Integral by parts $\int_0^\infty e^{-st}\frac{\sin(t)}{t} dt $ I want to solve the following integral by parts:
$$\int_0^\infty e^{-st}\frac{\sin(t)}{t} dt $$
I have been trying but I don't know what else to do. The result should be $\frac{\pi}{2}-\arctan\left(s\right) $. I took $\frac{\sin(t)}{t}$ as u and $ e^{-st} $ as dv, obtaining this:
$$ du = \bigg[\frac{\cos(t)}{t}-\frac{\sin(t)}{t^2}\bigg]dt $$
$$ v = -\frac{1}{s} e^{-st}$$
Applying the formula for definite integration by parts:
$$ uv\bigg|_0^\infty -\int_0^\infty vdu  $$
$$ -\frac{1}{s}e^{-st}\frac{\sin(t)}{t}\bigg|_0^\infty +\frac{1}{s}\int_0^\infty e^{-st}\bigg[\frac{\cos(t)}{t}-\frac{\sin(t)}{t^2}\bigg]dt  $$
From that point onwards, I am stuck. I would be most grateful if you may help me.
Thanks in advance.
 A: Per double integral
$$\begin{align}
\int_0^\infty \frac{e^{-st}}t \sin t \>dt
=\int_0^\infty \int_s^\infty e^{-xt}\sin t \>dx \>dt
= \int_s^\infty \frac1{1+x^2}dx = \tan^{-1}\frac1s\\
\end{align}$$
A: Your method doesn't work. Here is  small trick which makes the evaluation of the integral easy: Let $F(u)=\int_0^{\infty} e^{-st} \frac {\sin (ut)} t dt$. Then $F'(u)=\int_0^{\infty} e^{-st}\cos (ut) dt$. This is a standard integral. I will leave the evaluation (by two integration by parts) to you. Once you write down $F'(u)$ and observe that $F(0)=0$ you can write down $F(u)$ (in terms of $\arctan$). Put $u=1$ to finish. 
A: So, your integral is the Laplace transform of $\sin(t)/t$. Per an identity on this reference sheet,
$$\mathcal L \left\{ \frac{f(t)}{t} \right\} = \int_s^\infty F(u) du$$
where $F$ is the Laplace transform of $f$. Take $f(t) = \sin(t)$. Then by this identity and formula $(7)$ from the sheet with $a=1$,
$$\int_0^\infty e^{-st} \frac{\sin(t)}{t} dt = \int_s^\infty \frac{1}{u^2 + 1} du$$
The latter is a fairly standard integral.
$$\int_s^\infty \frac{1}{u^2 + 1} du = \lim_{u \to \infty} \arctan(u) - \arctan(s) = \frac \pi 2 - \arctan(s)$$
