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I encountered this scary integral $$\int_0^1\ln\ln\,_3F_2\left(\frac{1}{4},\frac{1}{2},\frac{3}{4};\frac{2}{3},\frac{4}{3};x\right)\,dx$$ where $_3F_2$ is a generalized hypergeometric function $$_3F_2\left(\frac{1}{4},\frac{1}{2},\frac{3}{4};\frac{2}{3},\frac{4}{3};x\right)=1+\frac{8\,\pi}{3\,\sqrt{3}}\sum _{n=1}^\infty\frac{2^{-8\,n}\ \Gamma(4 \,n)}{\Gamma(n)\ \Gamma(n+1)\ \Gamma\left(n+\frac{2}{3}\right)\ \Gamma\left(n+\frac{4}{3}\right)}x^n.$$ I do not hope much that anything can be done with it, but maybe somebody got an idea how to find a closed form for this integral.

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    $\begingroup$ Where did this integtal come from? $\endgroup$ Oct 4, 2013 at 18:36

1 Answer 1

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Let us denote $$\mathcal{F}(x)={_3F_2}\left(\frac14,\frac12,\frac34;\frac23,\frac43;x\right),$$ and the integral in question $$\mathcal{I}=\int_0^1\ln\ln\mathcal{F}(x)\,dx.$$ I am still working on some details in my proof, but quite amazingly, yes, there is a closed form for this integral, albeit a non-elementary one: $$\mathcal{I}=\frac{256}{27}\left(\operatorname{Ei}\left(4\ln\frac34\right)-\operatorname{Ei}\left(3\ln\frac34\right)-\ln\frac43\right)+\ln\ln\frac43,$$ where $\operatorname{Ei}(x)$ is the exponential integral: $$\operatorname{Ei}(x)=-\int_{-x}^\infty\frac{e^{-t}}t dt.$$


Proof sketch:

The crucial fact is that on the interval $x\in(0,\,1)$ the function $x\mapsto\mathcal{F}(x)$ is an inverse of the function $z\mapsto\frac{256\,(z-1)}{27\,z^4}$, so that the following equalities hold: $$\mathcal{F}\left(\frac{256\,(z-1)}{27\,z^4}\right)=z,$$ $$\frac{27\,x}{256}\mathcal{F}(x)^4+1=\mathcal{F}(x).$$ This can be seen from comparing coefficients in corresponding series expansions (for details please refer to the paper M.L. Glasser, Hypergeometric functions and the trinomial equation).

It means the function $\mathcal{F}(x)$ on the interval $x\in(0,\,1)$ can be expressed in radicals, as I mentioned in my comment above. But, actually, I do not need that expansion.

Note that the integrand $\ln\ln\mathcal{F}(x)$ is negative on the integration interval, and monotonically increasing from $-\infty$ to $\ln\ln\frac43\approx-1.245899...$, so we can think of an absolute value of $\mathcal{I}$ as the area above the function graph. Now, if we transpose this picture, we will see that it equals to the integral of the inverse function plus the rectangular area to the right of it: $$|\mathcal{I}|=\int_{-\infty}^{\ln\ln\frac43}\frac{256\,\left(e^{e^t}-1\right)}{27\,e^{4\,e^t}}dt+\int_{\ln\ln\frac43}^0dt.$$ Using the identity $$\int\frac{e^{e^t}-1}{e^{4\,e^t}}dt=\operatorname{Ei}\left(-3\,e^t\right)-\operatorname{Ei}\left(-4\,e^t\right)+C$$ (that can be checked by taking derivatives from both parts), we get the final result.

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    $\begingroup$ It is really amazing! $\endgroup$ Sep 28, 2013 at 1:24
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    $\begingroup$ This is a very helpful answer. Thanks! $\endgroup$
    – Nik Z.
    Oct 1, 2013 at 19:17

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