# proving a double integral indentity

Numerical integration suggests that the following identity holds for any $\lambda>0$: $$\tag{1} \int_0^1\int_0^\infty\frac{3\sqrt{2}}{2\pi\lambda(1-x)\sqrt{x}}\,\exp\left(-\frac{1}{\lambda(1-x)^\frac{3}{2}}\Big((1+y)^\frac{3}{2}-(1+xy)^\frac{3}{2}\Big)\right)\,dy\,dx=$$ $$\tag{2} \int_0^1\int_0^\infty\frac{\sqrt{2}}{\pi\sqrt{s(1-s)}}\Big(1+\lambda s^\frac{3}{2}z\Big)^{-\frac{2}{3}}\exp(-z)\,dz\,ds.$$

I've tried proving that the double integrals are equal by performing various substitutions in order to transform one integral into the other without success. For example, using the substitutions $$s=x$$ and $$z=\frac{1}{\lambda(1-x)^\frac{3}{2}}\Big((1+y)^\frac{3}{2}-(1+xy)^\frac{3}{2}\Big)$$ on integral (2) results in $$\int_0^1\int_0^\infty\frac{3\sqrt{2}}{2\pi\lambda(1-x)\sqrt{x}}\,\exp\left(-\frac{1}{\lambda(1-x)^\frac{3}{2}}\Big((1+y)^\frac{3}{2}-(1+xy)^\frac{3}{2}\Big)\right)A\,dy\,dx$$ where $$A=\frac{\sqrt{1+y}-x\sqrt{1+xy}}{1-x}\left(1+\left(\frac{x+xy}{1-x}\right)^\frac{3}{2}-\left(\frac{x+x^2 y}{1-x}\right)^\frac{3}{2}\right)^{-\frac{2}{3}}.$$

I've also tried taking Laplace transforms in $\lambda$ but the resulting double integrals don't seem any easier to deal with. Can anyone suggest a substitution that will work or perhaps another idea on how to prove the above identity?

• Out of curiosity, when you tried the substitution $$z = \frac{1}{\lambda(1-x)^{3/2}}\Big[(1+y)^{3/2} - (1+xy)^{3/2}\Big]$$ for $y$, what results? – Tom Apr 14 '17 at 1:42
• @Tom see updated post – Garimpeiro Apr 14 '17 at 4:22

This is not an answer but it is too long for a comment.

I would really like to know which CAS, which parameters, and so on you have been using for evaluating the integrals.

On my side, I had incredible difficulties with any working presision higher than $15$.

Naming $F(\lambda)$ the first integral and $G(\lambda)$ the second one; here are my results for a few values of $\lambda$ $$\left( \begin{array}{cccc} \lambda &F(\lambda) &G(\lambda) &F(\lambda) -G(\lambda) \\ 0.5 & 1.27886612084440 & 1.27892018433661 & -0.00005406349221 \\ 1.0 & 1.19862189059702 & 1.19867598859260 & -0.00005409799558 \\ 1.5 & 1.14063487165204 & 1.14068884924989 & -0.00005397759785 \\ 2.0 & 1.09517480195430 & 1.09522896034172 & -0.00005415838742 \\ 2.5 & 1.05782784648898 & 1.05788199826770 & -0.00005415177872 \\ 3.0 & 1.02618383885196 & 1.02623791849041 & -0.00005407963845 \\ 3.5 & 0.99877273954892 & 0.99882758718448 & -0.00005484763556 \\ 4.0 & 0.97463217310702 & 0.97468626325833 & -0.00005409015131 \\ 4.5 & 0.953091149810319 & 0.95314530223889 & -0.00005415242858 \\ 5.0 & 0.933667478956616 & 0.93372165627318 & -0.00005417731658 \end{array} \right)$$

• @Garimpeiro: And what reason might that be ? $($Feel free to share your intuition$).$ – Lucian Sep 1 '17 at 20:52