Question (The gamma function) The gamma function \( \Gamma(x) \) is defined by the improper integral \[ \Gamma(x)=\int_{0}^{\infty} t^{x-1} e^{-t} d t \] ( \( \Gamma \) is the Greek capital letter gamma.) (a) Show that the integral converges for \( x>0 \). (b) Use integration by parts to show that \( \Gamma(x+1)=x \Gamma(x) \) for \( x>0 \).>(c) Show that \( \Gamma(n+1)=n \) ! for \( n=0,1,2, \ldots \) (d) Given that \( \int_{0}^{\infty} e^{-x^{2}} d x=\frac{1}{2} \sqrt{\pi} \), show that \( \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi} \) and \( \Gamma\left(\frac{3}{2}\right)=\frac{1}{2} \sqrt{\pi} \). In view of (c), \( \Gamma(x+1) \) is often written \( x \) ! and regarded as a real-valued extension of the factorial function. Some scientific calculators (in particular, HP calculators) with the factorial function \( n \) ! built in actually calculate the gamma function rather than just the integral factorial. Check whether your calculator does this by asking it for \( 0.5 \) !. If you get an error message, it's not using the gamma function.
\[
\Gamma(x)=\int_{0}^{\infty} t^{x-1} e^{-t} d t
\]
( \( \Gamma \) is the Greek capital letter gamma.)
(a) Show that the integral converges for \( x>0 \).
(b) Use integration by parts to show that \( \Gamma(x+1)=x \Gamma(x) \) for \( x>0 \).>(c) Show that \( \Gamma(n+1)=n \) ! for \( n=0,1,2, \ldots \)
(d) Given that \( \int_{0}^{\infty} e^{-x^{2}} d x=\frac{1}{2} \sqrt{\pi} \), show that \( \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi} \) and \( \Gamma\left(\frac{3}{2}\right)=\frac{1}{2} \sqrt{\pi} \).
In view of (c), \( \Gamma(x+1) \) is often written \( x \) ! and regarded as a real-valued extension of the factorial function. Some scientific calculators (in particular, HP calculators) with the factorial function \( n \) ! built in actually calculate the gamma function rather than just the integral factorial. Check whether your calculator does this by asking it for \( 0.5 \) !. If you get an error message, it&qpos;s not using the gamma function.
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\Gamma(x)=\int_{0}^{\infty} t^{x-1} e^{-t} d ta) Since \lim _{t \rightarrow \infty} t^{x-1} e^{-t / 2}=0, there exists T>0 such that t^{x-1} e^{-t / 2} \leq 1 if t \geq T. Thus0 \leq \int_{T}^{\infty} t^{x-1} e^{-t} d t \leq \int_{T}^{\infty} e^{-t} d t=2 e^{-T / 2}and \int_{T}^{\infty} t^{x-1} e^{-t} d t converges by the comparison theorem. If x>0, then0 \leq \int_{0}^{T} t^{x-1} e^{-t} d t<\int_{0}^{T} t^{x-1} d tconverges by Theorem 2(b). Thus the integral defining \Gamma(x) converges.\text { b) } \begin{aligned}\Gamma(x+1)= & \int_{0}^{\infty} t^{x} e^{-t} d t \\= & \lim _{\substack{c \rightarrow 0+\\R \rightarrow \infty}} \int_{c}^{R} t^{x} e^{-t} d t \\& U=t^{x} \quad d V=e^{-t} d t \\& d U=x t^{x-1} d x \quad V=-e^{-t} \\= & \lim _{\substack{c \rightarrow 0+\\R \rightarrow \infty}}\left(-\left.t^{x} e^{-t}\right|_{c} ^{R}+x \int_{c}^{R} t^{x-1} e^{-t} d t\right) \\= & 0+x \int_{0}^{\infty} t^{x-1} e^{-t} d t=x \Gamma(x) .\end{aligned} c) \Gamma(1)=\int_{0}^{\infty} e^{-t} d t=1=0 !.By (b), \Gamma(2)=1 \Gamma(1)=1 \times 1=1=1.In general, if \Gamma(k+1)=k ! for some positive integer k, then\Gamma(k+2)=(k+1) \Gamma(k+1)=(k+1) k !=(k+1) ! .Hence \Gamma(n+1)=n ! for all integers n \geq 0, by induction.d)\begin{array}{l}\Gamma\left(\frac{1}{2}\right)=\int_{0}^{\infty} t^{-1 / 2} e^{-t} d t \quad \text { Let } t=x^{2} \\=\int_{0}^{\infty} \frac{1}{x} e^{-x^{2}} 2 x d x=2 \int_{0}^{\infty} e^{-x^{2}} d x=\sqrt{\pi} \\\Gamma\left(\frac{3}{2}\right)=\frac{1}{2} \Gamma\left(\frac{1}{2}\right)=\frac{1}{2} \sqrt{\pi} . \\\end{array} ...