How to solve this integration: $\int_0^1 \frac{x^{2012}}{1+e^x}dx$ I'm having troubles to solve this integration: $\int_0^1 \frac{x^{2012}}{1+e^x}dx$
I've tried a lot using so many techniques without success. I found $\int_{-1}^1 \frac{x^{2012}}{1+e^x}dx=1/2013$, but I couldn't solve from 0 to 1.
Thanks a lot.
 A: You have $$\int_{-1}^{1} \frac{x^{2012}}{1+e^{x}} \ dx =\underbrace{\int_{-1}^{0}\frac{x^{2012}}{1+e^{x}}}_{I_{1}} \ dx + \int_{0}^{1}\frac{x^{2012}}{1+e^{x}} \ dx \qquad \cdots (1)$$
In $I_{1}$ put $x=-t$, then you have $dx = -dt$, and so the limits range from $t=0$ to $t=1$. So you have $$I_{1}= -\int_{1}^{0} \frac{e^{t}\cdot t^{2012}}{1+e^{t}} \ dt = \int_{0}^{1} \frac{e^{x}\cdot x^{2012}}{1+e^{x}} \ dx$$
Put this in equation $(1)$ to get the value.
A: I doubt there is a simple expression for your integral.
One approach would be to write
$$\frac{1}{1+e^x} = \frac{1}{e^x(1+e^{-x})} = e^{-x} \sum_{k=0}^\infty (-1)^k e^{-kx} = \sum_{k=0}^\infty (-1)^k e^{-(k+1)x}$$
with uniform convergence on $x \ge c$ for every $c > 0$. Hence
\begin{align}
\int_0^1 \frac{x^{2012}}{1+e^x}\,dx &= \int_0^1 \left(  \sum_{k=0}^\infty (-1)^k e^{-(k+1)x} x^{2012} \right)\,dx  \\
&= \sum_{k=0}^\infty \left( (-1)^k \int_0^1 e^{-(k+1)x} x^{2012}\, dx \right)  \\
&= \sum_{k=0}^\infty \left( (-1)^k \int_0^{k+1} e^{-t} \left(\frac{x}{k+1}\right)^{2012}\frac{1}{k+1}\, dt \right) \\
&= \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)^{2013}}\gamma(2013,k+1)
\end{align}
where $\gamma$ is the lower incomplete Gamma function. Maple can't find a closed expression for this series (and neither can I). It's even tricky to evaluate numerically, so I'm not sure how much use it is.
A: Experimenting with wolfram alpha lead me to: 
$$ \int {x^n \over 1+e^x} = (-1)^{n+1} x^{n+1} + (-1)^n * log(e^x+1)+  \sum_{k = 0}^{n-1} x^{n-i-1} (-1)^n * {n! \over (n-i)!} * Li_{i+2}(-e^x)$$
I got this by trying out small values for n so wolfram alpha manages to do the computations and then generalized it. Can someone verify/correct that?
