I'm trying to prove that for $x,y,r>0$ the following identity (gotten via CAS) holds:
$$\int_0^x \frac{x^r}{x^r+y^r}\,dy=\frac{x}{2r}\left(\psi\left(\frac{1+r}{2r}\right)-\psi\left(\frac{1}{2r}\right)\right),$$
where $\psi(\cdot)=\frac{\Gamma'(\cdot)}{\Gamma(\cdot)}$ denotes the Digamma function. Does anyone has a clue, how integration can be done this way?
As an additional information following identity, holds also to be true:
$$\int_0^x \frac{y^r}{y^r+x^r}\,dy=\frac{x}{2r}\left(\psi\left(\frac{1}{2r}+1\right)-\psi\left(\frac{1}{2r}+\frac{1}{2}\right)\right),$$
Which was obtained using:
$$\psi(x)=H_{x-1} – \gamma,$$
where $\gamma$ denotes the Euler-Mascheroni constant and $H_x$ denotes the $x^{th}$ harmonic number.
Best Answer
First, substitute away the $x$ as in @Roman's answer. Then, $$\int_0^1 \frac{dz}{1+z^r} = \int_0^1 \sum_{k=0}^\infty (-1)^k z^{kr} dz = \sum_{k=0}^\infty (-1)^k \frac{z^{kr}}{kr+1}\Bigg|_0^1 = \sum_{k=0}^\infty \frac{(-1)^k}{rk+1}.$$ The result follows from the total convergence of the geometric series in an appropriate domain of $z$ and from the identity $$\sum_{k=0}^\infty \frac{(-1)^k}{rk+1} = \frac{1}{2r} \left( \psi\left(\frac{r+1}{2r} \right) - \psi\left( \frac{1}{2r}\right)\right),\tag{$*$} $$ which is listed in the MathWorld page for $\psi$. Identity $(*)$ can be proven through the following series representation of $\psi$, which links it to the harmonic series: $$\psi(s)=-\gamma+\sum_{k=0}^{\infty}\left(\frac{1}{k+1}-\frac{1}{k+s}\right);$$ through this, the RHS of $(*)$ becomes $$\sum_{k=0}^\infty\left( \frac{1}{2kr+1}-\frac{1}{2kr+r+1}\right), $$ which turns out to yield the series on the LHS. (Try it!)
This procedure concerns the first integral, but can in principle be used to calculate the second integral as well. I wouldn't advise it, though, seeing as there is a much faster way: notice that summing the two integrals (omitting the $x$ in front) yields $$\int_0^1 \frac{dz}{1+z^r} + \int_0^1 \frac{z^rdz}{1+z^r} = \int_0^1 \frac{1+z^r}{1+z^r} dz = 1. $$