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<math>\Gamma(z) = \int_{0}^{\infty}t^{z-1}e^{-t}dt</math>
<math>\Gamma(z) = \int_{0}^{\infty}t^{z-1}e^{-t}dt</math>
== Relations and values ==
===Factorial===
Using [[Integration_by_parts|integration by parts]],
Using [[Integration_by_parts|integration by parts]],
<math>\Gamma(z) = \int_{0}^{\infty}t^{z-1}d(e^{-t}dt)</math>
<math>\Gamma(z) = \int_{0}^{\infty}t^{z-1}d(e^{-t}dt)</math>
:<math>= \left[t^{z-1}(-e^{-t}) \right]_0^{\infty}- \int_{0}^{\infty}(z-1)t^{(z-2)}(-e^{-t})dt</math>
:<math>= \left[t^{z-1}(-e^{-t}) \right]_0^{\infty}- \int_{0}^{\infty}(z-1)t^{(z-2)}(-e^{-t})dt.</math>
At <math>t=0</math>, <math>t^{z-1}(-e^{-t})</math> goes to 0. Using [[L'Hopital's rule]], it's easy to show that <math>\lim_{t\to \infty} \frac{t^{z-1}}{e^{t}} = 0</math>.
So,
:<math>=(z-1) \int_{0}^{\infty}t^{(z-1)-1}e^{-t}dt</math>
:<math>=(z-1) \int_{0}^{\infty}t^{(z-1)-1}e^{-t}dt</math>
:<math>=(z-1) \Gamma(z-1).</math>
:<math>=(z-1) \Gamma(z-1).</math>
Thus,
Thus,
:<math> \Gamma (z) = (z-1) \Gamma(z-1)</math>
:<math> \Gamma (z) = (z-1) \Gamma(z-1).</math>
Using this, and the the fact that <math>\Gamma(0)=1</math>, then we can get the [[Factorial|factorial]]. For <math>n</math> a positive integer,
Using this, and the the fact that <math>\Gamma(0)=1</math>, then we can get the [[Factorial|factorial]]. For <math>n</math> a positive integer,
:<math> \Gamma(n) = (n-1)!. </math>
:<math> \Gamma(n) = (n-1)!. </math>
===z=1/2===
For z=1/2,
:<math> \Gamma(1/2) = \int_{0}^{\infty}t^{-1/2}e^{-t}dt </math>
Substituting <math>x=t^{1/2}</math>,
:<math>=\int_{0}^{\infty} (\frac{1}{x}) e^{-x^2} (2xdx )</math>
:<math>=2 \int_{0}^{\infty} e^{-x^2}dx </math>
:<math>= \int_{-\infty}^{\infty} e^{-x^2}dx </math>
where in the last line we used the fact that the function <math>e^{-x^2}</math> is even. The integral is called the Gaussian integral and has a well known value of <math>\sqrt\pi</math>.
Thus,
:<math> \Gamma(1/2) = \sqrt\pi. </math>
[[category:mathematics]]
[[category:mathematics]]