Prove that ∫∞0sin(2013x)x(cosx+coshx)dx=π4\int_0^{\infty} \frac{\sin(2013 x)}{x(\cos x+\cosh x)}dx=\frac{\pi}{4}

Prove that

Because the integrand is even, contour integration yields

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

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,