergent of above series? – Babak S. Jan 27 '13 at 13:35
@Babak Sorouh : Both of them should hav a closed form – gauss115 Jan 27 '13 at 13:39
$\frac{\sin(x)}{\cosh(x)+\cos(x)} = i (\frac{1}{e^{(1+i)x}+1} - \frac{1}{e^{(1-i)x}+1} )$ – achille hui Jan 28 '13 at 12:21
@achillehui : Thx for the comment :) – gauss115 Jan 28 '13 at 14:13
up vote 7 down vote accepted

I can address the second integral:

$$\int_{0}^{\infty }{dx \: \frac{x-\sin x}{\left( {{\pi }^{2}}+{{x}^{2}} \right){{x}^{3}}}}$$

Hint: We can use Parseval's Theorem

$$\int_{-\infty}^{\infty} dx \: f(x) \bar{g}(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \: \hat{f}(k) \bar{\hat{g}}(k) $$

where $f$ and $\hat{f}$ are Fourier transform pairs, and same for $g$ and $\bar{g}$. The FT of $1/(x^2+\pi^2)$ is easy, so we need the FT of the rest of the integrand, which turns out to be possible.


$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \: f(x) e^{i k x} $$

It is straightforward to show using the Residue Theorem that, when $f(x) = (x^2+a^2)^{-1}$, then

$$\hat{f}(k) = \frac{\pi}{a} e^{-a |k|} $$

Thus we need to compute, when $g(x) = (x-\sin{x})/x^3$,

$$\begin{align} \hat{g}(k) &= \int_{-\infty}^{\infty} dx \: \frac{x-\sin{x}}{x^3} e^{i k x} \\ &= \frac{\pi}{2}(k^2-2 |k|+1) \mathrm{rect}(k/2) \\ \end{align}$$


$$\mathrm{rect}(k) = \begin{cases} 1 & |k|<\frac{1}{2} \\ 0 & |k|>\frac{1}{2} \end{cases} $$

Then we can write, using the Parseval theorem,

$$\begin{align} \int_{0}^{\infty }{dx \: \frac{x-\sin x}{\left( {{\pi }^{2}}+{{x}^{2}} \right){{x}^{3}}}} &= \frac{1}{8} \int_{-1}^1 dk \: (k^2-2 |k|+1) e^{-\pi |k|} \\ &= \frac{\left(2-2 \pi +\pi ^2\right)}{4 \pi ^3}-\frac{ e^{-\pi }}{2 \pi ^3} \\ \end{align}$$


Deriving $\hat{g}(k)$ from scratch is challenging; nevertheless, it is straightforward (albeit, a bit messy) to prove that the expression is correct by performing the inverse transform on $\hat{g}(k)$ to obtain $g(x)$. I did this out and proved it to myself; I can provide the details to those that want to see them.

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THX! Very nice solution! The result matches :) – gauss115 Jan 28 '13 at 11:09
You're welcome; I like it. Plus, we get an interesting FT. – Ron Gordon Jan 28 '13 at 11:50
@gauss115: this result is even more interesting than I thought. If you notice, the quadratic in $\pi$ in the numerator of the first term represents the first 3 terms of the Taylor series for $2 e^{-\pi}$. This sort of cancellation mirrors the cancellation of the integrand, no? – Ron Gordon Jan 28 '13 at 16:26
Thx to mention that! I dont see it at first glance – gauss115 Jan 29 '13 at 10:21

Employing the generating function $$ \frac{\sin x}{\cosh x+\cos x}=\frac12\sum_{k=1}^\infty(-1)^{k-1}e^{-kx}\sin kx $$ and the classic result of $$ \int_0^\infty x^{m-1}e^{-ax} \sin bx \ dx = \frac{\Gamma(m)}{(a^{2} + b^{2})^{m/2}}\sin\left(m\tan^{-1}\left(\frac{b}{a}\right)\right) $$ The closed-form of the first integral is \begin{align} \int_{0}^{\infty}\frac{\sin x}{\cosh x+\cos x}\cdot \frac{x^n}{n!}\ dx&=\frac{2}{n!}\sum_{k=1}^\infty(-1)^{k-1}\int_{0}^{\infty}x^n\ e^{-kx}\sin kx\ dx\\[10pt] &=\frac{\left(1-2^{-n}\right) \zeta(n+1)}{\ \sqrt{2^{n-1}}}\sin\left(\frac{n+1}{4}\pi\right) \end{align}

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Maple says $$ \int_{0}^{\infty} \frac{x - \operatorname{sin} (x)}{\bigl(\pi^{2} + x^{2}\bigr) x^{3}} d x = \frac{2 \operatorname{sinh} (\pi) - 2 \pi - 2 \operatorname{cosh} (\pi) + \pi^{2} + 2}{4 \pi^{3}} \approx 0.04434578936 $$ but it didn't do the first one automatically

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