How to evaluate these integrals by hand

Question

How can I evaluate$$\int_{-\infty}^\infty \frac{\cos x}{\cosh x}\,\mathrm dx\text{ and }\int_0^\infty\frac{\sin x}{e^x-1}\,\mathrm dx.$$ Thanks in advance.

Answer 1

For the second one,

$$ \begin{align*} \int_{0}^{\infty} \frac{\sin x}{e^x - 1} \; dx &= \int_{0}^{\infty} \frac{\sin x \, e^{-x}}{1 - e^{-x}} \; dx \\ &= \int_{0}^{\infty} \left( \sum_{n=1}^{\infty} \sin x \, e^{-nx} \right) \; dx \\ &\stackrel{\ast}{=} \sum_{n=1}^{\infty} \int_{0}^{\infty} \sin x \, e^{-nx} \; dx \\ &= \sum_{n=1}^{\infty} \frac{1}{1+n^2}, \end{align*}$$

where the starred identity is justified by the following formula

$$ \begin{align*} \int_{0}^{\infty} \frac{\sin x}{e^x - 1} \; dx &= \int_{0}^{\infty} \left( \frac{1 - e^{-Nx}}{1 - e^{-x}} e^{-x} + \frac{e^{-Nx}}{e^x - 1} \right) \sin x \; dx \\ &= \sum_{n=1}^{N} \int_{0}^{\infty} \sin x \, e^{-nx} \; dx + \int_{0}^{\infty} \frac{\sin x \, e^{-Nx}}{e^x - 1} \; dx, \end{align*}$$

together with the dominated convergence theorem. Now the resulting infinite summation can be evaluated in numerous ways. For example, exploiting identities involving the digamma function,

$$ \sum_{n=1}^{\infty} \frac{1}{1+n^2} = \frac{1}{2i} \sum_{n=1}^{\infty} \left( \frac{1}{n-i} - \frac{1}{n+i} \right) = \frac{\psi_0(1+i) - \psi_0(1-i)}{2i} = -\frac{1}{2} + \frac{\pi}{2} \coth \pi. $$

Similar techniques apply to the first integral.

Answer 2

For the first one for example by using the rectangular contour $-R\to +R \to +R+2\pi i \to -R+2\pi i \to -R$
and since there are two simple poles from $\cosh(z)$ at $z=\frac {\pi i}2$ and $z=\frac {3\pi i}2$ of values $-ie^{-\frac {\pi i}2}$ and $ie^{-\frac {3\pi i}2}$ we have : $$2\pi i\ \left(\rm{Res}\left(\frac {e^{iz}}{cosh(z)}; \frac {\pi i}2\right)+\rm{Res}\left(\frac {e^{iz}}{cosh(z)}; \frac {3\pi i}2\right)\right)=2\pi i \left(-ie^{-\frac {\pi}2}+ie^{-\frac {3\pi}2}\right)$$

so that the integral over the rectangular contour of $\dfrac {e^{i x}}{\cosh(x)}$ will be : $$\int_{-R}^R \frac {e^{i x}}{\cosh(x)}\;dx+\int_0^{2\pi} \frac {e^{i R-y}}{\cosh (R+iy)}\;dy+\int_R^{-R} \frac {e^{-2\pi+i x}}{\cosh(2\pi i+x)}\;dx+\int_{2\pi}^0 \frac {e^{-i R-y}}{\cosh (-R+iy)}\;dy=2\pi \left(e^{-\frac {\pi}2}-e^{-\frac {3\pi}2}\right)$$

I'll let you prove that the two integrals in $y$ disappear as $R\to\infty$ so that only remains :

$$\lim_{R\to \infty} \int_{-R}^R \frac {e^{i x}}{\cosh(x)}\;dx-\lim_{R\to \infty}\int_{-R}^R \frac {e^{-2\pi-i x}}{\cosh(2\pi i-x)}\;dx=2\pi \left(e^{-\frac {\pi}2}-e^{-\frac {3\pi}2}\right)$$

it is easy to show that $\ \cosh(2\pi i+x)=\cosh(x)$ (use exponential notation) so that :

$$(1-e^{-2\pi})\int_{-\infty}^\infty \frac {e^{i x}}{\cosh(x)}\;dx=2\pi \left(e^{-\frac {\pi}2}-e^{-\frac {3\pi}2}\right)$$

and the result (keeping the real part) : $$\int_{-\infty}^\infty \frac {\cos x}{\cosh x}\;dx=2\pi \frac{e^{-\frac {\pi}2}}{1+e^{-\pi}}=\frac{\pi}{\cosh \frac{\pi}2}$$

(the second one may be solved the same way...)