Question

What is the value of r(1/2)?

Answer 1

$\displaystyle \sf{ \Gamma \bigg( \frac{1}{2} \bigg) } \: = \sqrt{\pi}$

Correct question :

$\displaystyle \sf{ What \: is \: the \: value \: of \: \: \: \Gamma \bigg( \frac{1}{2} \bigg) }$

Given :

$\displaystyle \sf{ \Gamma \bigg( \frac{1}{2} \bigg) }$

To find :

The value

Formula :

For m > 0 , n > 0

$\displaystyle \sf \Beta (m,n) = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m + n)}$

Solution :

Step 1 of 3 :

Define Beta function

We know that for m , n > 0

$\displaystyle \sf \Beta (m,n) =2 \int\limits_{0}^{ \frac{\pi}{2} } {sin}^{2m - 1} x \: {cos}^{2n - 1} x \: \, dx$

Step 2 of 3 :

Find the value of $\displaystyle \sf{ \Beta \bigg( \frac{1}{2} , \frac{1}{2} \bigg) }$

$\displaystyle \sf{ We \: put \: m = \frac{1}{2} \: \: and \: \: n = \frac{1}{2} }$

Thus we get

$\displaystyle \sf{ \Beta \bigg( \frac{1}{2} , \frac{1}{2} \bigg)}$

$\displaystyle \sf = 2\int\limits_{0}^{ \frac{\pi}{2} } \: \, dx$

$\displaystyle \sf = 2 \: x \bigg| _{0}^{ \frac{\pi}{2} }$

$\displaystyle \sf = 2 \bigg[ \frac{\pi}{2} - 0\bigg]$

$\displaystyle \sf = 2 \times \frac{\pi}{2}$

$\displaystyle \sf =\pi$

Step 3 of 3 :

$\displaystyle \sf{ Find\: the \: value \: of \: \: \: \Gamma \bigg( \frac{1}{2} \bigg) }$

We know that

$\displaystyle \sf \Beta (m,n) = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m + n)}$

$\displaystyle \sf{ We \: put \: m = \frac{1}{2} \: \: and \: \: n = \frac{1}{2} }$

$\displaystyle \sf{ \Beta \bigg( \frac{1}{2} , \frac{1}{2} \bigg)} = \frac{ \Gamma \bigg( \dfrac{1}{2} \bigg)\Gamma \bigg( \dfrac{1}{2} \bigg)}{\Gamma (1)}$

Putting the values we get

$\displaystyle \sf{ \pi= \frac{ {\Gamma \bigg( \dfrac{1}{2} \bigg)}^{2} }{1}}\: \: \: \bigg[ \: \because \:\Gamma (1) = 1\bigg]$

$\displaystyle \sf{ \implies {\Gamma \bigg( \dfrac{1}{2} \bigg)}^{2} = \pi}$

$\displaystyle \sf{ \implies {\Gamma \bigg( \dfrac{1}{2} \bigg)}^{} = \sqrt{\pi} }$

$\boxed{ \: \: \displaystyle \sf{ \Gamma \bigg( \dfrac{1}{2} \bigg) } = \sqrt{\pi} \: \: }$

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