Integral of the day XXVII

Integral involving the Gamma functon $\Gamma(\cdot)$

Today we show the proof of this integral posted by @Ali39342137 $$\int_{0}^{\infty} \left(\sqrt{1+x^4}-x^2\right)^nx^{k-1} dx = \frac{\Gamma\left(\frac{n}{2}-\frac{k}{4}\right)\Gamma\left(\frac{k}{2}\right)}{2^{\frac{k}{2}}\Gamma\left(\frac{n}{2}+\frac{k}{4}\right)}\frac{n}{2n+k}\quad n> \frac{k}{2}$$ This is a generalization for this other integral posted by @integralsbot with $n=s$ and $k=1$: $$\int_{0}^{\infty} \left(\sqrt{1+x^4}-x^2\right)^s dx = \frac{s}{2s+1}\sqrt{\frac{\pi}{2}} \frac{\Gamma\left(\frac{s}{2}-\frac{1}{4}\right)}{\Gamma\left(\frac{s}{2}+\frac{1}{4}\right)}$$ We previously posted another similar generalization here

Proof:

$$\begin{align*} I = \int_{0}^{\infty} \left(\sqrt{1+x^4}-x^2\right)^nx^{k-1} dx =& \frac{1}{2}\int_{0}^{\infty} \left(\sqrt{1+w^2}-w\right)^nw^{\frac{k}{2}-1} dw \quad ( w \mapsto x^2) \\ =& \frac{1}{2^{\frac{k}{2}+1}}\int_{0}^{\infty} t^{n-\frac{k}{2}-1}(1-t^2)^{\frac{k}{2}-1}(t^2+1) \quad (t \mapsto \sqrt{1+w^2}-w ) \\ =& \frac{1}{2^{\frac{k}{2}+1}}\left[\int_{0}^{\infty} t^{n-\frac{k}{2}+1}(1-t^2)^{\frac{k}{2}-1} dt +\int_{0}^{\infty} t^{n-\frac{k}{2}-1}(1-t^2)^{\frac{k}{2}-1} dt\right]\\ =& \frac{1}{2^{\frac{k}{2}+2}}\left[\int_{0}^{\infty} s^{\frac{n}{2}-\frac{k}{4}}(1-s)^{\frac{k}{2}-1} ds + \int_{0}^{\infty} s^{\frac{n}{2}-\frac{k}{4}-1}(1-s)^{\frac{k}{2}-1} ds\right] \quad ( s \mapsto t^2)\\ =& \frac{1}{2^{\frac{k}{2}+2}}\left[B\left(\frac{n}{2}-\frac{k}{4}+1, \frac{k}{2}\right)+ B\left(\frac{n}{2}-\frac{k}{4}, \frac{k}{2}\right)\right]\\ =& \frac{1}{2^{\frac{k}{2}+2}}\left[\frac{\Gamma\left(\frac{n}{2}-\frac{k}{4}+1\right)\Gamma\left(\frac{k}{2}\right)}{\Gamma\left(\frac{n}{2}+\frac{k}{4}+1\right)} + \frac{\Gamma\left(\frac{n}{2}-\frac{k}{4}\right)\Gamma\left(\frac{k}{2}\right)}{\Gamma\left(\frac{n}{2}+\frac{k}{4}\right)}\right]\\ =& \frac{1}{2^{\frac{k}{2}+2}}\left[\frac{\left(\frac{n}{2}-\frac{k}{4}\right)\Gamma\left(\frac{n}{2}-\frac{k}{4}\right)\Gamma\left(\frac{k}{2}\right)}{\left(\frac{n}{2}+\frac{k}{4}\right)\Gamma\left(\frac{n}{2}+\frac{k}{4}\right)} + \frac{\Gamma\left(\frac{n}{2}-\frac{k}{4}\right)\Gamma\left(\frac{k}{2}\right)}{\Gamma\left(\frac{n}{2}+\frac{k}{4}\right)}\right]\\ =& \frac{\Gamma\left(\frac{n}{2}-\frac{k}{4}\right)\Gamma\left(\frac{k}{2}\right)}{2^{\frac{k}{2}+2}\Gamma\left(\frac{n}{2}+\frac{k}{4}\right)}\left[\frac{\left(\frac{n}{2}-\frac{k}{4}\right)}{\left(\frac{n}{2}+\frac{k}{4}\right)} + 1\right]\\ =& \frac{\Gamma\left(\frac{n}{2}-\frac{k}{4}\right)\Gamma\left(\frac{k}{2}\right)}{2^{\frac{k}{2}}\Gamma\left(\frac{n}{2}+\frac{k}{4}\right)}\frac{n}{2n+k} \end{align*}$$ $$\boxed{ \int_{0}^{\infty} \left(\sqrt{1+x^4}-x^2\right)^nx^{k-1} dx = \frac{\Gamma\left(\frac{n}{2}-\frac{k}{4}\right)\Gamma\left(\frac{k}{2}\right)}{2^{\frac{k}{2}}\Gamma\left(\frac{n}{2}+\frac{k}{4}\right)}\frac{n}{2n+k}\quad n> \frac{k}{2} }$$