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Gamma function as solution to an integral

  1. Feb 26, 2007 #1
    1. The problem statement, all variables and given/known data

    Calculate [tex]\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1+\sin^2 x}} dx[/tex] expressing its solution in terms of the gamma function. It's suggested to first use the change of variable [itex]\sin x = t[/itex]

    2. Relevant equations

    The gamma function is defined as [tex]p>0, \Gamma(p)=\int_0^\infty x^{p-1} e^{-x} dx[/tex]

    3. The attempt at a solution

    After using the suggested change of variable the integral is transformed into [tex]\int_0^1 \frac{dt}{\sqrt{1-t^4}}[/tex]

    I then considered using [itex]1-t^4=e^{2y}[/itex] such that the integrand, without yet considering the factor from transforming the differential, would become [itex]e^{-y}[/itex], with the hope that the differential would provide the necessary factor to fit the definition of the gamma function. Unfortunately, the differential turns out to be [tex]dt=\frac{-e^{2y}dy}{2{\left(e^{2y}-1\right)}^\frac{3}{4}}[/tex] which obviously isn't simplifying things at all.

    Several more variations along this line of thinking followed with similar results.

    Also tried [itex]1-t^4=y^a e^{b y}[/itex] which results in [tex]\int \frac{-(a y^{a/2-1}+b y^{a/2})e^{by/2}dy}{4(1-y^ae^{by})^\frac{3}{4}}[/tex] which still isn't getting close to the form of the gamma function.
     
    Last edited: Feb 26, 2007
  2. jcsd
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