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While trying to answer the question "A closed form for $\displaystyle\int_0^1\frac{\ln(-\ln x)\ \operatorname{li}^2x}{x}dx$", I came up with a conjecture: $$\int_0^\infty\sqrt[3]z\ \operatorname{Ei}^2(-z)\,dz\ \stackrel?=\ \frac34\Gamma\left(\frac43\right)\left(\sqrt3\,\arccos\frac{5-3\,\sqrt[3]2}2-\ln\left(1+3\,\sqrt[3]2-3\,\sqrt[3]4\right)-3\,\sqrt[3]4\right)$$ where $\operatorname{Ei}z$ is the exponential integral: $$\operatorname{Ei}z=-\int_{-z}^\infty\frac{e^{-t}}t dt.$$ I have no proof whatsoever for it. It came up as a result of completely non-rigorous methods like lucky guessing, numerical fitting, inverse symbolic calculations and so on. But it holds numerically with a very high precision.

So, I invite you to try to prove it rigorously.

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With the substitution $t = zu^{2}$, we have

$$ \mathrm{Ei}(-z) = - \int_{z}^{\infty} \frac{e^{-t}}{t} \, dt = -2 \int_{1}^{\infty} \frac{e^{-zu^{2}}}{u} \, du. $$

Thus for $\Re s > 0$, we have

\begin{align*} \int_{0}^{\infty} z^{s-1} \mathrm{Ei}(-z)^{2} \, dz &= 4 \int_{0}^{\infty}\int_{1}^{\infty}\int_{1}^{\infty} \frac{z^{s-1} e^{-z(x^{2}+y^{2})}}{xy} \, dxdydz \\ &= 4 \Gamma(s) \int_{1}^{\infty}\int_{1}^{\infty} \frac{dxdy}{xy(x^{2}+y^{2})^{s}} \\ &= 4 \Gamma(s) \int_{0}^{1}\int_{0}^{1} \frac{x^{2s-1}y^{2s-1}}{(x^{2}+y^{2})^{s}} \, dxdy \tag{1} \\ &= 8 \Gamma(s) \iint\limits_{0 \leq y \leq x \leq 1} \frac{x^{2s-1}y^{2s-1}}{(x^{2}+y^{2})^{s}} \, dxdy, \tag{2} \end{align*}

where we utilized the substitution $(x, y) \mapsto (x^{-1}, y^{-1})$ at $(1)$ and exploited the symmetry of the domain of integration $\{ x \geq 1, y \geq 1 \}$ along $y = x$ to get $(2)$.

Now with the polar coordinate transform,

\begin{align*} \int_{0}^{\infty} z^{s-1} \mathrm{Ei}(-z)^{2} \, dz &= 8 \Gamma(s) \int_{0}^{\frac{\pi}{4}}\int_{0}^{\sec\theta} r^{2s-1}\cos^{2s-1}\theta\sin^{2s-1}\theta \, drd\theta \\ &= \frac{4 \Gamma(s)}{s} \int_{0}^{\frac{\pi}{4}} \frac{\sin^{2s-1}\theta}{\cos\theta}\,d\theta \\ &= \frac{2 \Gamma(s)}{s} \int_{0}^{\frac{\pi}{4}} \frac{\sin^{2s-2}\theta}{1 - \sin^{2}\theta} \cdot 2\cos\theta\sin\theta\,d\theta \\ &= \frac{2 \Gamma(s)}{s} \int_{0}^{\frac{1}{2}} \frac{u^{s-1}}{1 - u} \, du \tag{3} \\ &= \frac{2 \Gamma(s)}{s} \beta\left(\frac{1}{2} ; s, 0 \right) \end{align*}

where $\beta(x ; a, b)$ is the incomplete beta function and we used the substitution $u = \sin^{2}\theta$ at $(3)$. Now the rest is just a bit of calculus, yielding the desired result.


A slightly general result is as follows:

\begin{align*} \int_{0}^{\infty} z^{s-1} \Gamma(p, z) \Gamma(q, z) \, dz = \frac{\Gamma(s+p+q)}{s} \left\{ \beta\left( \frac{1}{2}; s+p, q\right) + \beta\left(\frac{1}{2}; s+q, p\right) \right\}, \end{align*}

where $\Gamma(s, z) = \int_{z}^{\infty} t^{s-1}e^{-t} \, dt$ is the incomplete gamma function.

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  • 3
    $\begingroup$ That was great, thanks! $\endgroup$ – Vladimir Reshetnikov Oct 4 '13 at 16:32
  • $\begingroup$ @VladimirReshetnikov, Thank you. I'm always enjoying both your virtuoso skills and calculation experiments. $\endgroup$ – Sangchul Lee Oct 5 '13 at 2:09
  • $\begingroup$ Would you mind to read this question and maybe write some thoughts. Thanks in advance! $\endgroup$ – Vladimir Reshetnikov Oct 5 '13 at 19:01
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You can have the following closed form for the integral

$$ \int_{0}^{\infty}z^{1/3}\,\rm{Ei(-z)}^2 \,dz = $$

$$-{\frac {\,\sqrt{3}\,\pi\left(\frac{9}{4}-\frac{\pi\,\sqrt{3}}{6}-\frac{3\,\ln( 3)}{2}\right) }{6\,\Gamma\left( 2/3 \right) }}+{\frac {8\sqrt{3}\pi}{63\Gamma(2/3)}}\,{ { {_4F_3(1,1,7/3,7/3;\,2,2,10/3;\,-1)}} }$$

$$ \sim 0.5786767028. $$

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