Improper Integral Question $ \int^{\infty}_0 \frac{\mathrm dx}{1+e^{2x}}$

Question

I want to check if it's improper integral or not $$ \int^{\infty}_0 \frac{\mathrm dx}{1+e^{2x}}.$$ What I did so far is :
set $t=e^{x} \rightarrow \mathrm dt=e^x\mathrm dx \rightarrow \frac{\mathrm dt}{t}=dx $ so the new integral is: $$ \int^{\infty}_0 \frac{\mathrm dt}{t(1+t^2)} = \int^{\infty}_0 \frac{\mathrm dt}{t}-\frac{\mathrm dt}{1+t^{2}}$$ now how I calculate the improper integral, I need to right the $F(x)$ of this integral and then to check the limit?
Thanks!

Answer 1

You made a couple of mistakes. Firstly, you forgot to change the limits of the integration, so your integral is actually $\displaystyle\int_1^\infty \frac{\mathrm{d}t}{t(1+t^2)}$. Furthermore, $\frac{1}{t(1+t^2)} \neq \frac{1}{t}-\frac{1}{1+t^2}$. Rather $\frac{1}{t(1+t^2)} = \frac{1}{t}-\frac{t}{1+t^2}$.

Hence your integral becomes $\displaystyle\int_1^\infty \frac{1}{t}-\frac{t}{1+t^2}\,\mathrm{d}t = \left[\log(t)-\frac{1}{2}\log(1+t^2)\right]_1^\infty = \left[\frac{1}{2}\log\left(\frac{t^2}{1+t^2}\right)\right]_1^\infty$ $$ = \frac{1}{2}\left[\log(1)-\log\left(\frac{1}{2}\right)\right] = \frac{1}{2}\log(2).$$

Answer 2

Hints:

First, it must be

$$\frac1{t(1+t^2)}=\frac1t-\frac t{1+t^2}\;\;,\;\;\text{then}:$$

$$\int\limits_1^\infty\left(\frac1 t-\frac t{1+t^2}\right)dt:=\lim_{b\to\infty}\left(\log\frac b{\sqrt{1+b^2}}+\log\sqrt 2\right)$$

Answer 3

Does this solution make sense too?

Let   $\displaystyle t=1+e^{2x},$  $x\in(0,\infty), t\in(2,\infty)$

So   $\displaystyle dx = \frac{1}{2(t-1)}dt$,

Then,

      $\displaystyle\int^{\infty}_0\frac{1}{1+e^{2x}}dx$

  $\displaystyle= \frac{1}{2}\int^{\infty}_2\frac{1}{t(t-1)}dt$   (Please pay attention to the changing interval)

  $\displaystyle= \frac{1}{2}\int^{\infty}_2(\frac{1}{t-1} - \frac{1}{t})dt$

  $\displaystyle= \frac{1}{2}\left(\left[\ln{\left(t-1\right)}\right]_2^{\infty} - \left[\ln{t}\right]_2^{\infty}\right)$

  $\displaystyle= \frac{1}{2}\lim_{t\to\infty}\ln{(t-1)} - \frac{1}{2}\lim_{t\to\infty}\ln{t} + \frac{1}{2}\ln{2}$

  $\displaystyle= \frac{1}{2}\lim_{t\to\infty}\ln{\frac{t-1}{t}} + \frac{1}{2}\ln{2}$

  $\displaystyle= \frac{1}{2}\ln{2}$

Answer 4

$$ \begin{aligned} \int_{0}^{\infty} \frac{d x}{1+e^{2 x}} &=\int_{0}^{\infty} \frac{e^{-2 x}}{e^{-2 x}+1} d x \\ &=-\frac{1}{2} \int_{0}^{\infty} \frac{d\left(e^{-2 x}+1\right)}{e^{-2 x}+1} \\ &=-\frac{1}{2}\left[\ln \left(e^{-2 x}+1\right)\right]_{0}^{\infty} \\ &=\frac{1}{2} \ln 2 \end{aligned} $$