Gamma Function

The relation between the beta function (Euler's integral of first kind) and the gamma function (Euler's integral of second kind) is

$B(m,n) = \dfrac{\Gamma (m) \Gamma (n)}{\Gamma (m+n)}$

The beta function can be represented as

$B(m,n) = 2 \int_0^{\pi/2} \sin^{2m-1} x \cos^{2n-1} x \,dx$

Putting $m=n=\dfrac 12$ gives,

$\begin{aligned} B\left(\dfrac 12 , \dfrac 12\right) &= 2 \int_0^{\pi/2} \,dx \\ &= \pi \end{aligned}$

This gives

$\begin{aligned} & B\left(\dfrac 12,\dfrac 12\right) &=& \dfrac{\Gamma \left(\dfrac 12\right) \Gamma \left(\dfrac 12\right)}{\Gamma \left(\dfrac 12+\dfrac 12\right)} \\ \implies & \pi &=& \dfrac{\Gamma \left(\dfrac 12\right) \Gamma \left(\dfrac 12\right)}{\Gamma \left(1\right)} \\ & &=& {\left[\Gamma \left(\dfrac 12\right)\right]}^2 \qquad \qquad \qquad \qquad \small \color{#3D99F6}{\text{As } \Gamma(1) = 1} \\ \implies & \Gamma \left(\dfrac 12\right) &=& \boxed{\sqrt{\pi}} \end{aligned}$

By the reflection principle: $\Gamma (x)\Gamma (1-x)=\frac { \pi }{ \sin { \pi x } }$ Plugging in one-half, we get: $\Gamma \left(\frac { 1 }{ 2 } \right)\Gamma \left(\frac { 1 }{ 2 } \right)=\frac { \pi }{ \sin { \pi (\frac { 1 }{ 2 } ) } } =\frac { \pi }{ 1 } =\pi$ Taking the square root of both sides: $\Gamma \left(\frac { 1 }{ 2 } \right)=\sqrt { \pi } \approx \color{#D61F06}{1.7724}$

I purposely chose to use the reflection principle because I could just take the square root of $\displaystyle{\Gamma \left(\frac { 1 }{ 2 } \right)\Gamma \left(\frac { 1 }{ 2 } \right)}$ to get $\displaystyle{\Gamma \left(\frac { 1 }{ 2 } \right)}$ . If it asked to calculate something else then the gamma of one-half, then I would have either used the definition of the gamma function or had to know the gamma of another fraction.