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Help with an intermediate integral

  1. Dec 31, 2014 #1
    1. The problem statement, all variables and given/known data
    I have been trying to evaluate an integral that has come up in the process of me solving a different problem, but am completely stuck. As I have confirmed with Wolfram Alpha that the integral once solved yields the correct solution to my problem. However, I am trying to better my ability to solve integrals without the help of tables or computer aids, so I am trying to find the correct solution to this one. Also, because my issue deals more with the math rather than the physics of the problem, I have included it here among the Math section. If this was inappropriate to put this here or if this is too little to prove that I have done enough work, please let me know and I will take this post down and include the full problem in the Physics section.


    The integral is ## \int_{-a}^{a}\frac{1}{(z^2+a^2)(z^2+2a^2)^{1/2}} dz ##

    The problem here simply is that every time I attempt this integral, the math becomes a "runaway train". I can use any help or hints with regards to how to fully evaluate this integral.


    2. Relevant equations

    ## \int_{-a}^{a}\frac{1}{(z^2+a^2)(z^2+2a^2)^{1/2}} dz ##

    3. The attempt at a solution

    I started off by using integration by parts and designating ##dv=[z^2+a^2]^{-1}## and ##u=[z^2+2a^2]^{-1/2}##. This led to the result of ##\frac{1}{a}(arctan(\frac{z}{a}))[z^2+2a^2]^{-1/2}\bigg|_{-a}^a +\int_{-a}^a\frac{z}{a}(arctan(\frac{z}{a}))[z^2+2a^2]^{-3/2} dz ##. When I try to evaluate the integral in this difference, that's when I go into a long never ending train of Integration by Parts. Any help with this integration would be must appreciated.
     
  2. jcsd
  3. Dec 31, 2014 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    Preliminary simplification is always worthwhile. If you call your integral ##F(a)##, changing variables to ##z = ay## gives
    [tex] F(a) = \frac{1}{a^2} F(1), \\
    \text{where } \;\; F(1) = \int_{-1}^1 \frac{dy}{(y^2+1)\sqrt{y^2+2}} = 2 \int_0^1 \frac{dy}{(y^2+1)\sqrt{y^2+2}} [/tex]
    Submit this last integration to Wolfram Alpha (a free on-line version of Mathematical lite) to see what you get.

    Alternatively, try the change of variable ##u = y/\sqrt{y^2+2}##.
     
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