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I have to evaluate this integral here

$$\int_x^\infty \frac{1}{1+e^t} dt$$

I've tried with the substitution $e^t=k$, but at the end I get $\log|k|-\log|1+k| $, which evaluated between the extrema doesn't converge... I hope that somebody can help!

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    $\begingroup$ Hint: $\frac{1}{1+e^t}=1-\frac{e^t}{1+e^t}$ $\endgroup$ – Damian Reding Sep 13 '15 at 13:51
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Using the substitution $e^t=u$ we have: $$ \int \dfrac{1}{1+e^t} dt=\int \dfrac{1}{u(1+u)} du $$ that, using partial fraction, become $$ \int \left(\dfrac{1}{u}-\dfrac{1}{1+u} \right) du = \log u -\log(1+u) +c = $$ $$ = \log \left( \dfrac{e^t}{1+e^t} \right) +c $$

Now taking the limit for $ t\rightarrow \infty$ we have:

$$ \lim_{t\rightarrow \infty}\left[ \log \left( \dfrac{e^t}{1+e^t} \right)\right]= \lim_{t\rightarrow \infty}\left[ \log \left( \dfrac{1}{1+1/e^t} \right) \right]=0 $$

and substituting also the other limit of integration $t=x$ we find: $$ \int_x^\infty \dfrac{1}{1+e^t} dt=f(x)=\log(e^x+1)-x $$

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  • $\begingroup$ Thanks a lot for your detailed answer. I've understood everything! $\endgroup$ – james42 Sep 13 '15 at 14:50
  • $\begingroup$ You are welcome :) $\endgroup$ – Emilio Novati Sep 13 '15 at 14:52
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Hint: $$\ln a-\ln b=\ln\frac ab.$$

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  • $\begingroup$ It doesn't seem to solve anything... $\endgroup$ – james42 Sep 13 '15 at 13:37
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    $\begingroup$ @ale42: Oh? What is the limit of $\frac{k}{1+k}$ as $k\to\infty$? And what is the logarithm of that? This shows that the evaluation between the extrema does converge. $\endgroup$ – Rory Daulton Sep 13 '15 at 13:50
  • $\begingroup$ It's only a hint, not a full solution. $\endgroup$ – Bernard Sep 13 '15 at 13:55
  • $\begingroup$ @RoryDaulton yup now I got it! $\endgroup$ – james42 Sep 13 '15 at 14:51
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Hint: $$\frac { 1 }{ 1+{ e }^{ t } } =\frac { 1 }{ { e }^{ t }\left( 1+{ e }^{ -t } \right) } =\frac { { e }^{ -t } }{ 1+{ e }^{ -t } } \\ \int { \frac { dt }{ 1+{ e }^{ t } } } =-\int { \frac { d\left( 1+{ e }^{ -t } \right) }{ 1+{ e }^{ -t } } } $$

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Notice, $$\int_{x}^{\infty}\frac{dt}{1+e^t}$$ $$=\int_{x}^{\infty}\frac{dt}{e^t(e^{-t}+1)}$$ $$=\int_{x}^{\infty}\frac{e^{-t} dt}{1+e^{-t}}$$ Let $1+e^{-t}=u\implies e^{-t}dt=-du$ $$=\int_{1+e^{-x}}^{1}\frac{(-du)}{u}$$ $$=-\int_{1+e^{-x}}^{1}\frac{du}{u}$$ $$=\int_{1}^{1+e^{-x}}\frac{du}{u}$$ $$=\left[\ln|u|\right]_{1}^{1+e^{-x}}$$ $$=\ln|1+e^{-x}|-\ln|1|$$ $$=\ln\left(1+e^{-x}\right)$$ $$=\ln\left(1+\frac{1}{e^{x}}\right)$$ $$=\ln\left(\frac{1+e^x}{e^{x}}\right)$$ $$=\ln\left(1+e^x\right)-\ln\left({e^{x}}\right)$$ $$=\color{red}{\ln(1+e^x)-x}$$

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The indefinite integral after you undo your substitution is $\displaystyle \ln \frac{e^t}{1+e^t} + C$.

The lower bound is easy, the value of the integral here is simply $\displaystyle \ln \frac{e^x}{1+e^x}$.

The value of integral at the upper bound is:

$$\lim_{x \to \infty} \ln \frac{e^x}{1+e^x}$$

and that's the natural log of an indeterminate form ($\frac{\infty}{\infty}$)

If you let $f(x) = e^x$ and $g(x) = 1+e^x$, you can use L' Hopital's rule to find that:

$$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{f'(x)}{g'(x)} = \lim_{x \to \infty} \frac{e^x}{e^x} = 1$$

So the value of the integral at the upper bound is $\ln 1 = 0$.

And so the definite integral is: $\displaystyle 0 - \ln \frac{e^x}{1+e^x} = \ln(1 + e^{-x})$.

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