when the following improper integral $\int_{-\infty}^\infty \frac{e^{-at}}{1+e^t} \mathrm{d}t$ converges according to $a$? I need to study the nature of the following integral (diverges or converges) according to $a$ :
$$I_a = \int_{-\infty}^\infty \frac{e^{-at}}{1+e^t} \mathrm{d}t$$
My approach was to dominate the integrand by a Riemann function and deduce $a$ but I couldn't even think of an example, I saw some examples on the website but they were using $u$-substitution, which is not a wise solution I think.
 A: Let's split this up into a few cases:

*

*Case: $a<-1$. Note that for $a<-1$, the Integral diverges because $$\lim_{x\to\infty}\frac{e^{-ax}}{1+e^x}= \infty$$


*Case: -1<a<0.For these values, the integral converges, which is easy to check with the substitution $$u=e^x$$ and the convergence of $$\int\limits_1^{\infty}\frac{1}{x^p}dx\text{ and }\int\limits_0^1\frac{1}{x^p}dx$$ for p>1 and p<1


*Case: a>0. Very similar to the 1st case,$$\lim_{x\to -\infty}\frac{e^{-ax}}{1+e^x}=\infty$$and thus the integral diverges


*Case: $a=-1$ $$\int\frac{e^x}{1+e^x}dx=\log{(e^x+1)}$$ and thus the integral also diverges for a=-1


*Case: $a=0$ $$\int\frac{1}{1+e^x}dx=x-\log{(e^x+1)}$$and this also diverges.
Summary: the integral only converges for $a\in(-1,0)\subset\mathbb{R}$
A: We have
$$
\frac{e^{-at}}{1+e^t}=e^{-at}\frac{e^{-t/2}}{e^{-t/2}+e^{t/2}}=\frac{e^{-(a+1/2)t}}{2}\mathrm{sech}(t/2).
$$
Since for $|t|$ large
$$
\mathrm{sech}(t/2)\sim \frac{2}{e^{|t|/2}},
$$
we obtain
$$
\frac{e^{-at}}{1+e^t} \sim \frac{e^{-(a+1/2)t}}{e^{|t|/2}}, \quad \text{for } |t|\text{ large}.
$$
Thus, at $t\to\infty$ this behaves as
$$
\frac{e^{-at}}{1+e^t} \sim e^{-(a+1)t}
$$
which is integrable in $(0,\infty)$ for $-1<a$. Similarly, at $t\to-\infty$ this behaves as
$$
\frac{e^{-at}}{1+e^t} \sim e^{-at}
$$
which is integrable in the interval $(-\infty,0)$ for $a<0$. Therefore
$$
\frac{e^{-at}}{1+e^t}
$$
is integrable in $(-\infty,\infty)$ for $-1<a<0$.
