# Solve the integral $\int_{0}^{1} \frac{x^\frac{1}{10}}{1+e^x}dx$

While solving a certain problem, I got stuck at this integral $$\displaystyle \int_{0}^{1} \frac{x^\frac{1}{10}}{1+e^x}dx$$ I tried elementary methods but nothing works. The last option would be a series solution. Can we solve this in closed form?

• If you gave some context as to where this came from, we'd maybe have a bit better of a shot at giving an answer. – Don Thousand Apr 18 at 1:21
• Wolframalpha says no, to a closed form, which is not surprising. – Cornman Apr 18 at 1:24
• Yes, Cornman. Already checked that. I wonder if it can be expressed in terms of some non-elementary functions if it resists to be in the closed form. You see it looks so cute and simple..but a monster! – ersh Apr 18 at 1:27
• why don't you use numerical methods if you just need to evaluate it? – Vasya Apr 18 at 1:43
• @Vasya Actually, I need to solve something general containing two parameters $\displaystyle \int_{0}^{1} \frac{x^{i+\frac{1}{10}}}{1+e^{nx}}dx$ where $i=0,1,2...$ and $n=1,2,3...$. I need to use it in analytic form somewhere not the numerical value. – ersh Apr 18 at 1:50

Considering $$I=\int \frac{x^a}{1+e^{nx}}\,dx$$ by Taylor $$\frac{1}{1+e^{nx}}=\frac 12+\sum_{p=0}^\infty n^{2 p+1}\frac{E_{2 p+1}(0)}{2 (2 p+1)!}x^{2p+1}$$ where appear Euler polynomials.
This makes $$J=\int_0^1 \frac{x^a}{1+e^{nx}}\,dx=\frac{1}{2 (a+1)}+\frac{1}{2}\sum_{p=0}^\infty \frac{E_{2 p+1}(0)\, n^{2 p+1}}{(2 p+1)! (a+2 p+2)}$$
At least for your case $$a=\frac 1 {10}$$ and $$n=1$$, this seems to converge very fast (considering $$\sum_{p=0}^q$$) $$\left( \begin{array}{cc} q & \text{result} \\ 0 & 0.33549783549783549784 \\ 1 & 0.34057913631084362792 \\ 2 & 0.34023760625619881917 \\ 3 & 0.34026363261324369474 \\ 4 & 0.34026151806647155721 \\ 5 & 0.34026169689918740783 \\ 6 & 0.34026168134981889845 \\ 7 & 0.34026168272958721037 \\ 8 & 0.34026168260523494156 \\ 9 & 0.34026168261658077711 \\ 10 & 0.34026168261553523742 \\ \cdots & \\ \infty &0.34026168261562407432 \end{array} \right)$$