How to evaluate integral: $ \int_{0}^{\infty} e^{-x}\left|\sin{x}\right| \ dx $ I try to evaluate integral below.I solved indefinite integral but after evaluating limit I get wrong result.I don't know where can be problem.Maybe I just use the wrong method?
$$ \int_{0}^{\infty} e^{-x}\left|\sin{x}\right| \ dx= $$
$$=  \left[ -\frac{1}{2}e^{-x}\operatorname{sgn}\left(\sin{x}\right)\left(\sin{x}+\cos{x}\right)\right]_0^\infty $$
 A: The problem comes from the fact that your antiderivative has discontinuities where $\sin x$ changes sign, and is not differentiable.
The correct integral can be found by summing the "jumps" required to restore continuity. (These jumps have amplitude $(-1)^ke^{-k\pi}$).
A: Integrating by parts twice, we get
$$
\int e^{-x}\sin(x)\,\mathrm{d}x=-\frac{\sin(x)+\cos(x)}2e^{-x}\tag1
$$
Thus,
$$
\int_{2k\pi}^{(2k+1)\pi} e^{-x}|\sin(x)|\,\mathrm{d}x=\frac12\left(e^{-2k\pi}+e^{-(2k+1)\pi}\right)\tag2
$$
and
$$
\int_{(2k+1)\pi}^{(2k+2)\pi} e^{-x}|\sin(x)|\,\mathrm{d}x=\frac12\left(e^{-(2k+1)\pi}+e^{-(2k+2)\pi}\right)\tag3
$$
Therefore,
$$
\begin{align}
\int_0^\infty e^{-x}|\sin(x)|\,\mathrm{d}x
&=\frac12+\sum_{k=1}^\infty e^{-k\pi}\\
&=\frac12+\frac{e^{-\pi}}{1-e^{-\pi}}\\
&=\frac12\frac{1+e^{-\pi}}{1-e^{-\pi}}\\[3pt]
&=\frac12\coth\left(\frac\pi2\right)
\end{align}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[15px,#ffd]{\int_{0}^{\infty}\expo{-x}\verts{\sin\pars{x}}\dd x = {1 \over 2}\,{\expo{\pi} + 1 \over \expo{\pi} - 1} =
{1 \over 2}\coth\pars{\pi \over 2}}:\ {\large ?}}$

\begin{align}
&\bbox[15px,#ffd]{\int_{0}^{\infty}\expo{-x}\verts{\sin\pars{x}}\dd x} =
\int_{0}^{\infty}\expo{-x}\mrm{sgn}\pars{\sin\pars{x}}\cos\pars{x}\,\dd x
\\[5mm] = &\
\int_{x\ =\ 0}^{x\ \to\ \infty}\mrm{sgn}\pars{\sin\pars{x}}
\,\dd\braces{{1 \over 2}\expo{-x}\bracks{\sin\pars{x} - \cos\pars{x}}}
\\[5mm] = &\
-\,{1 \over 2} - \int_{0}^{\infty}\braces{{1 \over 2}\expo{-x}
\bracks{\sin\pars{x} - \cos\pars{x}}}
\bracks{2\delta\pars{\sin\pars{x}}\cos\pars{x}}\,\dd x
\\[5mm] = &\
-\,{1 \over 2} + \int_{0}^{\infty}\expo{-x}\cos^{2}\pars{x}
\,\delta\pars{\sin\pars{x}}\,\dd x
\\[5mm] = &\
\sum_{n = -\infty}^{\infty}\int_{0}^{\infty}\expo{-x}\cos^{2}\pars{x}
\,\delta\pars{x - n\pi}\,\dd x
\\[5mm] = &\
-\,{1 \over 2} + \sum_{n = 0}^{\infty}\expo{-n\pi} =
-\,{1 \over 2} + {1 \over 1 - \expo{-\pi}} =
\bbox[15px,#ffd,border:1px solid navy]{{1 \over 2}\coth\pars{\pi \over 2}}\
\approx\ 0.5452
\end{align}
A: $$\int_{0}^{+\infty}e^{-x}|\sin x|\,dx =\sum_{k\geq 0}\int_{k\pi}^{(k+1)\pi}e^{-x}|\sin x|\,dx=\sum_{k\geq 0}(-1)^k\int_{k\pi}^{(k+1)\pi}e^{-x}\sin(x)\,dx$$
equals
$$ \sum_{k\geq 0}\int_{0}^{\pi}e^{-x-k\pi}\sin(x)\,dx =\int_{0}^{\pi}\sin(x)e^{-x}\sum_{k\geq 0}e^{-k\pi}\,dx=\frac{1}{1-e^{-\pi}}\int_{0}^{\pi}e^{-x}\sin(x)\,dx$$
or
$$ \frac{1}{1-e^{-\pi}}\,\text{Im}\int_{0}^{\pi}e^{(i-1)x}\,dx=\frac{1}{1-e^{-\pi}}\,\text{Im}\left[\frac{e^{(i-1)x}}{i-1}\right]_{0}^{\pi} =\frac{1}{1-e^{-\pi}}\,\text{Im}\left[\frac{-e^{-\pi}-1}{i-1}\right]=\frac{1}{2}\cdot\frac{1+e^{-\pi}}{1-e^{-\pi}}$$
that is $\color{blue}{\frac{1}{2}\coth\left(\frac{\pi}{2}\right)}$.
