# Thread: Triple integral take 2

1. ## Triple integral take 2

Evaluate $\displaystyle \int_0^{\infty} \int_0^{\infty} \int_0^{\infty} \frac{1}{(1+x^2+y^2+z^2)^2} \ dx \ dy \ dz$.
This is pretty swish! Here's what i've done:

$\displaystyle \int_0^{\infty} \int_0^{\infty} \int_0^{\infty} \frac{1}{(1+x^2+y^2+z^2)^2} \ dx \ dy \ dz=\int_0^{\infty} \int_0^{\infty} \int_0^{\infty}$$\displaystyle \frac{1}{\left( 1+y^2+z^2 \right)^2}. \frac{1}{\left( 1+\frac{x^2}{1+y^2+z^2} \right)^2} \ dx \ dy \ dz$

Let $\displaystyle \frac{x}{\sqrt{1+y^2+z^2}}=tan( \theta) \Rightarrow \frac{dx}{d \theta}=\sqrt{1+y^2+z^2} \ sec^2 (\theta)$

$\displaystyle \int_0^{\infty} \int_0^{\infty} \int_0^{\frac{\pi}{2}} \frac{1}{(1+y^2+z^2)^2}. \frac{1}{(1+tan^2 (\theta))^2} \sqrt{1+y^2+z^2} \ sec^2 (\theta) d \theta \ dy \ dz$

$\displaystyle \int_0^{\infty} \int_0^{\infty} \int_0^{\frac{\pi}{2}} \frac{1}{(1+y^2+z^2)^{\frac{3}{2}}} \ cos^2(\theta) \ d \theta \ dy \ dz$

$\displaystyle \frac{1}{2}\int_0^{\infty} \int_0^{\infty} \int_0^{\frac{\pi}{2}} \frac{1}{(1+y^2+z^2)^{\frac{3}{2}}} \ (1+cos(2 \theta)) \ d \theta \ dy \ dz$

$\displaystyle \frac{1}{2}\int_0^{\infty} \int_0^{\infty} \frac{1}{(1+y^2+z^2)^{\frac{3}{2}}}. \left[ \theta+\frac{1}{2}sin(2 \theta) \right]_0^{\frac{\pi}{2}} \ dy \ dz$

$\displaystyle \frac{\pi}{4}\int_0^{\infty} \int_0^{\infty} \frac{1}{(1+y^2+z^2)^{\frac{3}{2}}} \ dy \ dz$

$\displaystyle \frac{\pi}{4}\int_0^{\infty} \int_0^{\infty} \frac{1}{(1+z^2)^{\frac{3}{2}}}.\frac{1}{\left( 1+\frac{y^2}{1+z^2} \right)^{\frac{3}{2}}} \ dy \ dz$

Let $\displaystyle \frac{y}{\sqrt{1+z^2}}=tan( \phi) \Rightarrow \ \frac{dy}{d \phi}=\sqrt{1+z^2} \ sec^2 ( \phi)$

$\displaystyle \frac{\pi}{4}\int_0^{\infty} \int_0^{\frac{\pi}{2}} \frac{1}{(1+z^2)^{\frac{3}{2}}}.\frac{1}{\left( 1+tan^2 (\phi) \right)^{\frac{3}{2}}} \sqrt{1+z^2} \ sec^2 ( \phi) \ d \phi \ dz$

$\displaystyle \frac{\pi}{4} \int_0^{\infty} \int_0^{\frac{\pi}{2}} \frac{1}{1+z^2}. cos (\phi) \ d \phi \ dz$

$\displaystyle \frac{\pi}{4} \int_0^{\infty} \left[ \frac{1}{1+z^2} sin( \phi) \right] _0^{\frac{\pi}{2}} \ dz$

$\displaystyle \frac{\pi}{4} \int_0^{\infty} \frac{1}{1+z^2} \ dz$

Let $\displaystyle z=tan (\psi) \Rightarrow \ \frac{dz}{d \psi}=sec^2(\psi)$

$\displaystyle \frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{1}{1+tan^2(\psi)} sec^2(\psi) \ d \psi$

$\displaystyle \frac{\pi}{4} \int_0^{\frac{\pi}{2}} d \psi$

$\displaystyle =\frac{\pi^2}{8}$

WOW!!

My query is: Is there a shorter way to do this?

2. Yes there is !

Substituting
$\displaystyle x = r \cos \theta$ and $\displaystyle y = r \sin \theta$

$\displaystyle \int_0^{\infty} \int_0^{\infty} \int_0^{\infty} \frac{1}{(1+x^2+y^2+z^2)^2} \ dx \ dy \ dz = \int_0^{\infty} \int_0^{2\pi} \int_0^{\infty} \frac{1}{(1+r^2+z^2)^2} \ r \ dr \ d\theta \ dz$

$\displaystyle \int_0^{\infty} \int_0^{\infty} \int_0^{\infty} \frac{1}{(1+x^2+y^2+z^2)^2} \ dx \ dy \ dz = 2\:\pi\:\int_0^{\infty} \int_0^{\infty} \frac{1}{(1+r^2+z^2)^2} \ r \ dr \ dz$

$\displaystyle \int_0^{\infty} \int_0^{\infty} \int_0^{\infty} \frac{1}{(1+x^2+y^2+z^2)^2} \ dx \ dy \ dz = -\pi\:\int_0^{\infty} \left[\frac{1}{1+r^2+z^2}\right]_{r=0}^{r=\infty} \ dz$

$\displaystyle \int_0^{\infty} \int_0^{\infty} \int_0^{\infty} \frac{1}{(1+x^2+y^2+z^2)^2} \ dx \ dy \ dz = \pi\:\int_0^{\infty} \frac{dz}{1+z^2} = \frac{\pi^2}{2}$

I do not find the same result as you

3. hmm, well this is interesting.

I only worked out that integral in front of my computer just now. I can't see a mistake in what i've done (quickly checking what I have now!).

I'll try solving in tomorrow when i'm more refreshed, It's too late to do maths!

If there's someone who hasn't posted yet reading this, work out an answer! A third opinion would be very helpful!!

4. I have found my mistake !

Substituting
$\displaystyle x = r \cos \theta$ and $\displaystyle y = r \sin \theta$

$\displaystyle \int_0^{\infty} \int_0^{\infty} \int_0^{\infty} \frac{1}{(1+x^2+y^2+z^2)^2} \ dx \ dy \ dz = \int_0^{\infty} \int_0^{\frac{\pi}{2}} \int_0^{\infty} \frac{1}{(1+r^2+z^2)^2} \ r \ dr \ d\theta \ dz$

$\displaystyle \int_0^{\infty} \int_0^{\infty} \int_0^{\infty} \frac{1}{(1+x^2+y^2+z^2)^2} \ dx \ dy \ dz = \frac{\pi}{2}\:\int_0^{\infty} \int_0^{\infty} \frac{1}{(1+r^2+z^2)^2} \ r \ dr \ dz$

$\displaystyle \int_0^{\infty} \int_0^{\infty} \int_0^{\infty} \frac{1}{(1+x^2+y^2+z^2)^2} \ dx \ dy \ dz = -\frac{\pi}{4}\:\int_0^{\infty} \left[\frac{1}{1+r^2+z^2}\right]_{r=0}^{r=\infty} \ dz$

$\displaystyle \int_0^{\infty} \int_0^{\infty} \int_0^{\infty} \frac{1}{(1+x^2+y^2+z^2)^2} \ dx \ dy \ dz = \frac{\pi}{4}\:\int_0^{\infty} \frac{dz}{1+z^2} = \frac{\pi^2}{8}$

I do find the same result as you

5. magic!!!

I'll have to remember the method you use, it's soooooooo much faster than what I was doing!!

6. I had switched off my computer but when I realized my mistake I could not help coming back.

Now good night !

7. sleep tight!

Thanks for the help.

8. I've just had the opportunity to go over this in more detail and there's something I don't quite get.

Why do your $\displaystyle \theta$ limits range from $\displaystyle 0 \leq \theta \leq \frac{\pi}{2}$?

The range that was put before as $\displaystyle 0 \leq \theta \leq 2 \pi$ seemed a bit more reasonable.

9. Originally Posted by Showcase_22
I've just had the opportunity to go over this in more detail and there's something I don't quite get.

Why do your $\displaystyle \theta$ limits range from $\displaystyle 0 \leq \theta \leq \frac{\pi}{2}$?

The range that was put before as $\displaystyle 0 \leq \theta \leq 2 \pi$ seemed a bit more reasonable.
I believe it's because if $\displaystyle x, y, z > 0$ then you're only dealing with the first octant. That would mean that $\displaystyle 0 \leq \theta \leq \frac{\pi}{2}$.

10. Originally Posted by Prove It
I believe it's because if $\displaystyle x, y, z > 0$ then you're only dealing with the first octant. That would mean that $\displaystyle 0 \leq \theta \leq \frac{\pi}{2}$.
That's it !