Compute $\int_0^\infty \frac{e^{-tx}\sin(x)}{x}dx$ I have to compute$$\int_0^\infty \dfrac{e^{-tx}\cdot \sin(x)}{x}dx$$
This is following a helping problem $$\int_0^\infty e^{-tx}\cdot \sin(x)dx$$ which using IPB two times turned out to be $$\dfrac{1}{1+t^2}$$
I think there must be a substitution to get to the first problem, but I just cannot see it. Any hint appreciated.
 A: We can compute $\int_0^{\infty}e^{-tx}\sin x/x dx$ for a positive $t$. Indeed, define 
$$F(t,x):=\frac{\sin x}xe^{-tx}.$$
If $\delta$ is a positive number, then for any non-negative $x$, $$\sup_{s\geqslant \delta}\partial_tF(s,x)|\leqslant e^{-\delta x}.$$
Since the map $x\mapsto e^{-\delta x}$ is integrable on $[0,\infty)$, we can take the derivative under the integral.
A: Define
$$
I = \int \limits_0 ^\infty e^{-tx}\frac{\sin(x)}{x}dx
$$
then
$$
\dfrac{dI}{dt}=-\int \limits_0 ^\infty e^{-tx}\sin(x)dx = -\frac{1}{1+t^2}
$$
hence
$$
I = -\arctan(t) + D
$$
We fix $D$ using $I(0)=D=\dfrac\pi 2$, easily obtainable through complex analysis.
A: $\newcommand{\+}{^{\dagger}}%
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$\ds{\large\tt\mbox{With}\quad t > 0}$:

\begin{align}
&\color{#00f}{\large\int_{0}^{\infty}{\expo{-tx}\sin\pars{x} \over x}\,\dd x}
=\int_{0}^{\infty}\expo{-tx}\pars{\half\int_{-1}^{1}\expo{\ic kx}\,\dd k}
=\half\int_{-1}^{1}\int_{0}^{\infty}\expo{\pars{\ic k - t}x}\,\dd k
\\[3mm]&=\half\int_{-1}^{1}{1 \over t - \ic k}\,\dd k
=t\int_{0}^{1}{\dd k \over t^{2} + k^{2}}
=\int_{0}^{1/t}{\dd k \over 1 + k^{2}}= \arctan\pars{1 \over t}
\\[3mm]&=\color{#00f}{\Large{\pi \over 2} - \arctan\pars{t}}
\end{align}

A: $$ 
\begin{aligned}
\int_{0}^{\infty} e^{-tx} \frac{\sin x}{x} \, \mathrm{d}x &= \int_{0}^{\infty} e^{-tx} \sin x \int_{0}^{\infty} e^{-zx} \,\mathrm{d}z\,\mathrm{d}x\\
&= \int_{0}^{\infty}\!\!\int_{0}^{\infty} e^{-x\left(z+t\right)} \sin x\,\mathrm{d}x\,\mathrm{d}z\\
&= \int_{0}^{\infty} \frac{\mathrm{d}z}{\left(z+t\right)^2 +1}\\
&= \left[\lim_{N\to\infty}\arctan(N+t)\right] - \arctan t \\
&= \frac{\pi}{2} - \arctan t\\
&= \operatorname{arccot} t
\end{aligned}
 $$
