MHB Definite integral involving sine and hyperbolic sine

[MountEvariste]

Calculate $\displaystyle \int_0^{\infty} \frac{\sin x}{\cos x + \cosh x}\, \mathrm dx.$

 

[HOI]

If nothing else, you can express these functions as exponentials.

$sin(x)= \frac{e^{ix}- e^{-ix}}{2i}$
$cos(x)= \frac{e^{ix}+ e^{-ix}}{2}$
$cosh(x)= \frac{e^x+ e^{-x}}{2}$

 

[MountEvariste]

If nothing else, you can express these functions as exponentials.

$sin(x)= \frac{e^{ix}- e^{-ix}}{2i}$
$cos(x)= \frac{e^{ix}+ e^{-ix}}{2}$
$cosh(x)= \frac{e^x+ e^{-x}}{2}$

Spoiler: Good start

Can you derive an infinite series of the integrand using these definitions?

 

[MountEvariste]

Spoiler: Hint:

Show that $\displaystyle \frac{\sin x }{\cos x + \cosh x} = i \bigg( \frac{1}{1+e^{ix-x}}-\frac{1}{1+e^{-ix-x}}\bigg)$.

Expand the RHS into geometric series to get, for $x \ge 0$:

$\displaystyle \frac{\sin x}{\cos x + \cosh x}=2\sum_{n=1}^{\infty}(-1)^{n-1}\sin(nx)e^{-nx}.$

Source: Yaghoub Sharifi.

 

[MountEvariste]

Spoiler: Solution/link

Using the infinite sum in post #4, the answer follows by switching the order of sum and integral and using the result

$\displaystyle \int_0^{\infty} e^{-ax}\sin{bx} \, \mathrm dx =\frac{b}{a^2+b^2}$

Which can be derived via integration by parts or considering the fact that $\Im \left( e^{-ax+ibx} \right) = e^{-ax}\sin bx.$

See a detailed solution in this blog.