firstly , I recommend you to read these Gamma function to have fundamental understanding of the Gamma function.

G.ID.1: Differentiation (integral representation)

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

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

differentiate gamma function in term of x variable

${\Gamma }'(x)=\int _{0}^{\infty }t^{x-1}e^{-t}\ln (t) dt$

${\Gamma }^{n}(x)=\int _{0}^{\infty }t^{x-1}e^{-t}\ln ^{n}(t) dt$

if x=1

${\Gamma }'(1)=\int _{0}^{\infty }e^{-t}\ln (t) dt$

G.ID.2: integral representation

$\Gamma (z)=\int _{0}^{1}\ln (\frac{1}{x})^{z-1} dx$

proof

given that

$\Gamma (z)=\int _{0}^{\infty } t^{z-1} e^{-t} dt$

variable substitution

$t=\ln (\frac{1}{x})$

(1)

$dt=\frac{x}{-x^{2}}dx$

(2)

from (1), lower and upper integrals boundaries change

$t=0\rightarrow 0=\ln (\frac{1}{x})\rightarrow \frac{1}{x}=1\rightarrow x=1$

$t=\infty \rightarrow \infty=\ln (\frac{1}{x})\rightarrow \frac{1}{x}=\infty\rightarrow x=0$

(3)

$\Gamma (z)=\int _{1}^{0}\ln (\frac{1}{x})^{z-1}e^{-\ln (\frac{1}{x})}\frac{-1}{x} dx$

(4)

$\Gamma (z)=\int _{1}^{0}\ln (\frac{1}{x})^{z-1}x\frac{-1}{x} dx$

(5)

$\Gamma (z)=\int _{0}^{1}\ln (\frac{1}{x})^{z-1} dx$

G.ID.3: integral representation

$\Gamma (z)=2\int _{0}^{\infty } x^{2z-1} e^{-x^{2}} dx$

proof

given that

$\Gamma (z)=\int _{0}^{\infty } t^{z-1} e^{-t} dt$

variable substitution

$t=x^{2}$

$dt=2xdx$

$\Gamma (z)=\int _{0}^{\infty } x^{2(z-1)} e^{-x^{2}}2x dx$

$\Gamma (z)=2\int _{0}^{\infty } x^{2z-1} e^{-x^{2}} dx$

G.ID.4: Weierstrass Identity

$\Gamma (x)= e^{-\gamma x}\frac{1}{x}\prod _{n=1}^{\infty }\frac{e^{\frac{x}{n}}}{1+\frac{x}{n}}$

$\gamma$ = Euler–Mascheroni constant

Given

$\Gamma (x)=\lim n \to \infty \frac{n^{x}n!}{x(x+1)(x+2)(x+3)\cdot \cdot \cdot (x+n)}$

Proof

$\Gamma (x)= \frac{n^{x}n!}{x(x+1)(x+2)(x+3)\cdot \cdot \cdot (x+n)}$

(1)

$\Gamma (x)= \frac{n^{x}n!}{1\times 2\times 3\times 4\times 5\cdot \cdot \cdot \times n\, x(1+x)(1+\frac{x}{2})(1+\frac{x}{3})\cdot \cdot \cdot (1+\frac{x}{n})}$

(2)

$n^{x}=e^{\ln n^{x}}=e^{x\ln n}$

(3)

the $n!$ cancels out, and based on (3) , therefore,

$\Gamma (x)= \frac{e^{x\ln (n)}}{x(1+x)(1+\frac{x}{2})(1+\frac{x}{3})\cdot \cdot \cdot (1+\frac{x}{n})}$

(4)

$\left ( e^{x}e^{\frac{x}{2}}e^{\frac{x}{3}}e^{\frac{x}{4}}\cdot \cdot \cdote^{\frac{x}{n}} \right )\left ( e^{-x}e^{\frac{-x}{2}}e^{\frac{-x}{3}}e^{\frac{-x}{4}}\cdot \cdot \cdote^{\frac{-x}{n}} \right )=1$

$\left ( e^{x}e^{\frac{x}{2}}e^{\frac{x}{3}}e^{\frac{x}{4}}\cdot \cdot \cdote^{\frac{x}{n}} \right )\left ( e^{-x(1+\frac{1}{2}+\frac{1}{3}\cdot \cdot \cdot \frac{1}{n})} \right )=1$

(5)

multiply 4 by 5 , therefore

$\Gamma (x)= \frac{e^{x\ln (n)}\left ( e^{x}e^{\frac{x}{2}}e^{\frac{x}{3}}e^{\frac{x}{4}}\cdot \cdot \cdot e^{\frac{x}{n}} \right )\left ( e^{-x(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdot \cdot \cdot +\frac{1}{n})} \right )}{x(1+x)(1+\frac{x}{2})(1+\frac{x}{3})\cdot \cdot \cdot (1+\frac{x}{n})}$

(6)

(7)

$\gamma = \left ( (1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\cdot \cdot \cdot +\frac{1}{n})-\ln (n) \right )$

(8)

$\gamma$ = Euler–Mascheroni constant

therefore;

$\Gamma (x)=\frac{\left ( e^{x}e^{\frac{x}{2}} e^{\frac{x}{3}} e^{\frac{x}{4}} \cdot \cdot \cdot e^{\frac{x}{n}} \right )e^{-\gamma x}}{x(1+x)(1+\frac{x}{3})(1+\frac{x}{3})\cdot \cdot \cdot (1+\frac{x}{n})}$

(9)

$n \to \infty$

$\Gamma (x)= e^{-\gamma x}\frac{1}{x}\prod _{n=1}^{\infty }\frac{e^{\frac{x}{n}}}{1+\frac{x}{n}}$

G.ID.5: Reflection Formula

$\Gamma (x)\Gamma (1-x)=\frac{\pi}{\sin(\pi x)}$

$\Gamma (\frac{1}{2}+x)\Gamma (\frac{1}{2}-x)=\frac{\pi}{\cos(\pi x)}$

Given

$\frac{1}{\Gamma (x)}= e^{\gamma x}x\prod _{n=1}^{\infty }({1+\frac{x}{n})e^{\frac{-x}{n}}}$

(1)

$\frac{1}{\Gamma (-x)}= -e^{-\gamma x}x\prod _{n=1}^{\infty }({1-\frac{x}{n})e^{\frac{x}{n}}}$

(2)

Proof
multiplication of (1) and (2)

$\frac{1}{\Gamma (x)}\frac{1}{\Gamma (-x)}= -x^{2}\prod _{n=1}^{\infty }\left ( 1-\frac{x^{2}}{n^{2}} \right )$

(3)

sine identity

$\sin(\pi x)= \pi x \prod _{n=1}^{\infty }\left ( 1-\frac{x^{2}}{n^{2}} \right )$

(4)

Recurrence identity of gamma function

$\Gamma (1-x)=-x\Gamma (-x)$

(5)

based on (3) and (4) , (5)

$\frac{1}{\Gamma (x)}\frac{1}{\Gamma (1-x)}=\frac{\sin(\pi x)}{\pi}$

(5)

$\Gamma (x)\Gamma (1-x)=\frac{\pi}{\sin(\pi x)}$

(6)

replace x and (1-x) with the following identities

$\frac{1}{2}+z=x$

$\frac{1}{2}-z=1-x$

(7)

therefore

$\Gamma (\frac{1}{2}+z)\Gamma (\frac{1}{2}-z)=\frac{\pi}{\sin (\pi(\frac{1}{2}+z))}$

(8)

$\sin (\pi(\frac{1}{2}+z))=\sin(\frac{\pi}{2})\cos(\pi z) +\cos(\frac{\pi}{2})\sin(\pi z)$

$\sin (\pi(\frac{1}{2}+z))=\cos(\pi z)$

(9)

substituting (9) into (8) results in

$\Gamma (\frac{1}{2}+z)\Gamma (\frac{1}{2}-z)=\frac{\pi}{\cos(\pi z)}$

G.ID.6: Recurrence Identity

$\Gamma (s+1)=s\Gamma (s)$

given

$\Gamma (s)=\int _{0}^{\infty }x^{s-1}e^{-x} \, dx$

proof

$\Gamma (s)= \left | \frac{x^{s}}{s}e^{-x} \right |_{0}^{\infty} +\frac{1}{s}\int _{0}^{\infty }x^{s}e^{-x} \, dx$

$\Gamma (s)=\frac{1}{s}\Gamma (s+1)$

$\Gamma (s+1)=s\Gamma (s)$

if s=s+1/2

$\Gamma (s+\frac{1}{2})=\int _{0}^{\infty }x^{s-\frac{1}{2}}e^{-x} \, dx$

integrating by parts results:

$\Gamma (s+\frac{1}{2})=(s+\frac{1}{2})\int _{0}^{\infty } x^{s+\frac{1}{2}} e^{-x}\, dx$

after n times of integration by parts:

$\Gamma (s+\frac{1}{2})=\frac{1.3.5...(2n-1)}{2^{n}}\Gamma (\frac{1}{2})$

similarly for s+1/3 and s+1/4

$\Gamma (s+\frac{1}{3})=\frac{1.4.7...(3n-2)}{3^{n}}\Gamma (\frac{1}{3})$

$\Gamma (s+\frac{1}{4})=\frac{1.5.9...(4n-3)}{4^{n}}\Gamma (\frac{1}{4})$

G.ID.7:

$\int _{0}^{1}y^{q-1}log(\frac{1}{y})^{p-1}dy =\frac{\Gamma (p)}{q^{p}}$

proof

$\int _{0}^{1}y^{q-1}log(\frac{1}{y})^{p-1}dy = \int _{0 }^{\infty}e^{-(q-1)x}x^{p-1}(-e^{-x})dx$

(1)

$\int _{0}^{1}y^{q-1}log(\frac{1}{y})^{p-1}dy = \int _{0 }^{\infty}e^{-qx}x^{p-1}dx$

(2)

$qx=t \: ,\: dx= \frac{dt}{q}$

(3)

$\int _{0}^{1}y^{q-1}log(\frac{1}{y})^{p-1}dy = \int _{0 }^{\infty}e^{-t}(\frac{t}{q})^{p-1}\frac{1}{q}dt$

(4)

$\int _{0}^{1}y^{q-1}log(\frac{1}{y})^{p-1}dy = \frac{1}{q^{p}}\int _{0 }^{\infty}e^{-t}t^{p-1}dt$

(5)

$\int _{0}^{1}y^{q-1}log(\frac{1}{y})^{p-1}dy =\frac{\Gamma (p)}{q^{p}}$

(6)

Engineering Mathematics

Wednesday, May 4, 2011

Gamma Function

$\gamma$ = Euler–Mascheroni constant

$\gamma$ = Euler–Mascheroni constant

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