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?


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!


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.


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: $$\ln(2)-1+\int_{0}^{1}\frac{x^2}{x+1}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.