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.

  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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.

  • $\begingroup$ 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

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.