Explore convergence of improper integral $\int_{0}^{\infty} \frac{e^{-ax}}{(1+x)^2}dx$

Question

First I thought of Dirichlet's principle, but we don't have a bounded function. Then, I assumed, we can use it after decomposition in parts: $$\int_{0}^{\infty} \frac{e^{-ax}}{(1+x)^2}dx = \lim_{t\to\infty } e^{-ax} \arctan(x)\bigg|_0^t + \frac{1}{a} \lim_{t\to\infty } \int_{0}^{t}e^{-ax} \arctan(x)dx,$$but it seems pointless to me. What should be done?

Answer 1

Hint

For $a>0$ $$0<{e^{-ax}\over (1+x)^2}\le{e^{-ax}}.$$

Answer 2

$$\int_0^{\infty } \frac{e^{-a x}}{(x+1)^2} \, dx=\int_0^1 \frac{e^{-a x}}{(x+1)^2} \, dx+\int_1^{\infty } \frac{e^{-a x}}{(x+1)^2} \, dx=$$ $$=\left(1-\frac{e^{-a}}{2}-a e^a \Gamma (0,a)+a e^a \Gamma (0,2 a)\right)+\left(\frac{e^{-a}}{2}-a e^a \Gamma (0,2 a)\right)=1-a e^a \Gamma (0,a)$$ which exists only if $a>0$