How to integrate $\int_{0}^{\infty }{\frac{\sin x}{\cosh x+\cos x}\cdot \frac{{{x}^{n}}}{n!}\ \text{d}x} $? 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:
$$\begin{align*}
I&=\int_{0}^{\infty }{\frac{\sin x}{\cosh x+\cos x}\cdot \frac{{{x}^{n}}}{n!}\ \text{d}x}\tag1 \\ 
 J&= \int_{0}^{\infty }{\frac{x-\sin x}{\left( {{\pi }^{2}}+{{x}^{2}} \right){{x}^{3}}}\ \text{d}x}\tag2 \\ 
\end{align*}$$
 A: 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.
Define 
$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \: f(x) e^{i k x} $$
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 $g(x) = (x-\sin{x})/x^3$, 
$$\begin{align} \hat{g}(k) &= \int_{-\infty}^{\infty} dx \: \frac{x-\sin{x}}{x^3} e^{i k x} \\ &= \frac{\pi}{2}(k^2-2 |k|+1) \mathrm{rect}(k/2) \\ \end{align}$$
where 
$$\mathrm{rect}(k) = \begin{cases} 1 & |k|<\frac{1}{2} \\ 0 & |k|>\frac{1}{2} \end{cases} $$
Then we can write, using the Parseval theorem,
$$\begin{align} \int_{0}^{\infty }{dx \: \frac{x-\sin x}{\left( {{\pi }^{2}}+{{x}^{2}} \right){{x}^{3}}}} &= \frac{1}{8} \int_{-1}^1 dk \: (k^2-2 |k|+1) e^{-\pi |k|} \\ &= \frac{\left(2-2 \pi +\pi ^2\right)}{4 \pi
   ^3}-\frac{ e^{-\pi }}{2 \pi ^3} \\ \end{align}$$
NOTE
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.
A: 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}
A: 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
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{x - \sin\pars{x} \over \pars{\pi^{2} + x^{2}}x^{3}}\dd x} =
{1 \over 2}\int_{-\infty}^{\infty}{x - \sin\pars{x} \over \pars{\pi^{2} + x^{2}}x^{3}}\dd x
\\[5mm] = & \
{1 \over 2}\,\Im\int_{-\infty}^{\infty}{\ic x - \expo{\ic x} + 1 - x^{2}/2 \over \pars{\pi^{2} + x^{2}}x^{3}}\,\dd x
\\[5mm] = & \
{1 \over 2}\,\Im\braces{2\pi\ic\on{Res}
\bracks{{\ic x - \expo{\ic x} + 1 - x^{2}/2\over \pars{\pi^{2} + x^{2}}x^{3}}, x = \ic\pi}}
\\[5mm] = & \
\bbx{\color{#44f}{{2- 2\expo{-\pi} -2\pi + \pi^{2} \over 4\pi^{3}}}} \approx 0.04435 \\ &
\end{align}
