Question estion 16 The Gamma function is defined as I (x) = ſ tä-le-t dt. (I is the capital Greek letter 0 gamma). r Help Evaluate I (1) and I (2). f(1) = Number T(2) Number Next, use integration by parts to find a recursive formula which gives I (n + 1) = ſ t"et dt in terms of I (n). Indicate the recursive formula below: ~ 0 O I (n + 1) = (I (n))? O f(n + 1) = nf(n) Or(n + 1) = f(n)! Or(n + 1) = (f(n))"-1

\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} ...