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Technique to solve integral

  1. Jan 26, 2008 #1
    1. The problem statement, all variables and given/known data
    What technique would you use to do the integral:

    [tex]\int_0^1 \int_0^1 dy dz \frac{1}{(1+y^2+z^2)^{5/2}} [/tex]

    ?
    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 26, 2008 #2
    Change to polar coordinates. Then it's trivial.
     
  4. Jan 26, 2008 #3

    arildno

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    Not utterly trivial, Rainbow Child:
    Although the "radial" integration will go easy enough, the limits for the unit square, as represented in polar coordinates are somewhat nasty.
     
  5. Jan 26, 2008 #4
    Yes. But after the integration, he can change back to cartesian coordinates. :smile:
     
  6. Jan 26, 2008 #5

    arildno

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    Eeh, no.

    We get:
    [tex]\int_{0}^{2\pi}\int_{ugly}^{UGLY}\frac{rdrd\d\theta}{(1+r^{2})^{\frac{5}{2}}}=\int_{0}^{2\pi}(|_{ugly}^{UGLY}-\frac{1}{3}(1+r^{2})^{-\frac{3}{2}})d\theta[/tex]
    You can't shift this "back" into Cartesian coordinates in any simple, valid manner.
     
  7. Jan 26, 2008 #6
    I get

    [tex] (\theta_2 - \theta_1)(-1/3 (1+r^2)^{-3/2})|^{r_2}_{r_1} [/tex]

    How do I change back to rectangular coordinates?

    EDIT: arildno beat me
     
  8. Jan 26, 2008 #7

    malawi_glenn

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    I would treat 1 + z^2 as A

    The solving [tex] \int dy\frac{1}{(A+y^2)^{5/2}} = \frac{2y^3+3Ay}{3A^2(y^2+A)^{3/2}} [/tex] (used mathematica..) I think you can use integral by parts.
     
  9. Jan 26, 2008 #8

    arildno

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    Okay, I'll do a bit more:

    "ugly" is 0, so it wasn't too ugly after all.

    Here's UGLY:
    [tex]0\leq\theta\leq\frac{\pi}{4}, r=\frac{1}{\cos\theta}, \frac{\pi}{4}\leq\theta\frac{3\pi}{4}, r=\frac{1}{\sin\theta}[/tex]
    and so on around the unit square.

    This is not at all simple to go further with.
     
  10. Jan 26, 2008 #9
    Maybe there is no reasonable way to do it. Mathematica gave a nice simple answer so I thought there would be. But I was probably wrong.
     
  11. Jan 26, 2008 #10

    arildno

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    No, it is probably something far more cleverer than trivial coordinate change that is needed here.
     
  12. Jan 26, 2008 #11

    malawi_glenn

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    Where did you find this integral?

    I got by doing the whole in mathematica:

    [tex] \left[ 2z + \sqrt{z^2+1} \left( \frac{2z}{3(z^2+1)} + \frac{z}{3(z^2+1)^2} \right) \right] _{z=0}^1 [/tex]
     
  13. Jan 26, 2008 #12

    arildno

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    That must be wrong, malawi, since the given integral is clearly less than 1 in value, whereas yours is greater than 2.
     
  14. Jan 26, 2008 #13
    I see no substitution that makes this an easy one. However the direct calculation is not that extreme. However I have to admit I used a little help from the integrator. I changed the variables a bit, z became x, sorry for that. Call [tex]1+y^2=A^2[/tex], the inner integral becomes:

    [tex]\int_0^1 \frac{dx}{\left[A^2+x^2\right]^{\frac{5}{2}}}[/tex]

    This can be solved using:

    [tex]x=A \cdot sinh(t)[/tex]

    giving then:

    [tex]\int_0^{arcsinh \left(\frac{1}{A}\right)} \frac{dt}{cosh^4(t)}[/tex]

    I did this integral with the integrator. After filling in the limits and replacing A again with y,
    I got:

    [tex]\frac{1}{3 \cdot (1+y^2)^2 \sqrt{2+y^2}}\cdot \left(\frac{5+3y^2}{2+y^2}\right)[/tex]

    We are left with the following:

    [tex]I=\frac{1}{3}\int_0^1 \frac{5+3y^2} {(1+y^2)^2 \cdot \left(2+y^2\right)^{\frac{3}{2}}}dy[/tex]

    This integral can also be found using the integrator. After filling in the limits I got:

    [tex]I=\frac{1}{18}\left(2\sqrt{3}+\pi\right)[/tex]

    Sorry for the integrator use, but it's late and I don't have the time right now to solve them by hand. Can this result be confirmed by anyone?
     
  15. Jan 26, 2008 #14
    Yes. I cheated a bit too (used Maple 11) and obtained

    [tex]
    \frac{\pi}{18}+\frac{\sqrt{3}}{9}
    [/tex]
     
  16. Jan 27, 2008 #15
    You are right, this

    was stupid statement, since I didn't paid much attention to the limits of the integration. Even though the resulting can be evaluted without by hand (but not trivially :smile: !!!).

    The region of integration in polar coordinates becames

    [tex]\mathcal{D}_1=\left\{(r,\theta):0<\theta<\frac{\pi}{4},0<r<\frac{1}{\cos\theta}\right\},\, \mathcal{D}_2=\left\{(r,\theta):\frac{\pi}{4}<\theta<\frac{\pi}{2},0<r<\frac{1}{\sin\theta}\right\}[/tex]

    thus

    [tex]I=\int_0^1 \int_0^1 dx dy \frac{1}{(1+x^2+y^2)^{5/2}}=\int\int_{\mathcal{D}_1}d\theta\,dr\frac{r}{(1+r^2)^{5/2}}+\int\int_{\mathcal{D}_2}d\theta\,dr\frac{r}{(1+r^2)^{5/2}}=I_1+I_2[/tex]

    The first integral [tex]I_1[/tex] reads

    [tex]I_1=\int_0^{\frac{\pi}{4}}d\theta\int_0^{1/\cos\theta}dr\frac{r}{(1+r^2)^{5/2}}=-\frac{1}{3}\int_0^{\frac{\pi}{4}}d\theta\left(\frac{1}{(1+1/\cos^2\theta)^{3/2}}-1\right)=-\frac{1}{3}\int_0^{\frac{\pi}{4}}d\theta\frac{\cos^3\theta}{(1+\cos^2\theta)^{3/2}}+\frac{\pi}{12}=-\frac{1}{3}\,J_1+\frac{\pi}{12}[/tex]

    The integral [tex]J_1[/tex] reads

    [tex]J_1=\int_0^{\frac{\sqrt{2}}{2}}d(\sin\theta)\frac{\cos^2\theta}{(1+\cos^2\theta)^{3/2}}=\int_0^{\frac{\sqrt{2}}{2}}d(\sin\theta)\frac{1-\sin^2\theta}{(2-\sin^2\theta)^{3/2}}}}=\int_0^{\frac{\sqrt{2}}{2}}dt\frac{2-t^2}{(2-t^2)^{3/2}}-\int_0^{\frac{\sqrt{2}}{2}}dt\frac{1}{(2-t^2)^{3/2}}=\arcsin\frac{t}{\sqrt{2}}\Big|_0^{\frac{\sqrt{2}}{2}}-\frac{t}{2\,\sqrt{2-t^2}}\Big|_0^{\frac{\sqrt{2}}{2}}\Rightarrow[/tex]
    [tex]J_1=\frac{\pi}{6}-\frac{\sqrt{3}}{6}\Rightarrow I_1=\frac{\sqrt{3}}{18}+\frac{\pi}{36}[/tex]

    Similary for the 2nd integral [tex]I_2[/tex] we have

    [tex]I_2=\frac{\sqrt{3}}{18}+\frac{\pi}{36}[/tex]

    yielding to

    [tex]I=\frac{\pi}{18}+\frac{\sqrt{3}}{9}[/tex]

    as coomast and Troels posted.

    Obviously not trivial! :smile:
     
  17. Jan 27, 2008 #16

    arildno

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    Supercool, RainbowChild, tenacity's reward to you! :smile:
     
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