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

• Mayhem
I think I misunderstood what he read - now that I see what he means, I will try and integrate it into my...

#### Mayhem

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

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
dextercioby said:
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.

Ssnow said:
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.

Mayhem said:
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.

Mayhem said:
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.

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}##.

vela said:
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.

Mayhem said:
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.

vela said:
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.

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.