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.

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. 
