Integrating Absolutes

$\begin{aligned} I & = \int_0^\infty e^{-x} |\sin x| \ dx & \small \color{#3D99F6} \text{For }k \pi \le x < (k+1) \pi, |\sin x| = \begin{cases} \sin x & \text{if }k \text{ is even} \\ - \sin x & \text{if }k \text{ is odd} \end{cases} \\ & = \sum_{k=0}^\infty (-1)^k \int_{k\pi}^{(k+1)\pi} e^{-x} \sin x \ dx & \small \color{#3D99F6}{\text{By integration by parts, see Note}} \\ & = \sum_{k=0}^\infty \frac {e^{-k\pi} + e^{-(k+1)\pi}}2 \\ & = \frac 12 + \color{#3D99F6}{\sum_{k=1}^\infty e^{-k\pi}} & \small \color{#3D99F6}{\text{A sum of geometric progression}} \\ & = \frac 12 + \color{#3D99F6}{\frac {e^{-\pi}}{1-e^{-\pi}}} \\ & = \boxed{\dfrac {1+e^{-\pi}}{2(1-e^{-\pi})}} \end{aligned}$

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Integration by parts

$\begin{aligned} J & = \int_{k\pi}^{(k+1)\pi} e^{-x} \sin x \ dx & \small \color{#3D99F6}{f'(x) = e^{-x}, \ g(x) = \sin x} \\ & = e^{-x}\sin x\bigg| _{(k+1)\pi}^{k\pi} + \int_{k\pi}^{(k+1)\pi} e^{-x} \cos x \ dx & \small \color{#3D99F6}{f'(x) = e^{-x}, \ g(x) = \cos x} \\ & = 0 + e^{-x} \cos x \bigg| _{(k+1)\pi}^{k\pi} - \color{#3D99F6}{\int_{k\pi}^{(k+1)\pi} e^{-x} \sin x \ dx} \\ & = e^{-(k+1)\pi} + e^{-k\pi} - \color{#3D99F6}{J} \\ \implies (-1)^k J & = \frac {e^{-(k+1)\pi} + e^{-k\pi}}2 \end{aligned}$