2
$\begingroup$

Numerical integration suggests that the following identity holds for any $\lambda>0$: \begin{equation}\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= \end{equation} \begin{equation}\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. \end{equation}

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?

$\endgroup$
  • 1
    $\begingroup$ 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? $\endgroup$ – Tom Apr 14 '17 at 1:42
  • $\begingroup$ @Tom see updated post $\endgroup$ – Garimpeiro Apr 14 '17 at 4:22
2
$\begingroup$

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 $\lambda$ $$\left( \begin{array}{cccc} \lambda &F(\lambda) &G(\lambda) &F(\lambda) -G(\lambda) \\ 0.5 & 1.27886612084440 & 1.27892018433661 & -0.00005406349221 \\ 1.0 & 1.19862189059702 & 1.19867598859260 & -0.00005409799558 \\ 1.5 & 1.14063487165204 & 1.14068884924989 & -0.00005397759785 \\ 2.0 & 1.09517480195430 & 1.09522896034172 & -0.00005415838742 \\ 2.5 & 1.05782784648898 & 1.05788199826770 & -0.00005415177872 \\ 3.0 & 1.02618383885196 & 1.02623791849041 & -0.00005407963845 \\ 3.5 & 0.99877273954892 & 0.99882758718448 & -0.00005484763556 \\ 4.0 & 0.97463217310702 & 0.97468626325833 & -0.00005409015131 \\ 4.5 & 0.953091149810319 & 0.95314530223889 & -0.00005415242858 \\ 5.0 & 0.933667478956616 & 0.93372165627318 & -0.00005417731658 \end{array} \right)$$

Could you clarify, please ?

If possible, could you provide the codes ?

$\endgroup$
  • $\begingroup$ It seems my claim of "high precision" was an exaggeration since my numerical results are no better than yours. I have another, less direct reason to suspect the integrals are equal, though. $\endgroup$ – Garimpeiro Apr 14 '17 at 4:13
  • $\begingroup$ @Garimpeiro: And what reason might that be ? $($Feel free to share your intuition$).$ $\endgroup$ – Lucian Sep 1 '17 at 20:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.