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The beta integral can be evaluated elegantly using the convolution theorem for Laplace transforms.
Start with the following Laplace transform:
Since
 , the convolution theorem imples that
Writing out the definition of convolution, this becomes
Setting  and simplifying, we conclude that
QED
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![$\displaystyle s^{-\alpha} = {\cal L} \left[ {t^{\alpha - 1} \over \Gamma(\alpha)} \right] = \int_0^\infty e^{-st} {t^{\alpha - 1} \over \Gamma(\alpha)} dt $](http://images.planetmath.org:8080/cache/objects/6206/l2h/img1.png "$\displaystyle s^{-\alpha} = {\cal L} \left[ {t^{\alpha - 1} \over \Gamma(\alpha)} \right] = \int_0^\infty e^{-st} {t^{\alpha - 1} \over \Gamma(\alpha)} dt $")
