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Show that

$$ \int_{-\infty}^{\infty}\int_{-x}^{x}\int_{-y}^{y}\int_{-z}^{z}e^{-(x^2+y^2+z^2+w^2)}\dfrac{|zw|}{(1+x^2)(1+y^2)} \,dw \,dz\,dy\,dx\le\frac{\pi^2}{96} $$

I am not understanding how $\pi$ is coming into picture, and I'm not sure I can get the upper bound.

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    $\begingroup$ If you consider $I^2$, you may be able to change the coordinate system to polar coordinates, as is done when computing $$\int_{-\infty}^\infty e^{-x^2}dx.$$ That is probably where the $\pi$ is coming into play. $\endgroup$ – Joel Aug 20 '15 at 16:34
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    $\begingroup$ I'll paraphrase Lucian from here, 'pi appears in results because the formula or question asked involves a bounded sum of squares'. $\endgroup$ – Zach466920 Aug 20 '15 at 16:34
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By symmetry, then killing a couple of variables: $$\begin{eqnarray*} I &=& 16\int_{0}^{+\infty}\int_{0}^{x}\int_{0}^{y}\int_{0}^{z}\frac{zw e^{-(z^2+w^2)}}{(1+x^2)(1+y^2)}e^{-(x^2+y^2)}\,dw\,dz\,dy\,dx\\&=&8\int_{0}^{+\infty}\int_{0}^{x}\int_{0}^{y}\frac{z e^{-z^2}(1-e^{-z^2})}{(1+x^2)(1+y^2)}e^{-(x^2+y^2)}\,dz\,dy\,dx\\&=&2\int_{0}^{+\infty}\int_{0}^{x}\frac{(1-e^{-y^2})^2}{(1+x^2)(1+y^2)}e^{-(x^2+y^2)}\,dy\,dx\end{eqnarray*} $$ so by upper-bounding $(1-e^{-y^2})^2$ with $1$ and lower-bounding $e^{x^2}$ with $(1+x^2)$ we get: $$ I\leq \int_{0}^{+\infty}\int_{0}^{+\infty}\frac{e^{-x^2}\cdot e^{-y^2}}{(1+x^2)(1+y^2)}\,dx\,dy = \left(\int_{0}^{+\infty}\frac{dx}{e^{x^2}(1+x^2)}\right)^2 \leq \left(\int_{0}^{+\infty}\frac{dx}{(1+x^2)^2}\right)^2\leq \frac{\pi^2}{16}.$$ The missing $\frac{1}{6}$ factor probably comes from a more cunning management of the $(1-e^{-y^2})^2$ term.

Using the fact that $(1-e^{-y^2})^2 e^{-y^2}\leq\frac{4}{27}$ (that follows from the AM-GM inequality) we have: $$ I\leq \frac{8}{27}\int_{0}^{+\infty}\int_{0}^{x}\frac{1}{(1+y^2)(1+x^2)^2}\,dy\,dx = \frac{\pi^2-4}{54}$$ that is way closer to the inequality we want to prove. Using $(1-e^{-y^2})^2 e^{-y^2}\leq\min\left(y^4,\frac{4}{27}\right)$ we should improve the original inequality, so $\frac{\pi^2}{96}$ probably just comes from a probabilistic/Parseval/Cauchy-Schwarz argument buried in the thoughts of the problem poser.

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  • $\begingroup$ Do you forget to carry on the 2 from the last row of the aligned formula, when bounding $I$ in your second display math? $\endgroup$ – mickep Aug 20 '15 at 18:16
  • $\begingroup$ @mickep: no, it is right, when we replace $(1-e^{-y^2})^2$ with $1$ we have twice the integral over $0\leq y\leq x<+\infty$ of a symmetric function in $x,y$, hence that integral is just $\iint_{(0,+\infty)^2}f(x,y)\,dx\,dy$ by symmetry, again. $\endgroup$ – Jack D'Aurizio Aug 20 '15 at 19:38
  • $\begingroup$ Ah, OK! I just thought you replaced the $x$ in the $y$-integral by $+\infty$. It's a bit funny that this integral was not so easy to bound. The real value seems to be $\approx 0.011$, and $\pi^2/96\approx 0.103$, so there is some margin... $\endgroup$ – mickep Aug 21 '15 at 4:18
  • $\begingroup$ @mickep: a promising approach may be to use the AM-GM inequality to get rid of the term $(1-e^{-y^2})^2$, then apply Cauchy-Schwarz, separating the exponential part of the integral from the polynomial one. As you noticed, there is a wide margin for improvements. $\endgroup$ – Jack D'Aurizio Aug 21 '15 at 13:17

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