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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:

  1. Is $(1)$ is correct?
  2. Could $(1)$ have been derived by a more straightforward method?
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    $\begingroup$ Didn't read. Just upvoted for the enormous effort ;) $\endgroup$ – polfosol Jan 29 '17 at 6:49
  • $\begingroup$ @polfosol The latex is rather a job! $\endgroup$ – Anon Jan 29 '17 at 6:53
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Here is a way to go. Perform the substitution $y=8 x^2 +8 x +1$. Then your integral assumes the form $$\int_1^\infty\!dy\,\frac{1}{2y^{1/4}(y^2-1)^{1/2}}.$$

In a next step introduce $z=1/y^2$ and you obtain $$\frac{1}{4}\int_0^1\!dz\,z^{-7/8}(1-z)^{-1/2}.$$

This is solved by the Beta-function and you obtain the result $$ \frac14 B(1/8, 1/2) = \frac{\Gamma(1/8) \Gamma(1/2)}{4\Gamma(5/8)}. \tag{1} $$

Edit:

To show that (1) is equivalent to the expression given by the OP, we may employ the duplication formula $$2^{3/4} \pi^{1/2} \Gamma(1/4) = \Gamma(1/8) \Gamma(5/8)$$ to replace $\Gamma(5/8)$ in (1) and the fact that $\Gamma(1/2)=\pi^{1/2}$.

Edit 2:

Some more details for the first substitution:

$$\frac{dy}{dx} = 16 x + 8$$ and thus $$\frac{1}{\frac{dy}{dx} \sqrt{x^2+x}} =\frac{1}{\sqrt{(16 x+8)^2(x^2+x)}} = \frac{1}{\sqrt{4 [(8 x^2 +8 x +1)^2-1]}}.$$

For the second substitution, we have $$\left| \frac{dz}{dy}\right| = \frac{2}{y^3}$$and thus $$ \frac{1}{\left| \frac{dz}{dy}\right| 2 y^{1/4}(y^2-1)^{1/2}} =\frac{y^{11/4}}{4(y^2-1)^{1/2}} = \frac{1}{4 z^{7/8} \sqrt{1-z}}.$$

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  • $\begingroup$ @JackD'Aurizio: thanks. $\endgroup$ – Fabian Jan 29 '17 at 14:14
  • $\begingroup$ Really nice (+1) thank you. I guessed $B(x,y)$ might be involved but I just could not see it. $\endgroup$ – Anon Jan 29 '17 at 23:05

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