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
0sin(2013x)x(cosx+coshx)dx=π4

Answer

Because the integrand is even, contour integration yields
0sin(mx)x(cos(x)+cosh(x))dx=12sin(mx)x(cos(x)+cosh(x))dx=12iisin(mx)x(cos(x)+cosh(x))dx=14iγ+eimzz(cos(z)+cosh(z))dz14iγeimzz(cos(z)+cosh(z))dz=2πi4i12+22πi4ik=0(1)k+1sech(2k+12π)2k+12πcos(m2k+12π)exp(m2k+12π)=π4+πk=0(1)k+1sech(2k+12π)2k+12πcos(m2k+12π)exp(m2k+12π)=π4for odd m
where γ+ goes from Ri to +Ri then circles counterclockwise back in the upper half plane along |z+i|=R and γ+ goes from Ri to +Ri 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

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