Solve $\int\limits_0^x {x^n \over{1 + e^x}} dx$

integrationriemann-zeta

  1. The integral looks very similar to incomplete Riemann Zeta function, except for a $+$ in the denominator. WolframAlpha recognises incomplete Riemann Zeta function as Debye function and the answer has $\operatorname{Li}_n(e^{-x})$ terms.
  2. In case of my function WolframAlpha gives the answer which has $\operatorname{Li}_n(-e^{-x})$ terms. So it looks like it ignores the sign of $e^{x}$.

Although $\hspace{2mm} \int\limits_0^x {x^n \over{1 + e^x}} dx$ is (I guess) pretty close to Dirichlet Eta function, which can be expressed through Riemann Zeta, the equations connecting them are for complete integrals. I don't really think that the answer which Wolfram gives is correct. At least I cannot justify it for myself. And my question is:

  1. Is $\hspace{2mm} \int\limits_0^x {x^n \over{1 + e^x}} dx = \sum\limits_i\operatorname{Li}_i(-e^{-x}) + \Gamma(n)\left(1 – 2^{1 – n}\right)\zeta(n)$ correct and why?

  2. If it is not what can I do to solve it?

  3. Where do the coefficients before polylogarithms in Debye function come from? (I got mixed up with signs, but absolute values are ok) Where does $\zeta$ come from?

Best Answer

The "source" of $\zeta$ is $\int_0^\color{red}{\infty}t^n(1+e^t)^{-1}\,dt=n!\eta(n\color{red}{+1})=n!(1-2^{-n})\zeta(n+1)$. To prove, use $$\frac{1}{1+e^t}=\frac{e^{-t}}{1+e^{-t}}=\sum_{k=1}^\infty(-1)^{k-1}e^{-kt}$$ (then integrate termwise, etc.). For $x>0$, your original integral is treated the same way: $$\int_0^x\frac{t^n\,dt}{1+e^t}=\int_0^\infty\frac{t^n\,dt}{1+e^t}-\int_x^\infty\frac{t^n\,dt}{1+e^t}.$$

The first integral on the RHS is just considered, and the second is equal to $$\int_0^\infty\frac{(x+t)^n}{1+e^{x+t}}\,dt=\sum_{k=0}^n\binom{n}{k}x^{n-k}\int_0^\infty\frac{t^k\,dt}{1+e^{x+t}}.$$

The last integral is handled like the way above, and is equal to $$\sum_{j=1}^\infty(-1)^{j-1}e^{-jx}\int_0^\infty t^k e^{-jt}\,dt=-k!\sum_{j=1}^\infty\frac{(-e^{-x})^j}{j^{k+1}}=-k!\operatorname{Li}_{k+1}(-e^{-x}).$$

Finally, $$\int_0^x\frac{t^n\,dt}{1+e^t}=n!\left((1-2^{-n})\zeta(n+1)+\sum_{k=0}^{n}\frac{x^{n-k}}{(n-k)!}\operatorname{Li}_{k+1}(-e^{-x})\right).$$