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Math Help - Triple integral take 2

  1. #1
    Super Member Showcase_22's Avatar
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    Triple integral take 2

    Evaluate \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:

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

    \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

    \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

    \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

    \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

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

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

    \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

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

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

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

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

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

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

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

    WOW!!

    My query is: Is there a shorter way to do this?
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  2. #2
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    Yes there is !

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

    \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

    \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

    \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

    \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
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  3. #3
    Super Member Showcase_22's Avatar
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    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!!
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  4. #4
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    I have found my mistake !

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

    \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

    \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

    \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

    \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
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  5. #5
    Super Member Showcase_22's Avatar
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    magic!!!

    I'll have to remember the method you use, it's soooooooo much faster than what I was doing!!
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  6. #6
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    I had switched off my computer but when I realized my mistake I could not help coming back.

    Now good night !
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  7. #7
    Super Member Showcase_22's Avatar
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    sleep tight!

    Thanks for the help.
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  8. #8
    Super Member Showcase_22's Avatar
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    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 \theta limits range from 0 \leq \theta \leq \frac{\pi}{2}?

    The range that was put before as 0 \leq \theta \leq 2 \pi seemed a bit more reasonable.
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  9. #9
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    Quote Originally Posted by Showcase_22 View Post
    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 \theta limits range from 0 \leq \theta \leq \frac{\pi}{2}?

    The range that was put before as 0 \leq \theta \leq 2 \pi seemed a bit more reasonable.
    I believe it's because if x, y, z > 0 then you're only dealing with the first octant. That would mean that 0 \leq \theta \leq \frac{\pi}{2}.
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  10. #10
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    Quote Originally Posted by Prove It View Post
    I believe it's because if x, y, z > 0 then you're only dealing with the first octant. That would mean that 0 \leq \theta \leq \frac{\pi}{2}.
    That's it !
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