Integrating Logarithmic Functions with Binomial Terms

[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?

 

[Ssnow]

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

 

[dextercioby]

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.

 

[Mayhem]

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.

 

[Mayhem]

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 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 <br /> \ln(1 + x) = \sum_{n=1}^\infty \frac{(-1)^{n+1} x^n}{n} \qquad (|x| &lt; 1). So provided e^x &lt; 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 &lt; 1 throughout the domain of integration. Otherwise the sum doesn't converge and you can't integrate it term by term.

 

[Mayhem]

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?

 

[Mayhem]

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]

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]

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 $\frac{x^2}{2}$.

 

[Mayhem]

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: $$\begin{align*} \int \sum_{n=1}^{\infty} \left [ \frac{(-1)^{n-1}}{n} \sum_{k=0}^{n} \binom{n}{k} e^{x(n-k)} \right] dx
&=\sum_{n=1}^{\infty}\left[\frac{(-1)^{n-1}}{n}\int \sum_{k=0}^{n} \binom{n}{k}e^{x(n-k)} dx \right] \\
&= \sum_{n=1}^{\infty}\left[\frac{(-1)^{n-1}}{n} \sum_{k=0}^{n} \binom{n}{k} \int e^{x(n-k)} dx \right ] \\
&=\sum_{n=1}^{\infty}\left[\frac{(-1)^{n-1}}{n} \sum_{k=0}^{n} \binom{n}{k} \frac{e^{x(n-k)}}{n-k} \right] + C \end{align*}$$

 

[Tapias5000]

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]

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.

 

[Mayhem]

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]

For x &gt; 0 (e^x &gt; 1) you can use <br /> \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.