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How to prove \begin{align*} &\mathrel{\phantom{=}}6\int_0^1\mathrm{d}x\int_1^{\infty}\frac{x\left( x-1 \right)}{t}\exp \left( -\frac{t}{\sqrt{x\left( 1-x \right)}} \right) \,\mathrm{d}t\\ &=-\int_1^{\infty}e^{-2x}\frac{2x^2+1}{2x^4}\sqrt{x^2-1}\,\mathrm{d}x \end{align*} I have no idea how to deal with the LHS.Any help will be grateful.

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We can write \begin{align} I&=6\int_0^1\mathrm{d}x\int_1^{\infty}\frac{x\left( x-1 \right)}{t}\exp \left( -\frac{t}{\sqrt{x\left( 1-x \right)}} \right) \,\mathrm{d}t\\ &=6\int_0^1 x\left( x-1 \right)\mathrm{d}x\int_{\tfrac{1}{\sqrt{x\left( 1-x \right)}}}^{\infty}\frac{e^{-u}}{u} \,du\\ &=-12\int_0^{1/2} x\left( 1-x \right)\,dx\int_{\tfrac{1}{\sqrt{x\left( 1-x \right)}}}^{\infty}\frac{e^{-u}}{u} \,du\\ \end{align} We take advantage of the symmetry with respect to $x=1/2$ for the las expression. Now, with $y=\tfrac{1}{\sqrt{x\left( 1-x \right)}}$, one has $1-2x=\sqrt{1-\frac{4}{y^2}}$ and $dx=-\frac{2}{y^2\sqrt{y^2-4}}dy$, and thus \begin{align} I&=-24\int_2^\infty \frac{dy}{y^4\sqrt{y^2-4}}\int_{y}^{\infty}\frac{e^{-u}}{u} \,du\\ &=-\frac{3}{2}\int_1^\infty \frac{dz}{z^4\sqrt{z^2-1}}\int_{2z}^{\infty}\frac{e^{-u}}{u} \,du \end{align} As \begin{equation} \int \frac{1}{v^4\sqrt{v^2-1}}dv=\frac{1}{3}\frac{\left( 2v^2+1 \right)\sqrt{v^2-1}}{v^3} \end{equation} the last integral can be calculated by parts: \begin{equation} I=-\frac{1}{2}\int_1^\infty \frac{\left( 2z^2+1 \right)\sqrt{z^2-1}}{z^4}e^{-2z}\,dz \end{equation}

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  • $\begingroup$ Nice approach! Thx! $\endgroup$ – Renascence_5. Feb 7 '18 at 0:10

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