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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doraemonpaul
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Ofir Attia
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4 Answers
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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 $\displaystyle\int_1^\infty \frac{1}{t}-\frac{t}{1+t^2}\,\mathrm{d}t = \left[\log(t)-\frac{1}{2}\log(1+t^2)\right]_1^\infty = \left[\frac{1}{2}\log\left(\frac{t^2}{1+t^2}\right)\right]_1^\infty$ $$ = \frac{1}{2}\left[\log(1)-\log\left(\frac{1}{2}\right)\right] = \frac{1}{2}\log(2).$$
Abel
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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
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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
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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
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This is what I get: $$\lim_{b \to \infty}ln|t|-\frac{1}{2}ln|1+t^2|$$ and all this between 0 to $b$ after evaluating it I get $ln(0)$ in one of the expressions – Ofir Attia May 22 '13 at 13:43
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so how to calculate it? after set the values : $$(ln|b|-\frac{1}{2}ln|1+b^2|)-(ln|0|-\frac{1}{2}ln|1|)$$ there is and error there, or I write its a not improper integral. – Ofir Attia May 22 '13 at 13:51
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The lower end of integration is $t=1$, not $t=0$ remember? Also, it might be convenient to write $\ln |b| - \frac{1}{2}\ln|1+b^2| = \frac{1}{2}\ln\left|\frac{b^2}{1+b^2}\right|$. – Abel May 22 '13 at 13:56
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yes, thank you, but it still no improper integral. you helped me to understand few things! – Ofir Attia May 22 '13 at 15:01
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Hints:
First, it must be
$$\frac1{t(1+t^2)}=\frac1t-\frac t{1+t^2}\;\;,\;\;\text{then}:$$
$$\int\limits_1^\infty\left(\frac1 t-\frac t{1+t^2}\right)dt:=\lim_{b\to\infty}\left(\log\frac b{\sqrt{1+b^2}}+\log\sqrt 2\right)$$
DonAntonio
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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
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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
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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
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I just realized there was also a mistake with the lower bound...and the downvoter rushed in. Oh, well. – DonAntonio May 22 '13 at 13:03
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I see it now, I changed it, I wrote it fine on my paper but here something else.thanks. – Ofir Attia May 22 '13 at 13:03
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Dear Don, people in glass houses. I seem to get random downvotes on old questions of mine every time I comment negatively on one of your answers. Either you have a stalking defender, or you are personally punitively downvoting other users. – Thomas Andrews May 22 '13 at 13:10
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Good you realize that, @ThomasAndrews: either you "forget" stuff or else these are your first, and most probably not the last, times you "get" what many (I'd say "most") of us get from time to time: serial downvoters (and sometimes: just downvoters). Take care. – DonAntonio May 22 '13 at 13:50
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I don't mind random downvotes, I just note that in the last few months, almost all of the random downvotes I've gotten, without comment, come after I've commented on one of your posts. – Thomas Andrews May 22 '13 at 14:02
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Incidentally, was that an admission to punitive downvoting, @DonAntonio? – Thomas Andrews May 22 '13 at 14:42
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Apparently you didn't even understand what I wrote, @ThomasAndrews, but don't worry and do what I do: don't care about points. They aren't money and they don't give you happiness. :) – DonAntonio May 22 '13 at 14:47
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Was just trying to understand the phrase "Good you realize that." It wasn't clear what you were suggesting I realized. I don't know why I get downvoted when I comment negatively on your posts, but I do know it happens. Punative downvoting is not the same as serial downvoting - a single downvote can be punative, but it isn't serial and won't be caught automatically by the system. – Thomas Andrews May 22 '13 at 15:00
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And you most definitely care about points, or you wouldn't have complained above that people were downvoting your wrong answer. (I removed my downvote when you fixed it, by the way.) – Thomas Andrews May 22 '13 at 15:04
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Oh, I see...so you did downvote the answer? And about complaining: I wonder what were you doing when you wrote your "Dear Don" comment... – DonAntonio May 22 '13 at 15:31
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Either you have a short memory, or you don't recall complaining, "I just realized there was also a mistake with the lower bound...and the downvoter rushed in. Oh, well." BTW, I never said I didn't downvote. I downvoted it because, after your first edit, it was still wrong, and if you had made any effort, you'd have seen that your error was already covered in comments to the question. (When my first error is trusting the work of the OP, I tend to re-read the question and comments, when I edit.) @DonAntonio – Thomas Andrews May 22 '13 at 16:50
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Had you had some patience, you could have either asked or waited until it was clear the mistakes stem from following what the OP wrote. Not a justification, just an explanation: he both had wrong the partial fraction development and the integral's lower limit. Once this was mentioned by someone else and I had the opportunity to note that the due corrections were made. You though rushed to downvote. Well, it's your privilege, I guess...as is anyone else's. – DonAntonio May 22 '13 at 16:59
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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)
$\displaystyle= \frac{1}{2}\int^{\infty}_2(\frac{1}{t-1} - \frac{1}{t})dt$
$\displaystyle= \frac{1}{2}\left(\left[\ln{\left(t-1\right)}\right]_2^{\infty} - \left[\ln{t}\right]_2^{\infty}\right)$
$\displaystyle= \frac{1}{2}\lim_{t\to\infty}\ln{(t-1)} - \frac{1}{2}\lim_{t\to\infty}\ln{t} + \frac{1}{2}\ln{2}$
$\displaystyle= \frac{1}{2}\lim_{t\to\infty}\ln{\frac{t-1}{t}} + \frac{1}{2}\ln{2}$
$\displaystyle= \frac{1}{2}\ln{2}$
Tianyu Zheng
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$$ \begin{aligned} \int_{0}^{\infty} \frac{d x}{1+e^{2 x}} &=\int_{0}^{\infty} \frac{e^{-2 x}}{e^{-2 x}+1} d x \\ &=-\frac{1}{2} \int_{0}^{\infty} \frac{d\left(e^{-2 x}+1\right)}{e^{-2 x}+1} \\ &=-\frac{1}{2}\left[\ln \left(e^{-2 x}+1\right)\right]_{0}^{\infty} \\ &=\frac{1}{2} \ln 2 \end{aligned} $$
Lai
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$$\int_1^\infty\cdots$$ – RETAS May 22 '13 at 12:51