How to evaluate this integral ?

Question

$$ I(\alpha)=\displaystyle \int_{0}^{\infty}\dfrac{\arctan(x)}{e^{\alpha x}} \,dx $$ for all alpha>=0

I tried to do it with Laplace transform , but having trouble with finding LT of arctan(x), any other suggestions ?

Answer 1

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 $$ \text{Ci}(z)=- \int_{z}^{\infty}\frac{\cos t}{t}\:dt,\quad\text{Re}\:z>0, \qquad \text{si}(z)=-\int_{z}^{\infty}\frac{\sin t}{t}\:dt, \quad z \in \mathbb{C}. $$

Answer 2

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?