When does this integral converge? $\;\;\int_0^\infty {\frac{e^{-ax}}{x^2+1}}\,\mathrm{d}x$ $$\int_0^\infty \frac{e^{-ax}}{x^2+1}\,\mathrm{d}x$$  - $a$ real, for which a does this converge? (The final answer is $a\ge 0$)
I've tried doing this by parts and it seems to work at first, but then everything cancels and I get $0=0$ at the 2nd passage. I can't come up with any manipulations, any hints?
Thanks.

Hint


*

*For $a\ge0$ we have


$$\frac{e^{-ax}}{1+x^2}\le \frac1{1+x^2}\in L^1(0,\infty)$$
and


*

*For $a<0$ we have


$$\frac{e^{-ax}}{1+x^2}\ge x,\quad\text{for $x$ sufficently large}$$
