I am wondering how to evaluate this integral. Wolfram Alpha says it is $\frac{\pi^{3/2}}2$ but I have no idea how get there.

$$\\ \int_0^\infty \int_0^\infty \int_0^\infty e^{-(xy+yz+zx)}\ dx\ dy\ dz\\ $$

I guess it may have some connection with $$\int_0^\infty \int_0^\infty \int_0^\infty e^{-(x^2+y^2+z^2)}\ dx\ dy\ dz=\left(\int_0^\infty e^{-x^2}\ dx\right)^3=\frac{\pi^{3/2}}8$$ and/or $$\int_0^\infty \int_0^\infty \int_0^\infty e^{-(x+y+z)^2}\ dx\ dy\ dz=\dots=\frac{\sqrt{\pi}}8$$ but I am not sure. Thanks in advance.

Taking @ersh's suggestion, I did: \begin{align} &\int_0^\infty \int_0^\infty \int_0^\infty e^{-(xy+yz+zx)}\ dx\ dy\ dz\\ &=\int_0^\infty \int_0^\infty e^{-yz}\int_0^\infty e^{-(y+z)x}\ dx\ dy\ dz\\ &=\int_0^\infty \int_0^\infty \frac{e^{-yz}}{y+z}\ dy\ dz=\int_0^\infty \int_z^\infty \frac{e^{-(y-z)z}}{y}\ dy\ dz\\ &=\int_0^\infty e^{z^2}\int_z^\infty \frac{e^{-yz}}{y}\ dy\ dz\\ \end{align} Now let me work on this integral: \begin{align} I(a)&=\int_z^\infty \frac{e^{-ayz}}{y}\ dy\\ \frac{dI(a)}{da}&=\int_z^\infty ze^{-ayz}\ dy=\left[-\frac{e^{-ayz}}a\right]_z^\infty=\frac{e^{-az^2}}{a}\\ \lim_{a\to0}I(a)&=-z^2\\ \therefore I(a)&= \end{align} Wait. I'm in the loop!

  • $\begingroup$ @ZangMingJie Actually that's how I got to this question... $\endgroup$ – Kay K. Feb 10 at 5:41
  • 1
    $\begingroup$ I think you can write $e^{-(xy+yz+zx)}=e^{-(y+z)x}.e^{-yz}$. Now, you can integrate directly with respect to $x$ to find the inner integral and so on. Note $\int e^{-(y+z)x}dx=\frac{-e^{-(y+z)x}}{y+z}$. $\endgroup$ – ersh Feb 10 at 5:43
  • $\begingroup$ @ersh Thanks for your suggestion. I tried (see above), but I couldn't get the second integral. $\endgroup$ – Kay K. Feb 10 at 6:07

Here's another approach: By making substitution $$ (u,v,w) = (xy,yz,zx), $$ we have$$\mathrm du \mathrm dv\mathrm dw = 2xyz \mathrm dx\mathrm dy\mathrm dz\implies \frac{\mathrm du \mathrm dv\mathrm dw}{2\sqrt{uvw}}=\mathrm dx\mathrm dy\mathrm dz.$$ Hence, $$\begin{align*} \int_0^\infty \int_0^\infty \int_0^\infty e^{-(xy+yz+zx)}\mathrm dx\mathrm dy\mathrm dz&=\frac12\int_0^\infty \int_0^\infty \int_0^\infty \frac{e^{-u-v-w}\mathrm du \mathrm dv\mathrm dw}{\sqrt{uvw}}\\&=\frac12\left( \int_0^\infty \frac{e^{-u}\mathrm du}{\sqrt{u}}\right)^3\\&=\frac12\left(2 \int_0^\infty e^{-v^2}\mathrm dv\right)^3\\ &=\frac{\pi^{\frac32}}{2}. \end{align*}$$

  • $\begingroup$ Wow. I like this better. Thanks! $\endgroup$ – Kay K. Feb 10 at 6:23
  • $\begingroup$ @KayK. You're welcome :) $\endgroup$ – Song Feb 10 at 6:27
  • $\begingroup$ Umm.. is $\mathrm dxyz = \mathrm dx\mathrm dy\mathrm dz$..? $\endgroup$ – Kay K. Feb 10 at 7:38
  • $\begingroup$ I think that you mean $(x\mathrm dy+y\mathrm dx)(y\mathrm dz+z\mathrm dy)(z\mathrm dx+x\mathrm dz)\approx 2xyz\mathrm dx\mathrm dy\mathrm dz$? If then, I understood. $\endgroup$ – Kay K. Feb 10 at 7:47
  • $\begingroup$ @KayK. That is what I meant exactly. $\endgroup$ – Song Feb 10 at 12:17

A chain of mostly elementary manipulations: \begin{align*}I &= \int_0^{\infty}\int_0^{\infty}\int_0^{\infty}e^{-(xy+yz+zx)}\,dx\,dy\,dz\\ &= \int_0^{\infty}\int_0^{\infty}\int_0^{\infty}e^{-yz}e^{-x(y+z)}\,dx\,dy\,dz\\ &= \int_0^{\infty}\int_0^{\infty}\frac{e^{-yz}}{y+z}\,dy\,dz\end{align*} First, we evaluate the inner integral with elementary techniques. There's some clutter that amounts to constant multipliers, but it's just $e^{ax}$ at its core. Next, to handle $y$, we substitute $y=zu$, $dy=z\,du$: \begin{align*}I &= \int_0^{\infty}\int_0^{\infty} \frac{e^{-z^2u}}{z(1+u)}\cdot z\,du\,dz\\ &= \int_0^{\infty}\int_0^{\infty} \frac{e^{-uz^2}}{1+u}\,dz\,du\\ &= \int_0^{\infty}\frac{\frac{\sqrt{\pi}}{2\sqrt{u}}}{1+u}\,du\end{align*} After the substitution, the $u$ integral isn't something we want to deal with - so we switch the order and do the $z$ integral next. That's just a Gaussian - no elementary antiderivative, but we know that $\int_0^{\infty} e^{-az^2}\,dz=\frac{\sqrt{\pi}}{2\sqrt{a}}$. This leaves us with a $u$ integral that's algebraic; we substitute $v=\sqrt{u}$, $dv=\frac1{2\sqrt{u}}\,du$ to rationalize it: $$I = \sqrt{\pi}\int_0^{\infty}\frac1{1+v^2}\,dv = \sqrt{\pi}\cdot\frac{\pi}{2}=\frac{\pi^{\frac32}}{2}$$ Done.

There's probably a way to connect this to a three-dimensional Gaussian, but I found it easier to break it down this way.

  • $\begingroup$ Wow. That's awesome. I appreciate. $\endgroup$ – Kay K. Feb 10 at 6:09
  • 1
    $\begingroup$ The trickiest step was that first substitution $y=zu$. The point of that was to make the denominator nicer. It took some trial and error to find it. $\endgroup$ – jmerry Feb 10 at 6:15

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.