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.


2 Answers 2


Let's split this up into a few cases:

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

  2. 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

  3. 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

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

  5. 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 $a\in(-1,0)\subset\mathbb{R}$

  • $\begingroup$ (+1) Does $\lim_{x\to \infty} f(x)=\infty$ means that $\int f$ diverges (I've never used this property before.) $\endgroup$ Dec 29, 2022 at 17:44
  • 1
    $\begingroup$ it doesn't always mean that, but we can easily find a divergent lower bound. Note that $$\int\limits_0^1\frac{1}{\sqrt{x}}dx$$ converges even though the function goes to $\infty$ at $x\to 0$. Also if f is strictly increasing and positive on $(a,\infty)$ and $\lim_{x\to \infty}f(x)=\infty$, then $$\int_a^{\infty}f(x)dx$$ diverges $\endgroup$
    – Fix
    Dec 29, 2022 at 17:51
  • $\begingroup$ Thanks it makes sense now ! $\endgroup$ Dec 29, 2022 at 18:15

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 $-1<a<0$.


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