How to solve $\int\limits_0^x {x^n \over{1 + e^x}} dx$?

Question

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?

Answer 1

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

Answer 2

Let me denote the incomplete Riemann zeta function by $$ A(s,z) = \frac{1}{{\Gamma (s)}}\int_0^z {\frac{{t^{s - 1} }}{{e^t - 1}}dt} . $$ Then \begin{align*} \int_0^x {\frac{{t^n }}{{e^t + 1}}dt} & = \int_0^x {\frac{{t^n }}{{e^t + 1}}dt} - n!A(n + 1,x) + n!A(n + 1,x) \\ & = \int_0^x {\frac{{t^n }}{{e^t + 1}}dt} - \int_0^x {\frac{{t^n }}{{e^t - 1}}dt} + n!A(n + 1,x) \\ & = - 2\int_0^x {\frac{{t^n }}{{e^{2t} - 1}}dt} + n!A(n + 1,x) \\ & = - \frac{1}{{2^n }}\int_0^{2x} {\frac{{s^n }}{{e^s - 1}}ds} + n!A(n + 1,x) \\ & = - \frac{1}{{2^n }}n!A(n + 1,2x) + n!A(n + 1,x). \end{align*}

Answer 3

First, a nice relation: $$I(t)=\int_0^t\frac{x^n}{1+e^x}dx$$ $$u=-x\Rightarrow dx=-du$$ $$I(t)=-\int_0^{-t}\frac{(-1)^nu^n}{1+e^{-u}}du=(-1)^n\int_{-t}^0\frac{x^n}{1+e^{-x}}dx=(-1)^n\int_{-t}^0x^n\frac{e^x}{1+e^x}dx$$ $$I(t)=(-1)^n\left[x^n\ln(1+e^x)\right]_{x=-t}^0+(-1)^{n+1}n\int_{-t}^0x^{n-1}\ln(1+e^x)dx$$ $$=-t^n\ln(1+e^{-t})+n\int_0^tx^{n-1}\sum_{k=1}^\infty\frac{(-1)^k}{k}e^{-kx}dx$$ $$=-t^n\ln(1+e^{-t})+n\sum_{k=1}^\infty\frac{(-1)^k}{k}\int_0^tx^{n-1}e^{-kx}dx$$ which looks a lot like the incomplete gamma function