By Kay K.

2019-02-10 05:20:17 8 Comments

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!


@Song 2019-02-10 06:22:25

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, $$\begin{align*} \int_0^\infty \int_0^\infty \int_0^\infty e^{-(xy+yz+zx)}\mathrm dx\mathrm dy\mathrm dz&=\frac12\int_0^\infty \int_0^\infty \int_0^\infty \frac{e^{-u-v-w}\mathrm du \mathrm dv\mathrm dw}{\sqrt{uvw}}\\&=\frac12\left( \int_0^\infty \frac{e^{-u}\mathrm du}{\sqrt{u}}\right)^3\\&=\frac12\left(2 \int_0^\infty e^{-v^2}\mathrm dv\right)^3\\ &=\frac{\pi^{\frac32}}{2}. \end{align*}$$

@Kay K. 2019-02-10 06:23:56

Wow. I like this better. Thanks!

@Song 2019-02-10 06:27:10

@KayK. You're welcome :)

@Kay K. 2019-02-10 07:38:15

Umm.. is $\mathrm dxyz = \mathrm dx\mathrm dy\mathrm dz$..?

@Kay K. 2019-02-10 07:47:08

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.

@Song 2019-02-10 12:17:20

@KayK. That is what I meant exactly.

@Kay K. 2019-02-10 14:07:56

Okay, thanks a lot.

@jmerry 2019-02-10 06:05:39

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.

@Kay K. 2019-02-10 06:09:58

Wow. That's awesome. I appreciate.

@jmerry 2019-02-10 06:15:44

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.

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