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?

  • $\begingroup$ You are correct. So the integral of 1/t is a standard result, log(t), and then to integrate 1/(1+t^2) think about making a tan substitution. $\endgroup$ – Wooster May 22 '13 at 12:46
  • $\begingroup$ Perhaps you mean "improper integral" = "אינטגרל לא-אמיתי" ? $\endgroup$ – DonAntonio May 22 '13 at 12:47
  • 2
    $\begingroup$ You need to change the integration boundaries. Also, there is an error in the partial fractions step. The original integral is convergent, and your answer should be a finite number. $\endgroup$ – Hans Engler May 22 '13 at 12:51
  • 2
    $\begingroup$ You have to change the limits of integration to be $$\int_1^\infty\cdots$$ $\endgroup$ – RETAS May 22 '13 at 12:51
  • 1
    $\begingroup$ The integral is improper. If you just need to determine if the integral converges, you can use the comparison test ($0<{1\over 1+e^{2x}}<{1\over e^{2x}}$) to show it is convergent. $\endgroup$ – David Mitra May 22 '13 at 12:57

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

  • $\begingroup$ I guess this was homework? $\endgroup$ – Hans Engler May 22 '13 at 12:58
  • $\begingroup$ I know why you are changing $0 \rightarrow 1$ but where the values between 0 to 1 goes? $\endgroup$ – Ofir Attia May 22 '13 at 13:14
  • $\begingroup$ I'm not changing $0$ into $1$. It's just that $e^0 = 1$ and $e^\infty = \infty$. Hence if $x\in[0,\infty)$, then $t\in[1,\infty)$. $\endgroup$ – Abel May 22 '13 at 13:16
  • $\begingroup$ Ok, I got it, now I write the F(x) then I evaluate the limit between 0 to $\infty$ of him? $\endgroup$ – Ofir Attia May 22 '13 at 13:30
  • $\begingroup$ Indeed. ${{{}}}$ $\endgroup$ – Abel May 22 '13 at 13:32


First, it must be

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

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

  • $\begingroup$ There is no arctan in the antiderivative. The OP made an algebra error after the substitution step.. $\endgroup$ – Hans Engler May 22 '13 at 12:53
  • $\begingroup$ True, I relied on what he wrote. Thanks $\endgroup$ – DonAntonio May 22 '13 at 12:55
  • $\begingroup$ Where is the algebra error? this is what you get if you will make partial fraction substitution $\endgroup$ – Ofir Attia May 22 '13 at 12:57
  • $\begingroup$ No Offir, check it carefully. $\endgroup$ – DonAntonio May 22 '13 at 13:00
  • $\begingroup$ The bounds also need to change with the change of variables. And I don't think that notation $\lim_{b\to\infty,\epsilon\to 0}$ is particularly clear, since the expression does not include either $b$ or $\epsilon$. $\endgroup$ – Thomas Andrews May 22 '13 at 13:01

Does this solution make sense too?

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

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



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

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

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

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

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

  $\displaystyle= \frac{1}{2}\ln{2}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.