How to evaluate this integral ? $$ I(\alpha)=\displaystyle \int_{0}^{\infty}\dfrac{\arctan(x)}{e^{\alpha x}} \,dx     $$  for all alpha>=0 
I tried to do it with Laplace transform , but having trouble with finding LT of arctan(x), any other suggestions ?
 A: Hint. We assume $\alpha>0$. By an integration by parts (as proved by @the_candyman), one has
$$
I(\alpha) = \left[\frac{e^{-\alpha x}}{-\alpha}\cdot\arctan x\right]_0^\infty+\frac{1}{\alpha} \int_0^{\infty} \frac{e^{-\alpha x}}{1+x^2}\:dx=0+\frac{1}{\alpha} \int_0^{\infty} \frac{e^{-\alpha x}}{1+x^2}\:dx
$$ then one deduces
$$
\begin{align}
I(\alpha) &=\frac{i}{2\alpha}  \int_0^{\infty} \left(\frac{1}{x+i}-\frac{1}{x-i}\right)e^{-\alpha x}dx
\\\\I(\alpha)&=-\frac{1}{\alpha}\cdot\text{Im}\int_0^{\infty} \frac{e^{-\alpha x}}{x+i}\:dx
\end{align}
$$giving

$$
\begin{align}
I(\alpha)=\int_{0}^{\infty}e^{-\alpha x}\arctan(x)\,dx  &=\frac{\sin \alpha}{\alpha}\cdot \text{Ci}(\alpha)+\frac{\cos \alpha}{\alpha}\cdot \text{si}(\alpha), \qquad \alpha>0,
\end{align}
$$ 

where we have used the special functions $\text{Ci}$ and $\text{si}$ given by
$$
\text{Ci}(z)=- \int_{z}^{\infty}\frac{\cos t}{t}\:dt,\quad\text{Re}\:z>0, \qquad \text{si}(z)=-\int_{z}^{\infty}\frac{\sin t}{t}\:dt, \quad z \in \mathbb{C}.
$$
A: You can write your integral as follows:
$$I(\alpha)=\displaystyle \int_{0}^{\infty}\arctan(x)e^{-\alpha x} \,dx.$$
Recall that you can use the integration by parts: $$\int f(x)g'(x)dx = f(x)g(x) - \int f'(x) g(x) dx.$$
Using this fact, you can pose:
$$\begin{cases}
f(x) = \arctan(x)\\
g'(x) = e^{-\alpha x}
\end{cases} \Rightarrow 
\begin{cases}
f'(x) = \frac{1}{1+x^2}\\
g(x) = -\frac{1}{\alpha}e^{-\alpha x}
\end{cases}.$$
Then:
$$I(\alpha) = \left[-\frac{1}{\alpha}\arctan(x)e^{-\alpha x}\right]_{x=0}^{x=+\infty} - \int_0^{+\infty} -\frac{1}{\alpha}e^{-\alpha x}\frac{1}{1+x^2}dx .$$
Can you go on from this point?
