Gamma function

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Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} \sin x=?\,} This article/section deals with mathematical concepts appropriate for late high school or early college.

The Gamma funciton is defined as

Relations and values

Factorial

Using integration by parts,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Gamma(z) = \int_{0}^{\infty}t^{z-1}d(e^{-t}dt)}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \left[t^{z-1}(-e^{-t}) \right]_0^{\infty}- \int_{0}^{\infty}(z-1)t^{(z-2)}(-e^{-t})dt.}

At Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=0} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t^{z-1}(-e^{-t})} goes to 0. Using L'Hopital's rule, it's easy to show that .

So,

Thus,

Using this, and the fact that , then we can get the factorial function. For a positive integer,

z=1/2

For z=1/2,

Substituting ,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\int_{0}^{\infty} (\frac{1}{x}) e^{-x^2} (2xdx )}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =2 \int_{0}^{\infty} e^{-x^2}dx }
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \int_{-\infty}^{\infty} e^{-x^2}dx }

where in the last line we used the fact that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{-x^2}} is an even function. The integral is called the Gaussian integral and has a well known value of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt\pi} .

Thus,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Gamma(1/2) = \sqrt\pi. }