Question

111. Gamma function The gamma function is defined by (Gamma(p)=int_{0}^{infty} x^{p-1} e^{-x} d x), for (p) not equal to zero or a negative integer. a. Use the reduction formula [int_{0}^{infty} x^{p} e^{-x} d x=p int_{0}^{infty} x^{p-1} e^{-x} d x ext{ for } p=1, 2, 3, ...] to show that (Gamma(p+1)=p!) (p factorial). b. Use the substitution (x=u^{2}) and the fact that (int_{0}^{infty} e^{-u^{2}} d u=frac{sqrt{pi}}{2}) to show that (Gammaleft(frac{1}{2} ight)=sqrt{pi}). (Gamma(p)=int_{0}^{infty} x^{p-1} e^{-x} d x ightarrow Gamma(p+1)=p!) (=int_{0}^{infty} x^{p} e^{-x} d x) (u = x^{p} quad v = -e^{-x}) (u' = px^{p-1} quad v' = e^{-x}) (int uv' = uv - int v u') (= x^{p} cdot -e^{-x} vert_{0}^{infty} - int_{0}^{infty} -e^{-x} cdot px^{p-1} d x) (= [x^{p} cdot -e^{-x} vert_{0}^{infty}] + p int_{0}^{infty} e^{-x} cdot x^{p-1} d x) (= [lim_{b ightarrow infty} b^{p} cdot -e^{-b} - (0)^{p} cdot -e^{-(0)}]) (= [0 - 0] + p int_{0}^{infty} e^{-x} cdot x^{p-1} d x) (= p int_{0}^{infty} e^{-x} cdot x^{p-1} d x) (int uv' = uv - int v u') How to pick u... 1st Log Inverse Algebraic (+, -, x, (div, sqrt[n]{}, x^{n})) Trig Exponential

          111. Gamma function The gamma function is defined by (Gamma(p)=int_{0}^{infty} x^{p-1} e^{-x} d x), for (p) not equal to zero or a negative integer.
a. Use the reduction formula
[int_{0}^{infty} x^{p} e^{-x} d x=p int_{0}^{infty} x^{p-1} e^{-x} d x 	ext{ for } p=1, 2, 3, ...]
to show that (Gamma(p+1)=p!) (p factorial).
b. Use the substitution (x=u^{2}) and the fact that (int_{0}^{infty} e^{-u^{2}} d u=frac{sqrt{pi}}{2}) to show that (Gammaleft(frac{1}{2}
ight)=sqrt{pi}).
(Gamma(p)=int_{0}^{infty} x^{p-1} e^{-x} d x 
ightarrow Gamma(p+1)=p!)
(=int_{0}^{infty} x^{p} e^{-x} d x)
(u = x^{p} quad v = -e^{-x})
(u' = px^{p-1} quad v' = e^{-x})
(int uv' = uv - int v u')
(= x^{p} cdot -e^{-x} vert_{0}^{infty} - int_{0}^{infty} -e^{-x} cdot px^{p-1} d x)
(= [x^{p} cdot -e^{-x} vert_{0}^{infty}] + p int_{0}^{infty} e^{-x} cdot x^{p-1} d x)
(= [lim_{b 
ightarrow infty} b^{p} cdot -e^{-b} - (0)^{p} cdot -e^{-(0)}])
(= [0 - 0] + p int_{0}^{infty} e^{-x} cdot x^{p-1} d x)
(= p int_{0}^{infty} e^{-x} cdot x^{p-1} d x)
(int uv' = uv - int v u')
How to pick u...
1st Log
Inverse
Algebraic (+, -, x, (div, sqrt[n]{}, x^{n}))
Trig
Exponential

$111. Gamma function The gamma function is defined by (Gamma(p)=int0^infty x^p-1 e^-x d x), for (p) not equal to zero or a negative integer. a. Use the reduction formula [int0^infty x^p e^-x d x=p int0^infty x^p-1 e^-x d x ext for p=1, 2, 3, ...] to show that (Gamma(p+1)=p!) (p factorial). b. Use the substitution (x=u^2) and the fact that (int0^infty e^-u^2 d u=fracsqrtpi2) to show that (Gammaleft(frac12 ight)=sqrtpi). (Gamma(p)=int0^infty x^p-1 e^-x d x ightarrow Gamma(p+1)=p!) (=int0^infty x^p e^-x d x) (u = x^p quad v = -e^-x) (u' = px^p-1 quad v' = e^-x) (int uv' = uv - int v u') (= x^p cdot -e^-x vert0^infty - int0^infty -e^-x cdot px^p-1 d x) (= [x^p cdot -e^-x vert0^infty] + p int0^infty e^-x cdot x^p-1 d x) (= [limb ightarrow infty b^p cdot -e^-b - (0)^p cdot -e^-(0)]) (= [0 - 0] + p int0^infty e^-x cdot x^p-1 d x) (= p int0^infty e^-x cdot x^p-1 d x) (int uv' = uv - int v u') How to pick u... 1st Log Inverse Algebraic (+, -, x, (div, sqrt[n], x^n)) Trig Exponential$

Added by Anthony L.

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