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.

If $a=0$ then evaluating the integral is routine.
If $a>0$ then we have $\displaystyle \int_0^\infty \frac{e^{-ax}}{x^2+1}\le\int_0^\infty e^{-ax}\,dx$, and that converges, so the original integral converges by comparison.
If $x<0$ then $\dfrac{e^{-ax}}{x^2+1}\to\infty$ as $x\to\infty$, so the integral diverges.
You're main mistake is trying to actually evaluate the integral.  You don't need to do that in order to answer the question.
