Gamma Function by Euler’s Second Integral

$Gamma Function and Euler's Second Integral$ ……..(1)

1). Recurrence Formula :

$Recurrence Formula and Euler's Second Integral$ n > 0

Proof – Put value (n+1) in equation no.1

$https://latex.codecogs.com/svg.image?large Gamma (n+1)=int_{0}^{infty }x^{n}e^{-x}dx =displaystyle lim_{x to 0}int_{0}^{a}x^{n}e^{-x}dx$

$https://latex.codecogs.com/svg.image?large = displaystyle lim_{x to 0}left [ left{x^{n}(-e^{-x}) right}_{0}^{a} -int_{0}^{a}(nx^{n-1})(-e^{-x})dxright ]$

$https://latex.codecogs.com/svg.image?large = displaystyle lim_{x to infty }(-a^{n}e^{-a})+ndisplaystyle lim_{ xto infty }int_{0}^{a}x^{n-1}e^{-x}dx$

$https://latex.codecogs.com/svg.image?large = 0+nint_{0}^{infty }x^{n-1}e^{-x}dx$

Or, $\Gamma (n+1)=n\Gamma (n),$ if n > 0

2). Relation between Gamma Function and Factorial :

$Relation between Gamma Function and Factorial and Euler's Second Integral$ $n \epsilon N$

Proof –

$https://latex.codecogs.com/svg.image?large Gamma (1)= int_{0}^{infty }e^{-x}dx=displaystyle lim_{a to infty }int_{0}^{a}e^{-x}dx$

$https://latex.codecogs.com/svg.image?large =displaystyle lim_{a to infty }[e^{-x}]_{0}^{a}=displaystyle lim_{a to infty }[1-e^{-a}]$

$\therefore\Gamma (1)=1!$

Substituting n = 1, 2, 3, ……….. in the recurrence relations of Gamma Function ,

Γ(2) = 1Γ(1) = 1!

Γ(3) = 2Γ(2) = 2*1. = 2!

Γ(4) = 3Γ(3) = 3*(2)! = 3!

Γ(5) = 4Γ(4) = 4*(3)! = 4!

If general, $\Gamma (n+1)=n!$ if $n\epsilon N$

Remark : $\Gamma 0 =\infty ,$ $\Gamma (n+1) =\infty ,$ for $n\epsilon N$

3). Standard Integral :

$Standard Integral and Euler's Second Integrals$

Proof – Substituting x = az in the definition of the Gamma Function ,

$https://latex.codecogs.com/svg.image?large Gamma(n)=int_{0}^{infty } (az)^{n-1}e^{-az}a.dz$

$https://latex.codecogs.com/svg.image?large =a^{n}int_{0}^{infty } z^{n-1}e^{-az}dz=a^{n}int_{0}^{infty }x^{n-1}e^{-ax}dx$

$https://latex.codecogs.com/svg.image?large therefore int_{0}^{infty }x^{n-1}e^{-ax}dx = frac{Gamma (n)}{a^{n}}$

4). Gamma Formula :

$Gamma Formula and Euler's Second Integrals$

Proof –

$https://latex.codecogs.com/svg.image?large B(m,n)=2int_{0}^{pi /2}sin^{2m-1}theta cos^{2n-1}theta dtheta$ ……………(2)

$https://latex.codecogs.com/svg.image?large B(m,n)=frac{Gamma (n)Gamma (m)}{Gamma (m+n)}$ ………………(3)

From equation 2 & 3 ,

$https://latex.codecogs.com/svg.image?large 2int_{0}^{pi /2}sin^{2m-1}theta cos^{2n-1}theta dtheta =frac{Gamma (m).Gamma (n)}{Gamma (m+n)}$

Replacing m by (m+1)/2 and n by (n+1)/2 ,

$https://latex.codecogs.com/svg.image?large int_{0}^{pi /2}sin^{m}theta cos^{n}theta dtheta =frac{Gamma left{ (m+1)/2right}.Gamma left{ (n+1)/2right}}{2Gamma left{ (m+n+2)/2right}}$

👉 Properties of Beta Function

Gamma Function by Euler’s Second Integral

Gamma Function by Euler’s Second Integral

$Gamma Function and Euler's Second Integral$ ……..(1)

1). Recurrence Formula :

$Recurrence Formula and Euler's Second Integral$ n > 0

2). Relation between Gamma Function and Factorial :

Remark : $\Gamma 0 =\infty ,$ $\Gamma (n+1) =\infty ,$ for $n\epsilon N$

3). Standard Integral :

4). Gamma Formula :

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