Improper integral with module

Question

faced with a problem when calculating the value of the integral

$$ \int_{0}^{\infty} e^{-x}|\sin(x)|\, \mathrm{d}x$$

Is there any idea how?

Answer 1

Break up the integral into the sum of integrals

$$\begin{align} \sum_{n=0}^\infty \int_{n\pi}^{(n+1)\pi} e^{-x}|\sin(x)|\,dx&=\sum_{n=0}^\infty (-1)^n \int_{n\pi}^{(n+1)\pi} e^{-x}\sin(x)\,dx\\\\ &=\frac12 (1+e^{-\pi})\sum_{n=0}^\infty e^{-n\pi}\\\\ &=\frac{1+e^{-\pi}}{2(1-e^{-\pi})}\\\\ &=\frac12 \coth(\pi/2) \end{align}$$

Answer 2

Our integral equals:

$$ \sum_{n\geq 0}\int_{0}^{\pi}e^{-x-n\pi}\sin(x)\,dx = \frac{e^{\pi}}{e^{\pi}-1}\int_{0}^{\pi}e^{-x}\sin x\,dx =\color{red}{\frac{1}{2}\coth\frac{\pi}{2}}.$$