$\int_{0}^{\infty} \frac{e^{-x} \sin(x)}{x} dx$ Evaluate Integral Compute the following integral:
$$\int_{0}^{\infty} \frac{e^{-x} \sin(x)}{x} dx$$
Any hint, suggestion is welcome. 
 A: If you know a bit about Fourier theory. You could Parseval's theorem
$$\int \!dx \,f(x) g(x)^* = \int \!d\xi\,\hat f(\xi) \hat g(\xi)^* $$
with $f(x) = \sin(x)/x$, $g(x) = \Theta(x) e^{-x}$ and $\hat{f}$, $\hat{g}$ their Fourier transforms and $\Theta(x)$ the Heaviside step function.
Hint: $\hat{f}(\xi) = \tfrac12\sqrt{\frac{\pi}{2}} [\Theta(1-\xi) + \Theta(1+\xi)] =\sqrt{\frac{\pi}{2}} \mathop{\rm rect}(\xi) $.
A: Let 
$$f(z) = \frac{e^{-z+iz}}{z}$$
and let $C$ be the contour that travels along $0$ to $R$, makes a quarter of a circle around to $iR$ and back to $0$, properly indented around $0$ with a quarter circle of radius $\delta$ to avoid the pole.  

As $R \to \infty$, the integral over rounded part of the contour tends to $0$ and the part around $0$ tends to $-i\frac{\pi}{2}$ (N.B. this is $-i\frac{\pi}{2}$ of the residue at $z=0$) as $\delta \to 0$.  Then by Cauchy's theorem:
$$
0=\oint_C f(z)\,dz =\\
\int_0^\infty \frac{e^{-z+iz}}{z}\,dz -\int_0^{\infty} \frac{e^{-iz-z}}{z}\,dz - i\frac{\pi}{2}
$$
And upon taking imaginary parts and solving:
$$
\frac{\pi}{4}=\int_0^\infty \frac{e^{-z}\sin(z)}{z}\,dx
$$
A: Yet a different approach: parametric integration. Let
$$
F(\lambda)=\int_{0}^{\infty} \frac{e^{-\lambda x} \sin(x)}{x}\,dx,\qquad\lambda>0.
$$
Then
$$
F'(\lambda)=-\int_{0}^{\infty} e^{-\lambda x} \sin(x)\,dx=-\frac{1}{1+\lambda^2}.
$$
Integrating and taking into account that $\lim_{\lambda\to\infty}F(\lambda)=0$ we have
$$
F(\lambda)=\frac\pi2-\arctan\lambda
$$
and
$$
\int_{0}^{\infty} \frac{e^{-x} \sin(x)}{x}\,dx=F(1)=\frac\pi4.
$$
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
&\color{#00f}{\large\int_{0}^{\infty}{\expo{-x}\sin\pars{x} \over x}\,\dd x}
=\int_{0}^{\infty}\expo{-x}\pars{\half\int_{-1}^{1}\expo{\ic kx}\,\dd k}
=\half\int_{-1}^{1}\int_{0}^{\infty}\expo{\pars{\ic k - 1}x}\,\dd k
\\[3mm]&=\half\int_{-1}^{1}{1 \over 1 - \ic k}\,\dd k
=\int_{0}^{1}{\dd k \over 1 + k^{2}} = \arctan\pars{1}
=\color{#00f}{\Large{\pi \over 4}}\ \ \ \
\end{align}
A: $$\int_0^{\infty} \frac{e^{-x}\sin x}{x}\,dx=\int_0^{\infty}\int_0^{\infty} e^{-x}\sin x \,e^{-xy}\,dy\,dx=\int_0^{\infty} \int_0^{\infty} e^{-x(1+y)}\sin x\,dx\,dy$$
From integration by parts or otherwise, one can show that:
$$\int_0^{\infty}e^{-x(1+y)}\sin x\,dx=\frac{1}{1+(1+y)^2}$$
Hence
$$\int_0^{\infty} \int_0^{\infty} e^{-x(1+y)}\sin x\,dx\,dy=\int_0^{\infty} \frac{dy}{1+(1+y)^2}=\left(\arctan(1+y)\right|_0^{\infty}=\boxed{\dfrac{\pi}{4}}$$
A: Using Laplace Transform,
$$\mathcal{L}(\sin(x)) = \frac{1}{s^2 + 1}$$
$$\mathcal{L}\left(\frac{\sin(x)}{x}\right) = \int_r^\infty \frac{1}{s^2 + 1} ds = \frac{\pi}{2} - \arctan(r)$$
Therefore,
$$\int_0^\infty e^{-rx} \frac{\sin(x)}{x} dx = \frac{\pi}{2} - \arctan(r)$$
Substituting r = 1,
$$\int_0^\infty e^{-x} \frac{\sin(x)}{x} dx = \frac{\pi}{4}$$
A: Another approach: 
$$\begin{eqnarray*}
\int_{0}^{\infty} dx\, \frac{e^{-x} \sin(x)}{x} 
&=& \int_{0}^{\infty}dx\, \frac{e^{-x}}{x}  \sum_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{(2k+1)!}  \\
&=& \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}
\int_{0}^{\infty}dx\, x^{2k} e^{-x}    \\
&=& \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}(2k)! \\
&=& \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} 
    \hspace{5ex} \textrm{(Leibniz series for $\pi$)}\\
&=& \frac{\pi}{4}.
\end{eqnarray*}$$
A: Write this as
$$
\lim_{\epsilon\to0}\int_\epsilon^{1/\epsilon}\frac{e^{-(1-i)x}-e^{-(1+i)x}}{2ix}\,\mathrm{d}x\tag{1}
$$
and then consider the path integral
$$
\frac{1}{2i}\int_{\gamma_\epsilon} e^{-z}\,\frac{\mathrm{d}z}{z}\tag{2}
$$
where $\gamma_\epsilon$ comes in along the line $(1+i)x$, makes a quarter circle clockwise along $|z|=\epsilon$, goes out along the line $(1-i)x$ and then back a quarter circle counter-clockwise along $|z|=1/\epsilon$. There are no poles inside this path, so the integral in $(2)$ is $0$.
The part along $|z|=1/\epsilon$ dies away exponentially as $\epsilon\to0$. The two parts along the lines sum to our integral, $(1)$, and the part along $|z|=\epsilon$ tends to $\frac14$ of the integral of $\frac{1}{2iz}$ clockwise around the origin; that is, $-\pi/4$. Since the sum of these parts is $0$, the limit in $(1)$ must be $\pi/4$. That is,
$$
\int_0^\infty\frac{e^{-x}\sin(x)}{x}\mathrm{d}x=\frac{\pi}{4}\tag{3}
$$
