$\int_0^1 \frac{x^{p}}{x^{p+1}+(1-x)^{p+1}} dx=?$

Question

$$\int_0^1 \frac{x^{p}}{x^{p+1}+(1-x)^{p+1}} dx=?$$ I tried to use $$\int_0^1 \frac{x^{p+1}}{x^{p+1}+(1-x)^{p+1}} dx=\frac{1}{2}$$ and integration by parts. I do not know if there is any restriction on p.in original question p=2014,Question from Jalil Hajimir.

Answer 1

Assume $p$ is a positive integer. Of course, you don't need it this strong.

First, observe that \begin{align} I(p) =& \int^1_0 \frac{x^p}{x^{p+1}+(1-x)^{p+1}}\ dx = \int^1_0 \frac{x^{p+1}}{x^{p+1}+(1-x)^{p+1}}\ \frac{dx}{x}\\ =& \int^1_0 \frac{x^{p+1}}{x^{p+1}+(1-x)^{p+1}}\ \frac{(1-x)}{x} dx+ \int^1_0 \frac{x^{p+1}}{x^{p+1}+(1-x)^{p+1}}\ dx\ \ \ \ \ (1)\\ =&\ \int^1_0 \frac{(1-x)^{p+1}}{x^{p+1}+(1-x)^{p+1}}\ \frac{x}{1-x} dx+ \int^1_0 \frac{(1-x)^{p+1}}{x^{p+1}+(1-x)^{p+1}}\ dx\ \ \ \ \ (2) \end{align}

which means \begin{align} 2I(p) = \int^1_0 f(x)+f(1-x)\ dx +1 = 2\int^1_0 f(x)\ dx +1 \end{align} where \begin{align} f(x) = \frac{x^{p+1}}{x^{p+1}+(1-x)^{p+1}}\ \frac{(1-x)}{x} = \frac{(\frac{1-x}{x})}{1+(\frac{1-x}{x})^{p+1}}. \end{align}

Hence \begin{align} I(p) = \int^\infty_0 \frac{u}{(1+u^{p+1})(1+u)^2} du+\frac{1}{2}. \end{align}

You can finish the rest using contour integration.

Edit: I realize I did a lot of unnecessary calculations. In fact, we have

\begin{align} I(p) =& \int^1_0 \frac{1}{(1+(\frac{1-x}{x})^{p+1})}\frac{dx}{x} = \int^\infty_0 \frac{du}{(1+u^{p+1})(1+u)} \end{align}

and

\begin{align} \lim_{p\rightarrow \infty}I(p)=\lim_{p\rightarrow \infty}\int^\infty_0 \frac{du}{(1+u^{p+1})(1+u)} =\int^1_0 \frac{du}{1+u} = \ln(2). \end{align}