Prove that $\int_0^{\infty} \frac{\sin(2013 x)}{x(\cos x+\cosh x)}dx=\frac{\pi}{4}$ Prove that
$$\int_0^{\infty} \frac{\sin(2013 x)}{x(\cos x+\cosh x)}dx=\frac{\pi}{4}$$
 A: Because the integrand is even, contour integration yields
$$
\begin{align}
&\int_0^\infty\frac{\sin(mx)}{x(\cos(x)+\cosh(x))}\mathrm{d}x\\
&=\frac12\int_{-\infty}^\infty\frac{\sin(mx)}{x(\cos(x)+\cosh(x))}\mathrm{d}x\\
&=\frac12\int_{-\infty-i}^{\infty-i}\frac{\sin(mx)}{x(\cos(x)+\cosh(x))}\mathrm{d}x\\
&=\frac1{4i}\int_{\gamma^+}\frac{e^{imz}}{z(\cos(z)+\cosh(z))}\mathrm{d}z
-\frac1{4i}\int_{\gamma^-}\frac{e^{-imz}}{z(\cos(z)+\cosh(z))}\mathrm{d}z\\
&=\frac{2\pi i}{4i}\frac12
+2\frac{2\pi i}{4i}\sum_{k=0}^\infty(-1)^{k+1}\frac{\mathrm{sech}\left(\frac{2k+1}{2}\pi\right)}{\frac{2k+1}{2}\pi}\frac{\cos\left(m\frac{2k+1}{2}\pi\right)}{\exp\left(m\frac{2k+1}{2}\pi\right)}\\
&=\frac\pi4+\pi\sum_{k=0}^\infty(-1)^{k+1}\frac{\mathrm{sech}\left(\frac{2k+1}{2}\pi\right)}{\frac{2k+1}{2}\pi}\frac{\cos\left(m\frac{2k+1}{2}\pi\right)}{\exp\left(m\frac{2k+1}{2}\pi\right)}\\
&=\frac\pi4\qquad\text{for odd $m$}
\end{align}
$$
where $\gamma^+$ goes from $-R-i$ to $+R-i$ then circles counterclockwise back in the upper half plane along $|z+i|=R$ and $\gamma^+$ goes from $-R-i$ to $+R-i$ then circles clockwise back in the lower half plane along $|z+i|=R$.
The residue of $f(z)=\dfrac{e^{imz}}{z(\cos(z)+\cosh(z))}$ at $z=0$ is $\frac12$.
Let $\alpha=\frac{1+i}{2}$. As noted in this answer, $f$ has singularities at $\pm(2k+1)\pi\alpha$ and $\pm(2k+1)\pi\overline{\alpha}$.
The sum of the residues in the upper half plane at $(2k+1)\pi\alpha$ and $-(2k+1)\pi\overline{\alpha}$ is the same as the sum of the residues in the lower half plane at $(2k+1)\pi\overline{\alpha}$ and $-(2k+1)\pi\alpha$. Both are equal to
$$
(-1)^{k+1}\frac{\mathrm{sech}\left(\frac{2k+1}{2}\pi\right)}{\frac{2k+1}{2}\pi}\frac{\cos\left(m\frac{2k+1}{2}\pi\right)}{\exp\left(m\frac{2k+1}{2}\pi\right)}
$$
Note that for odd $m$, these are all $0$; $x=m\frac{2k+1}{2}\pi\equiv\frac\pi2\pmod{\pi}\Rightarrow\cos(x)=0$. This is not so for even $m$. Thus,
$$
\int_0^\infty\frac{\sin(2013x)}{x(\cos(x)+\cosh(x))}\mathrm{d}x=\frac\pi4
$$
A: Consider a general case:  
$$I(m)=\int_0^{\infty} \frac{\sin(m x)}{x(\cos x+\cosh x)}dx$$ where $m$ is a positive integer.  
At first, let's try with m=1  
$$I(1)=\int_0^{\infty} \frac{\sin x}{x(\cos x+\cosh x)}dx$$ As a first step we note that  
$$\frac{\sin x}{\cos x+\cosh x}=-2\sum_{k=1}^{\infty}(-1)^ke^{-xk}\sin(xk)$$ This relationship can be derived from the following geometric series sum:
$$\sum_{k=1}^{\infty}(-1)^ke^{-xk}e^{ixk}=-\frac{e^{-x}e^{ix}}{1+e^{-x}e^{ix}}$$ Imaginary part of this gives the relationship above. Thus, the integral:  
$$I(1)=-2\sum_{k=1}^{\infty}(-1)^k\int_{0}^{\infty}e^{-xk}\sin(xk)\frac{dx}{x}$$ But  
$$\int_{0}^{\infty}e^{-xk}\sin(xk)\frac{dx}{x}=\frac{\pi}{4}$$ This result follows from the well known integral  
$$\int_{0}^{\infty}e^{-xa}\sin(xk)dx=\frac{k}{a^2+k^2}$$ if we integrate the last with respect to $a$ from $a=k$ to $a=\infty$.  We get: 
$$I(1)=-2\sum_{k=1}^{\infty}(-1)^k\frac{\pi}{4}=\frac{\pi}{2}\sum_{k=1}^{\infty}(-1)^{k-1}$$ The sum  
$$\sum_{k=1}^{\infty}(-1)^{k-1}=1-1+1-1+...=\frac{1}{2}$$  was already known in the times of Leibniz.   
Finally:  
$$I(1)=\frac{\pi}{4}$$ But even in the next case of $m=2$ i have no clue, how to evaluate the integral. Numerical calculations suggest that the result holds for every $m$ including $m=2013$
