Let $a \in \mathbb{C}$. Consider the integral $$\int_{-\infty}^{+\infty} \frac{e^{-ax}}{1 + e^x} dx,$$ for which values of $a$ is this convergent? Is it right to say that $a$ has to be purely imaginary?
I have been staring at this for way too long, so I need some reassurance. Thanks in advance!
On
As $x$ approaches $\infty$, we have
$$\frac{e^{-ax}}{1+e^x}\sim e^{-(a+1)x}$$
Thus, at the upper limit, we require that $\text{Re}(a+1)>0 \implies \text{Re}(a)>-1$ in order for the integral to converge.
As $x$ approaches $-\infty$, we have
$$\frac{e^{-ax}}{1+e^x}\sim e^{-ax}$$
Thus, at the lower limit, we require that $\text{Re}(a)<0$.
Thus, the integral converges for
$$\bbox[5px,border:2px solid #C0A000]{0>\text{Re}(a)>-1}$$
It will look more familiar after a change of variables: set $y=e^x$. Then the limits become $0$ and $\infty$, and $dy/y = dx$, so the integral is $$ \int_0^{\infty} \frac{y^{-a-1}}{1+y} \, dy $$ Near zero, this integrand looks like $y^{-a-1}$, which has finite integral when $\Re(a)<0$. Near $\infty$, the integrand looks like $y^{-a-2}$, which has finite integral when $\Re(a)>-1$. Therefore the whole integral converges when $-1<\Re(a)<0$.