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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.

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Perhaps the trouble is the missing $dx$..... (: – JavaMan Feb 13 '13 at 6:01
@JavaMan thank you – user42912 Feb 13 '13 at 6:05
up vote 2 down vote accepted

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.

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If the sum is tricky to evaluate numerically, perhaps the integral isn't... – vonbrand Feb 13 '13 at 14:54
@vonbrand Actually, the integral is a little tricky to evaluate numerically as well (but not quite as tricky). The remark was not meant to imply that a numerical approximation to the integral can't be found. – mrf Feb 13 '13 at 14:57

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.

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But the problem is when we substitute in the equation (1), we don't find the integration from 0 to 1. Thank you for your answer. – user42912 Feb 13 '13 at 6:07

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?

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