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Integral substitution?

  1. Apr 5, 2014 #1
    1. The problem statement, all variables and given/known data

    sorry if question is unclear can't draw the integal sign out

    Show that
    Integral infinity-0 dz/((e^2z) - 1)^1/2 = integral 1- 0 dx/(1-x^2)^1/2 = pi/2



    3. The attempt at a solution

    I can get from the second integral to pi/2, as the second integral is sin^1(1) = pi/2

    However, I do not understand how to go between these two integrals

    infinity-0 dz/((e^2z) - 1)^1/2 = 1- 0 dx/(1-x^2)^1/2

    I tried substituing things like x = z, but doesn't work. Can't see how you would change the limits from infinity to 0, to 1 to 0?

    Maths is not my strong point so this could well be quite simple

    any help much appreciated
     
  2. jcsd
  3. Apr 5, 2014 #2

    Curious3141

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    Hint: the denominator can be expressed as ##\displaystyle (e^z + 1)^{\frac{1}{2}}.(e^z - 1)^{\frac{1}{2}} = e^z.(1 + e^{-z})^{\frac{1}{2}}.(1 - e^{-z})^{\frac{1}{2}}##

    Does that help?
     
  4. Apr 5, 2014 #3
    Thank you for the help!

    I've subsituted x = e^-z

    so dz = dx/-e^-z

    integral becomes

    1/e^z(1+x)^1/2 (1-x^1/2) * dx/-e^-z

    e^z*-e^-z = 1

    so 1/(1+x)^1/2 (1-x^1/2) dx = 1/(1-x^2)^1/2

    and then e^-z = x e^-infinity = 1 hence new limits 1 to 0

    I think that works?
     
  5. Apr 5, 2014 #4

    Curious3141

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    Very difficult to read your post without LaTex.

    But this: "e^z*-e^-z = 1" is an error, because the result should be negative one.

    And that negative sign is important when you transform the bounds.
     
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