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}

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,
$$\int_0^\infty\frac{\sin(2013x)}{x(\cos(x)+\cosh(x))}\mathrm{d}x=\frac\pi4$$