Question

Q2) Prove that 

2

𝑛Γ(𝑛+

1

2

)

√𝜋

= 1 ∙ 3 ∙ 5 ⋯ ⋯ ⋯ (2𝑛 −

Answer 1

$$\varGamma(x)=\displaystyle\int_{0}^{\infin}t^{x-1}e^{-t}dt, x>0$$

Then

$$\varGamma(n+\dfrac{1}{2})=\displaystyle\int_{0}^{\infin}t^{n+\tfrac{1}{2}-1}e^{-t}dt$$

$$=\displaystyle\int_{0}^{\infin}t^{n-\tfrac{1}{2}}e^{-t}dt$$

$$\varGamma(n+\dfrac{1}{2})=\varGamma((n-\dfrac{1}{2})+1)=$$

$$=(n-\dfrac{1}{2})\varGamma(n-\dfrac{1}{2})$$

$$=(n-\dfrac{1}{2})(n-\dfrac{3}{2})\varGamma(n-\dfrac{3}{2})$$

$$=(n-\dfrac{1}{2})(n-\dfrac{3}{2})...(\dfrac{1}{2})\varGamma(\dfrac{1}{2})$$

$$\varGamma(\dfrac{1}{2})=\sqrt{\pi}$$

$$\varGamma(n+\dfrac{1}{2})=\dfrac{(2n-1)(2n-3)...(1)}{2^n}\sqrt{\pi}$$

$$=\dfrac{(2n-1)(2n-2)(2n-3)(2n-4)...(1)}{2^n(2n-2)(2n-4)...(2)}\sqrt{\pi}$$

$$=\dfrac{(2n-1)!}{2^n\cdot2^{n-1}(n-1)!}\sqrt{\pi}$$

$$=\dfrac{(2n-1)!(2n)}{2^{2n-1}(n-1)!(2n)}\sqrt{\pi}$$

$$=\dfrac{(2n)!}{2^{2n}n!}\sqrt{\pi}$$

$$2^{2n}n!\varGamma(n+\dfrac{1}{2})=(2n)!\sqrt{\pi}$$