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Integrate exp(-(x^2)) using the substitution u=tanh(x)

  1. Sep 6, 2008 #1
    The question is to convert the infinity limits of the integral [tex]\int^\infty_{-\infty} e^{{-x}^2} dx[/tex] to finite limits [tex]\int^{u_a}_{u_b} g(u) du[/tex] using the substitution [tex]u = tanh(x)[/tex].

    How do I go about it?
     
    Last edited: Sep 6, 2008
  2. jcsd
  3. Sep 6, 2008 #2

    tiny-tim

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    Hi sai2020! :smile:

    Hint: tanhx = sinhx/coshx = (ex - e-x)/(ex + e-x) :smile:
     
  4. Sep 6, 2008 #3
    I did that and I ended up with [tex]\int^1_{-1} log_e(\frac{1+u}{1-u})[/tex].

    Is that correct?
     
  5. Sep 6, 2008 #4

    tiny-tim

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    Hi sai2020! :smile:

    [tex]\int^1_{-1}[/tex] is correct …

    but log((1+u)/(1-u)) = log((coshx + sinhx)/(coshx - sinhx)) = 2x,

    and you haven't converted dx into du. :frown:

    … but didn't the question only ask for the limits? :confused:
     
  6. Sep 6, 2008 #5
    Hi Tim! :)

    Well this is a part of a bigger problem and I need to find g(u) as well. Here's what I did

    Sorry I made a mistake. Is it

    [tex] exp(\frac{-1}{2} (log_e \frac{1+u}{1-u})^2) [/tex]

    How do I simplify further?

    Thanks a lot :)
     
  7. Sep 6, 2008 #6
    yeah that would be [tex]dx = (1-u^2) du[/tex]
     
  8. Sep 6, 2008 #7

    tiny-tim

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    No … [tex]dx = \frac{du}{1-u^2}[/tex] :smile:
    i suppose … [tex]\left(\frac{1\,+\,u}{1\,-\,u}\right)^{\frac{1}{2}\,log\frac{1\,+\,u}{1\,-\,u}}[/tex]

    but I don't see where you go from there … :confused:

    This isn't the usual way of solving [tex]\int^\infty_{-\infty} e^{{-x}^2} dx[/tex] :frown:
     
  9. Sep 7, 2008 #8
    Also wondering why on earth do we want to use the substitution u=tanh(x) ?

    If I could recall correctly, we evaluate the expression using either
    i) gamma function
    ii) normal distribution pdf
    iii) polar coordinates.
     
  10. Sep 8, 2008 #9

    tiny-tim

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    Hi matematikawan! :smile:

    I think it's just an exercise in converting limits :smile:

    which is, after all, what most people on this forum seem to find the difficult part of calculating an integral by substitution! :wink:
     
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