Prove that

∫∞0sin(2013x)x(cosx+coshx)dx=π4

**Answer**

Because the integrand is even, contour integration yields

∫∞0sin(mx)x(cos(x)+cosh(x))dx=12∫∞−∞sin(mx)x(cos(x)+cosh(x))dx=12∫∞−i−∞−isin(mx)x(cos(x)+cosh(x))dx=14i∫γ+eimzz(cos(z)+cosh(z))dz−14i∫γ−e−imzz(cos(z)+cosh(z))dz=2πi4i12+22πi4i∞∑k=0(−1)k+1sech(2k+12π)2k+12πcos(m2k+12π)exp(m2k+12π)=π4+π∞∑k=0(−1)k+1sech(2k+12π)2k+12πcos(m2k+12π)exp(m2k+12π)=π4for odd m

where γ+ goes from −R−i to +R−i then circles counterclockwise back in the upper half plane along |z+i|=R and γ+ 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)=eimzz(cos(z)+cosh(z)) at z=0 is 12.

Let α=1+i2. As noted in this answer, f has singularities at ±(2k+1)πα and ±(2k+1)π¯α.

The sum of the residues in the upper half plane at (2k+1)πα and −(2k+1)π¯α is the same as the sum of the residues in the lower half plane at (2k+1)π¯α and −(2k+1)πα. Both are equal to

(−1)k+1sech(2k+12π)2k+12πcos(m2k+12π)exp(m2k+12π)

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

**Attribution***Source : Link , Question Author : user 1591719 , Answer Author : Community*