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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!

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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. –  Wooster May 22 '13 at 12:46
    
Perhaps you mean "improper integral" = "אינטגרל לא-אמיתי" ? –  DonAntonio May 22 '13 at 12:47
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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. –  Hans Engler May 22 '13 at 12:51
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You have to change the limits of integration to be $$\int_1^\infty\cdots$$ –  RETAS May 22 '13 at 12:51
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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. –  David Mitra May 22 '13 at 12:57

3 Answers 3

up vote 5 down vote accepted

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

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

Hints:

First, it must be

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

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

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There is no arctan in the antiderivative. The OP made an algebra error after the substitution step.. –  Hans Engler May 22 '13 at 12:53
    
True, I relied on what he wrote. Thanks –  DonAntonio May 22 '13 at 12:55
    
Where is the algebra error? this is what you get if you will make partial fraction substitution –  Ofir Attia May 22 '13 at 12:57
    
No Offir, check it carefully. –  DonAntonio May 22 '13 at 13:00
    
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 $\epsilon$. –  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   $\displaystyle dx = \frac{1}{2(t-1)}dt$,

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)

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

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

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

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

  $\displaystyle= \frac{1}{2}\ln{2}$

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