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I have the following, very messy integral: $$\int_{0}^{1}\sum_{k=1}^{\infty} \frac{(-\frac{e^x}{x+1}+1)^k}{k} dx $$

Neither WolframAlpha nor Mathematica was of any help. I'm not even sure where to start on this integral - I don't believe that it even has a closed form, at least in terms of elementary functions. But I can't prove that. I've been able to manipulate the integrand into a variety of forms, none of which has helped me at all.

Is there a closed form for this integral? If so, how can I find it?

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Let $I$ denote the integral given by

$$I=\int_{0}^{1}\sum_{k=1}^{\infty} \frac{(1-\frac{e^x}{x+1})^k}{k} \,dx$$

Note that the series representation for $-\log(1-y)$ is given by

$$-\log(1-y)=\sum_{k=1}^\infty \frac{y^k}{k}$$

for $-1\le y<1$. Therefore, setting $y=1-\frac{e^x}{x+1}$ we find that

$$\begin{align} I&=-\int_0^1 \log\left(\frac{e^x}{x+1}\right)\,dx\\\\ &=\int_0^1 \left(\log(1+x)-x\right)\,dx\\\\ &=2\log(2)-\frac32 \end{align}$$

And we are done!

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Since $1-\frac {e}{2}<1-\frac {e^x}{x+1}<0$, so you can use this identity $x+\frac {x^2}{2}+\frac {x^3}{3}........\infty=-ln(1-x)$. So your summation will become $S=-ln(1-(1-\frac {e^x}{x+1}))=-ln(\frac {e^x}{x+1})$. Now you have to integrate $S$ from 0 to 1. So integral would become $$I=\int_0^1 (ln(x+1)-x)dx=2ln2-\frac {3}{2}$$ Which is the final answer.

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The infinite sum in the integrand is equal to the polylogarithm of $-\frac{e^x}{x+1}+1$ of order 1 (See the infinite sum here). So now we have: $$\int_{0}^{1}\text{Li}_{1}(-\frac{e^x}{x+1}+1)$$ Which simplifies things a lot. Remember that: $$\text{Li}_{1}(z)=-\ln(1-z)$$ And this seems to be the direction your teacher/the person who gave you this integral wanted you to take, because $$-\ln(1+\frac{e^x}{x+1}+1)=-\ln(\frac{e^x}{x+1})$$ So now we have manipulated the integral into the following form: $$\int_{0}^{1}-\ln(\frac{e^x}{x+1})$$ Which is easy enough to solve, because by integrating by parts and simplifying we get: $$\ln(2)-1+\int_{0}^{1}\frac{x^2}{x+1}$$

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