Community Answer

Honor CodeSolved 1 Answer

See More Answers for FREE

Enhance your learning with StudyX

Receive support from our dedicated community users and experts

See up to 20 answers per week for free

Experience reliable customer service

Get Started

\Gamma(x)=\int_{0}^{\infty} t^{x-1} e^{-t} d tplug x=1 and 2\begin{array}{l}\Gamma(1)=\int_{0}^{\infty} t^{1-1} e^{-t} d t=\int_{0}^{\infty} e^{-t} d t \\=-\left[e^{-t}\right]_{0}^{\infty}=-\left[e^{-\infty}-e^{-\infty}\right]=-[0-1] \\\Gamma(1)=1 \\\Gamma(2)=\int_{0}^{\infty} 4^{2-1} e^{-t} d t=\int_{0}^{\infty} t e^{-t} d t \\\end{array}Apbly the prradach ruate integretion by parts\begin{array}{l} \Gamma(2)\left.=[t] e^{-t} d t\right]_{0}^{\infty}-\int \frac{d t}{d t}\left(\int e^{-t} d t\right) d t \\=\left[-t e^{-t}\right]^{\infty}+\int_{0}^{\infty} e^{-t} d t=\lim _{t \rightarrow \infty}\left(-t e^{-t}\right)-0+1 \\=0 \div 0+1=1 \\\left\{\because \lim _{t \rightarrow \infty}\left(-t^{n} e^{-t}\right)=0\right.\end{array}Now\Gamma(n+1)=\int_{0}^{\infty} t^{n} e^{-t} d tAbbly integration by parts\begin{array}{l}\Gamma(n+1)=\left[t^{4} \int_{e^{-t} d t}\right]_{0}^{\infty}-\int_{0}^{\infty} \frac{d}{d t} t^{4}\left(\int e^{-t} d t\right) d t \\=\left[-t^{n} e^{-t}\right]_{0}^{\infty}+\int_{0}^{\infty} n t^{n-1} e^{-t} d t \\\begin{array}{l}=\lim _{t \rightarrow \infty}\left(-t^{n} e^{-t}\right)-\sin ^{0}(-0)+n \int^{\infty} t^{n-1} e^{-t} d t \\=0+0+n\end{array} \\=0+0+n \Gamma(n) \quad\left\{\quad \left\{\begin{array}{l}=n \Gamma(n) \\0\end{array} \lim _{t \rightarrow \infty}\left(t^{n} e^{-t}\right)=0\right.\right. \\\Gamma(n+1)=n \Gamma(n) \\\end{array} ...