Convergence of $\int^{\infty}_{0} \frac{1}{e^x+1}dx$

Question

I'm trying to prove the convergence of

$$\int^{\infty}_{0} \frac{1}{e^x+1}dx$$

Any suggestion or hint for how to deal with it?

Thanks in advance!

Answer 1

What if you compare the integrand to $\mathrm{e}^{-x}$ because we know how to integrate that. We also know that it is always greater than $0$, so use the squeeze theorem.

Answer 2

Use the substitute $e^x+1=t$. Then your integral becomes $\displaystyle \int_2^{\infty}\frac{1}{t(t-1)}dt.$

For $t>2$, $t^2>2t$. Hence, $\displaystyle \frac{1}{t(t-1)}=\frac{1}{t^2-t}\leq \frac{1}{t^2-t^2/2}=\frac{2}{t^2}.$

$\displaystyle \int_2^{\infty}\frac{1}{t^2}dt $ converges since $2>1$. So by comparison test, the integral converges.

Answer 3

It is sufficient to show that the integral from $0$ to $\infty$ of $\exp (-x)$ converges. But this is merely that the integral from - $\infty$ to $0$ of $\exp(x) $converges. That is, $\exp(x)$ from $-\infty$ to $0$ converges, which is just $1-0=1$.

Answer 4

Rewrite the integral as \begin{equation} \lim_{n \to \infty} \int_{0}^n \frac{1}{1+e^x} dx = \lim_{n \to \infty} \int_{0}^n \left(1- \frac{e^x}{1+e^x}\right) dx \end{equation}

Now, evaluate the integral above to obtain \begin{equation} n - \ln(1+e^n) + \ln2 = \ln \left( \frac{e^n}{1+e^n}\right) + \ln2 \end{equation}

Now, send $n \to \infty$ and use the continuity of $t \mapsto \ln(t)$ to push the limit inside, you will immediately obtain the value of integral as $\ln2$.