Integrating ##\ln{(e^x+1)}##

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  • #1
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Homework Statement:
Evaluate ##\int{\ln{(e^x+1)}}##
Relevant Equations:
##\ln{x} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^n}{n}##

##(x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k}y^k##
Given the integral $$\int \ln{(e^x+1)} dx$$ we can rewrite this as the integral of the Taylor expansion of ##\ln{(e^x+1)}##. $$\int \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(e^x+1)^n}{n} dx$$ Which can then be rewritten using the binomial theorem: $$\int \sum_{n=1}^{\infty} \left [ \frac{(-1)^{n-1}}{n} \sum_{k=0}^{n} \binom{n}{k} e^{x(n-k)} \right] dx $$ What I want: a way to rewrite this such that I can directly integrate the binomial term, as this is simply a linear combination of ##e^{x(n-k)}## which is trivial to integrate.

Any help?
 

Answers and Replies

  • #2
Ssnow
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First you must to see if it is possible to change the integral with the infinite sum ...
Ssnow
 
  • #3
dextercioby
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It can't be a homework in exactly the same form as you posted it (as an indefinite integral), because special functions such as dilogarithms are not standard high-school or university curriculum.
 
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It can't be a homework in exactly the same form as you posted it (as an indefinite integral), because special functions such as dilogarithms are not standard high-school or university curriculum.
It's not exactly homework, just self study that is phrased as homework.
 
  • #5
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First you must to see if it is possible to change the integral with the infinite sum ...
Ssnow
Any help? I can't exactly see how it can be done here since the sum is nested.
 
  • #6
pasmith
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Homework Statement:: Evaluate ##\int{\ln{(e^x+1)}}##
Relevant Equations:: ##\ln{x} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^n}{n}##

I think you mean [tex]
\ln(1 + x) = \sum_{n=1}^\infty \frac{(-1)^{n+1} x^n}{n} \qquad (|x| < 1).[/tex] So provided [itex]e^x < 1[/itex] you can have [tex]\ln(1 + e^x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}e^{nx}}{n}[/tex] which is easily integrated term by term provided [itex]e^x < 1[/itex] throughout the domain of integration. Otherwise the sum doesn't converge and you can't integrate it term by term.
 
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  • #7
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If I want to integrate term-by-term, can I just enterchange the infinite sum with the integral sign, factor out the ##(-1)^{n-1}/n## term and integrate from there?
 
  • #8
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Then I obtain the final expression $$\sum_{n=1}^{\infty}\left[ \frac{(-1)^{n-1}}{n}\sum_{k=0}^{n} \frac{e^{x(n-k)}}{n-k} \right] + C$$ Which might be wrong. But if it isn't, then I've found a way to find a good approximation if we let the Taylor sum's limit be finite.
 
  • #9
vela
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Then I obtain the final expression $$\sum_{n=1}^{\infty}\left[ \frac{(-1)^{n-1}}{n}\sum_{k=0}^{n} \frac{e^{x(n-k)}}{n-k} \right] + C$$ Which might be wrong. But if it isn't, then I've found a way to find a good approximation if we let the Taylor sum's limit be finite.
Are you integrating the expression you gave in your original post, or the one @pasmith gave in post #6? Either way, your answer doesn't look correct.
 
  • #10
Delta2
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I think its worth to note that for large x (large ##e^x##) it is ##e^x+1\approx e^x##, so the integral approximates to ##\frac{x^2}{2}##.
 
  • #11
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Are you integrating the expression you gave in your original post, or the one @pasmith gave in post #6? Either way, your answer doesn't look correct.
Using the final expression in the OP, this is my work: $$\begin{align*} \int \sum_{n=1}^{\infty} \left [ \frac{(-1)^{n-1}}{n} \sum_{k=0}^{n} \binom{n}{k} e^{x(n-k)} \right] dx
&=\sum_{n=1}^{\infty}\left[\frac{(-1)^{n-1}}{n}\int \sum_{k=0}^{n} \binom{n}{k}e^{x(n-k)} dx \right] \\
&= \sum_{n=1}^{\infty}\left[\frac{(-1)^{n-1}}{n} \sum_{k=0}^{n} \binom{n}{k} \int e^{x(n-k)} dx \right ] \\
&=\sum_{n=1}^{\infty}\left[\frac{(-1)^{n-1}}{n} \sum_{k=0}^{n} \binom{n}{k} \frac{e^{x(n-k)}}{n-k} \right] + C \end{align*}$$
 
  • #12
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I think that doing it that way would be a fairly complex expression to evaluate, the most effective thing would be to arrive at the expression of a dilogarithm.

Of course unless you want to continue down that route.
 
  • #13
vela
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Using the final expression in the OP, this is my work: $$\begin{align*} \int \sum_{n=1}^{\infty} \left [ \frac{(-1)^{n-1}}{n} \sum_{k=0}^{n} \binom{n}{k} e^{x(n-k)} \right] dx
&=\sum_{n=1}^{\infty}\left[\frac{(-1)^{n-1}}{n}\int \sum_{k=0}^{n} \binom{n}{k}e^{x(n-k)} dx \right] \\
&= \sum_{n=1}^{\infty}\left[\frac{(-1)^{n-1}}{n} \sum_{k=0}^{n} \binom{n}{k} \int e^{x(n-k)} dx \right ] \\
&=\sum_{n=1}^{\infty}\left[\frac{(-1)^{n-1}}{n} \sum_{k=0}^{n} \binom{n}{k} \frac{e^{x(n-k)}}{n-k} \right] + C \end{align*}$$
Did you understand @pasmith's post? He was pointing out an error you made deriving the series.
 
  • #14
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Did you understand @pasmith's post? He was pointing out an error you made deriving the series.
I think I misunderstood what he read - now that I see what he means, I will try and integrate it into my problem.
 
  • #15
pasmith
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For [itex]x > 0[/itex] ([itex]e^x > 1[/itex]) you can use [tex]
\ln(1 + e^x) = \ln(e^x(1 + e^{-x})) = x + \ln(1 + e^{-x})[/tex] which can be integrated term-by-term as in my pervious post.
 

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