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

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?

Delta2

Ssnow
Gold Member
First you must to see if it is possible to change the integral with the infinite sum ...
Ssnow

dextercioby
Homework Helper
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.

George Jones
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.

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.

pasmith
Homework Helper
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 $$\ln(1 + x) = \sum_{n=1}^\infty \frac{(-1)^{n+1} x^n}{n} \qquad (|x| < 1).$$ So provided $e^x < 1$ you can have $$\ln(1 + e^x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}e^{nx}}{n}$$ which is easily integrated term by term provided $e^x < 1$ throughout the domain of integration. Otherwise the sum doesn't converge and you can't integrate it term by term.

Mayhem, Delta2 and vela
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?

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.

vela
Staff Emeritus
Homework Helper
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.

Delta2
Homework Helper
Gold Member
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}##.

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*}

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.

vela
Staff Emeritus
Homework Helper
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.

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.

pasmith
Homework Helper
For $x > 0$ ($e^x > 1$) you can use $$\ln(1 + e^x) = \ln(e^x(1 + e^{-x})) = x + \ln(1 + e^{-x})$$ which can be integrated term-by-term as in my pervious post.