Some time ago, I used a fairly formal method (in the second sense of this answer) to derive the following integral, and am wondering whether it is correct or not:
$$\int_0^\infty\frac{1}{\sqrt{x^2+x}\sqrt[4]{8x^2+8x+1}}\;dx=\frac{\Gamma\left(\frac{1}{8}\right)^2}{2^{\frac{11}{4}}\Gamma\left(\frac{1}{4}\right)}\tag{1}$$
Wolfram Alpha doesn't evaluate this integral, but gives the first 5 decimal places of the transformed integral $\int_0^1\frac{1}{\sqrt{x-x^2}\sqrt[4]{x^2-8x+8}}\;dx$ which are correct. The method I used was formal and convoluted, with much rearrangement of integrals and series whose convergence I did not know; such methods have given me incorrect results in the past (e.g. here and here). Nevertheless, the integral appears simple and perhaps there is a change of variables which would enable easy evaluation to the RHS (in particular I wonder whether elliptic integrals are involved) but I cannot find one. My question is: is the above result correct, and if so is there a simpler way to prove it than that below (which feels like I might have almost come in a circle)?
Derivation: I began with the fact that $\int_0^\infty e^{-x^n}dx=\frac{\Gamma(\frac{1}{n})}{n}$ and then proceeded as follows:
$$\left[\int_0^\infty e^{-x^8}\;dx\right]^2=\frac{\Gamma(\frac{1}{8})^2}{64}=\int_0^\infty\int_0^\infty e^{-x^8-y^8}\;dx\;dy=\int_0^\infty\int_0^\frac{\pi}{2}re^{-r^8(\cos^8\theta\;+\;\sin^8\theta)}\;d\theta\;dr={\int_0^\infty\int_0^\frac{\pi}{2}re^{-r^8(1\;-\;\sin^2\theta\;+\;\frac{\sin^42\theta}{8})}\;d\theta\;dr}={\int_0^\infty re^{-r^8}\int_0^\frac{\pi}{2}e^{r^8(\sin^2\theta\;-\;\frac{\sin^42\theta}{8})}\;d\theta\;dr}$$
In order to evaluate the inner integral I used Maclaurin series, the binomial theorem, and freely rearranged integrals and sums as follows:
$$\int_0^\frac{\pi}{2}e^{a(\sin^2\theta\;-\;\frac{\sin^42\theta}{8})}\;d\theta=\int_0^\frac{\pi}{2}\sum_{n=0}^\infty \frac{a^n}{n!}\left(\sin^2\theta-\frac{\sin^42\theta}{8}\right)^n\;d\theta={\int_0^\frac{\pi}{2}\sum_{n=0}^\infty \frac{a^n}{n!}\sum_{m=0}^n\frac{n!}{m!(n-m)!}\sin^{2n-2m}2\theta\;\left(\frac{-1}{8}\right)^m\sin^{4m}2\theta\;d\theta}={\sum_{n=0}^\infty\sum_{m=0}^n\frac{a^n\left(\frac{-1}{8}\right)^m}{m!(n-m)!}\int_0^\frac{\pi}{2}\sin^{2n+2m}2\theta\;d\theta}$$
Using $\int_0^\frac{\pi}{2}\sin^{2k}2\theta\;d\theta=\frac{1}{2}\int_0^\pi\sin^{2k}\theta\;d\theta=\int_0^\frac{\pi}{2}\sin^{2k}\theta\;d\theta=\frac{\pi(2k)!}{2^{2k+1}(k!)^2}$, I got:
$${\sum_{n=0}^\infty\sum_{m=0}^n\frac{a^n\left(\frac{-1}{8}\right)^m}{m!(n-m)!}\int_0^\frac{\pi}{2}\sin^{2n+2m}2\theta\;d\theta}={\frac{\pi}{2}\sum_{n=0}^\infty\sum_{m=0}^n\left(\frac{a}{4}\right)^n\left(\frac{-1}{32}\right)^m\frac{(2n+2m)!}{m!(n-m)!(n+m)!^2}}$$
So that I had:
$$\int_0^\frac{\pi}{2}e^{a(\sin^2\theta\;-\;\frac{\sin^42\theta}{8})}\;d\theta=\frac{\pi}{2}\sum_{m=0}^\infty\sum_{n=0}^\infty\left(\frac{a}{4}\right)^n\left(\frac{-a}{128}\right)^m\frac{(2n+4m)!}{m!\;n!\;(n+2m)!^2}\tag{2}$$
Substituting $(2)$ into our original integral, using the series and integral definitions of the hypergeometric function, and rearranging integrals and sums gave:
$$\frac{\Gamma(\frac{1}{8})^2}{64}=\frac{\pi}{2}\int_0^\infty re^{-r^8}\sum_{m=0}^\infty\sum_{n=0}^\infty\left(\frac{r^8}{4}\right)^n\left(\frac{-r^8}{128}\right)^m\frac{(2n+4m)!}{m!\;n!\;(n+2m)!^2}\;dr={\frac{\pi}{2}\sum_{m=0}^\infty\sum_{n=0}^\infty\left(\frac{1}{4}\right)^n\left(\frac{-1}{128}\right)^m\frac{(2n+4m)!}{m!\;n!\;(n+2m)!^2}\int_0^\infty r^{8n+8m+1}e^{-r^8}\;dr}={\frac{\pi}{16}\sum_{m=0}^\infty\sum_{n=0}^\infty\left(\frac{1}{4}\right)^n\left(\frac{-1}{128}\right)^m\frac{(2n+4m)!}{m!\;n!\;(n+2m)!^2}\Gamma\left(m+n+\frac{1}{4}\right)}={\frac{\sqrt{\pi}}{16}\sum_{m=0}^\infty\left(\frac{-1}{8}\right)^m\frac{\Gamma(m+\frac{1}{4})\;\Gamma(2m+\frac{1}{2})}{m!\;\Gamma(2m+1)}\sum_{n=0}^\infty\frac{\Gamma(m+n+\frac{1}{4})}{\Gamma(m+\frac{1}{4})}\frac{\Gamma(2m+1)}{\Gamma(n+2m+1)}\frac{\Gamma(n+2m+\frac{1}{2})}{\Gamma(2m+\frac{1}{2})}}={\frac{\sqrt{\pi}}{16}\sum_{m=0}^\infty\left(\frac{-1}{8}\right)^m\frac{\Gamma(m+\frac{1}{4})\;\Gamma(2m+\frac{1}{2})}{m!\;\Gamma(2m+1)}\;_2F_1\left(m+\frac{1}{4},2m+\frac{1}{2};2m+1;1\right)}={\frac{1}{16}\sum_{m=0}^\infty\left(\frac{-1}{8}\right)^m\frac{\Gamma(m+\frac{1}{4})}{m!}\int_0^1 x^{2m-\frac{1}{2}}(1-x)^{-\frac{1}{2}}(1-x)^{-m-\frac{1}{4}}\;dx}={\frac{1}{16}\int^1 x^{-\frac{1}{2}}(1-x)^{-\frac{3}{4}}\sum_{m=0}^\infty\frac{\Gamma(m+\frac{1}{4})}{m!}\left(-\frac{x^2}{8(1-x)}\right)^m\;dx}$$
In order to evaluate a sum of the form present in the integrand I used basic Laplace transform identities and interchanged sum and Laplace transform to get:
$$\sum_{n=0}^\infty\frac{1}{n!}\frac{\Gamma(n+\frac{1}{4})}{s^{n+\frac{1}{4}}}=\sum_{n=0}^\infty\frac{1}{n!}L\left[t^{-\frac{3}{4}}\right](s)=L\left[t^{-\frac{3}{4}}e^t\right](s)=\frac{\Gamma(\frac{1}{4})}{(s-1)^\frac{1}{4}}$$
and hence $\sum\limits_{n=0}^\infty\frac{x^n}{n!}\Gamma\left(n+\frac{1}{4}\right)=\frac{\Gamma(\frac{1}{4})}{(1-x)^\frac{1}{4}}$ so that I had:
$$\frac{\Gamma(\frac{1}{8})^2}{64}={\frac{\Gamma(\frac{1}{4})}{16}\int_0^1 x^{-\frac{1}{2}}(1-x)^{-\frac{3}{4}}\left(1+\frac{x^2}{8(1-x)}\right)^{-\frac{1}{4}}\;dx}={\frac{8^{\frac{1}{4}}\Gamma(\frac{1}{4})}{16}\int_0^1 x^{-\frac{1}{2}}(1-x)^{-\frac{1}{2}}(8-8x+x^2)^{-\frac{1}{4}}\;dx}$$
From which $(1)$ follows immediately by the substitution $u=\frac{1}{x}-1$. My questions are:
- Is $(1)$ is correct?
- Could $(1)$ have been derived by a more straightforward method?
