Here's something basic on the transformation of Gamma function and I'm wondering if someone could explain to me how the following works:

we know the function has two formulations:

$\Gamma(x) = \int_{0}^{\infty}t^{x-1}e^{-t}dt$


$\Gamma(x) = \int_{0}^{1}(-\log t)^{x-1}dt$

but how can we transform one into the other?

I've tried the method https://math.stackexchange.com/a/360687/573671 via user17762 but failed.

Thanks in advance for your time and help!


They used the transformation $ t = -\log(u)$. Then you should note that $$ e^{-t}\,\mathrm{d}t = - \,\mathrm{d} u $$ Moreover, the integration bounds change from $(0,\infty) \mapsto (1,0)$. Now, you should reverse the order of integration to get the bounds $(0,1)$. This gives a minus sign which cancels against the one we had above.

I hope this helps. If you have any further questions, feel free to ask them!

  • $\begingroup$ No problem. I am glad to help $\endgroup$ – Stan Tendijck Jul 3 '18 at 8:36

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