prove that Gamma(n+1)=nGamma(n)

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Introduction : In this solution, we write formula of Gamma function and prove that prove that Gamma(n+1)=nGamma(n) in the following

Question : Prove that

\[\mathrm{\Gamma}\left(n+1\right)=n\mathrm{\Gamma}\left(n\right)\]

Solution :

To prove

\[\mathrm{\Gamma}\left(n+1\right)=n\mathrm{\Gamma}\left(n\right)\]

We general formula of Gamma function

\[\mathrm{\Gamma}\left(n\right)=\int_{0}^{\infty}{e^{-x}x^{n-1}dx}\]

replacing n by n+1 here we get

\[\mathrm{\Gamma}\left(n+1\right)=\int_{0}^{\infty}{e^{-x}x^{n+1-1}dx}\]

which becomes

\[=\int_{0}^{\infty}{e^{-x}x^ndx}\]

now integrating by parts

\[={\left[-e^{-x}x^n\right]^\infty}_0-\int_{0}^{\infty}\left\{\frac{d}{dx}\left(x^n\right)\int{e^{-x}dx}\right\}dx\]

\[=-\left(limit_{x\rightarrow\infty}\frac{x^n}{e^x}-0\right)+n\int_{0}^{\infty}{{x}^{n-1}e^{-x}dx}\]

\[=-\left(0-0\right)+n\int_{0}^{\infty}{e^{-x}x^{n-1}dx}\]

\[\mathrm{\Gamma}\left(n+1\right)=n\int_{0}^{\infty}{e^{-x}x^{n-1}dx}\]

using Gamma of n in the right hand side

\[=n\mathrm{\Gamma}\left(n\right)\]

This is proved that

\[\mathrm{\Gamma}\left(n+1\right)=n\mathrm{\Gamma}\left(n\right)\]


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