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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.

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    $\begingroup$ How do you know identity should hold? Also are you sure the integral on the right is correct? $\endgroup$
    – DMcMor
    Commented Mar 19, 2021 at 19:00
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    $\begingroup$ It was solved using computer algebra (Wolfram Mathematica). Yes im shure. $\endgroup$
    – oyy
    Commented Mar 19, 2021 at 19:03

3 Answers 3

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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 $$\int_0^1 \frac{dz}{1+z^r} + \int_0^1 \frac{z^rdz}{1+z^r} = \int_0^1 \frac{1+z^r}{1+z^r} dz = 1. $$

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Assuming $x\neq0$, substitute $z=y/x$: $$ \int_0^x\frac{x^r}{x^r+y^r}dy=x\int_0^1\frac{dz}{1+z^r} $$ Then, using Mathematica as an integral-table lookup device:

Integrate[1/(1 + z^r), {z, 0, 1}]

$$ \frac{\psi\left(\frac{1+r}{2 r}\right)-\psi\left(\frac{1}{2 r}\right)}{2 r} \qquad \text{if $\left(\Re\left((-1)^{\frac{1}{r}}\right)\geq 1\lor \Re\left((-1)^{\frac{1}{r}}\right)\leq 0\lor (-1)^{\frac{1}{r}}\notin \mathbb{R}\right)\land \Re(r)>0$} $$ The conditions on $r$ are always satisfied for $r>0$.

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  • $\begingroup$ Thank you, I've already tried this transformation. Well I'm interested in the proof of this, unfortunatly an integral-table lookup doesn't helps too much, since I want to understand this identity. $\endgroup$
    – oyy
    Commented Mar 19, 2021 at 19:10
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Just to add a bit to the evaluation provided by @giobrach and @Roman - another way to get Digamma function explicitly. $$I=\int_0^x \frac{x^r}{x^r+y^r}\,dy\overset{z=\frac{y}{x}}=x\int_0^1 \frac{1}{1+z^r}\,dz=x\int_0^1 \frac{1-z^r}{(1+z^{r})(1-z^r)}\,dz=x\int_0^1 \frac{1-z^r}{1-z^{2r}}\,dz$$ Next we make regularisation: $$I=lim_{\epsilon\to0} \,I_\epsilon $$ where

$$I_\epsilon=x\int_0^1 \frac{1-z^r}{(1-z^{2r})^{1-\epsilon}}\,dz\overset{t=z^{2r}}=\frac{x}{2r}\int_0^1 \frac{t^{(\frac{1}{2r}-1)}-t^{(\frac{1}{2r}-\frac{1}{2})}}{(1-t)^{1-\epsilon}}\,dt=\frac{x}{2r}\Bigr(B(\frac{1}{2r},\epsilon)-B(\frac{1}{2r}+\frac{1}{2},\epsilon)\Bigr)$$ $B(a,b)$ - Beta function: $B(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$ $$I_\epsilon=\frac{x}{2r}\biggl(\frac{\Gamma(\epsilon)\Gamma(\frac{1}{2r})}{\Gamma(\frac{1}{2r}+\epsilon)}-\frac{\Gamma(\epsilon)\Gamma(\frac{1}{2r}+\frac{1}{2})}{\Gamma(\frac{1}{2r}+\frac{1}{2}+\epsilon)}\biggr)$$

Taking into consideration that $\Gamma(\epsilon)=\frac{1}{\epsilon}+O(1)$ at $\epsilon\to0$ we get $$I=lim_{\epsilon\to0} \,I_\epsilon = lim_{\epsilon\to0}\,\frac{x}{2r}\,\frac{1}{\epsilon}\biggl(\frac{\Gamma(\frac{1}{2r})}{\Gamma(\frac{1}{2r}+\epsilon)}-\frac{\Gamma(\frac{1}{2r}+\frac{1}{2})}{\Gamma(\frac{1}{2r}+\frac{1}{2}+\epsilon)}\biggr)$$ $$=lim_{\epsilon\to0}\,\frac{x}{2r}\,\frac{1}{\epsilon}\biggl(\frac{\Gamma(\frac{1}{2r})}{\Gamma(\frac{1}{2r})+\Gamma \,'(\frac{1}{2r})\epsilon + O(\epsilon^2)}-\frac{\Gamma(\frac{1}{2r}+\frac{1}{2})}{\Gamma(\frac{1}{2r}+\frac{1}{2})+\Gamma\,'(\frac{1}{2r}+\frac{1}{2})\epsilon)+O(\epsilon^2)}\biggr)$$ $$=lim_{\epsilon\to0}\,\frac{x}{2r}\,\frac{1}{\epsilon}\biggl(\frac{1}{1+\log'(\Gamma(\frac{1}{2r}))\epsilon + O(\epsilon^2)}-\frac{1}{1+\log'(\Gamma(\frac{1}{2r}+\frac{1}{2}))\epsilon+O(\epsilon^2)}\biggr)$$ $$=\frac{x}{2r}\,\biggl(\log'(\Gamma(\frac{1}{2r}+\frac{1}{2}))-\log'(\Gamma(\frac{1}{2r}))\biggr)$$ Finally

$$I=\frac{x}{2r}\,\biggl(\Psi(\frac{1}{2r}+\frac{1}{2})-\Psi(\frac{1}{2r})\biggr)$$

were $\Psi(t)=\frac{d}{dt}\log(\Gamma(t))$ - Digamma function

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