Easy Gamma Function

Ok, let us derive the value of $\Gamma\left(\dfrac{1}{2}\right)$ from its definition. The gamma function can be represented by a definite Euler integral of the second kind. From definition, we have,

$\Gamma\left(z\right)=\int_0^\infty t^{z-1}e^{-t}\,dt$

Now, for $z=\dfrac{1}{2}$ , we have,

$\Gamma\left(\dfrac{1}{2}\right)=\int_0^\infty \frac{e^{-t}}{\sqrt{t}}\,dt$

We make the substitution $t=x^2 \implies \,dt = 2x \,dx$ . After simplifying, it comes out as follows:

$\Gamma\left(\dfrac{1}{2}\right)=2\cdot \int_0^\infty e^{-x^2} \,dx$

Since $f(x)=e^{-x^2}$ is an even function, the integral can also be written as,

$\Gamma\left(\frac{1}{2}\right)=\int_{-\infty}^{+\infty} e^{-x^2} \,dx$

And behold, we have before us the famous Gaussian integral which has the value $\sqrt{\pi}$ . $\therefore \quad \Gamma\left(\dfrac{1}{2}\right)=\sqrt{\pi}$

I was waiting for someone to post a solution using Euler's Reflection Principle . But since there were no takes, I'll do the honours myself :)

$\Gamma(s)\cdot \Gamma(1-s)=\frac{\pi}{\sin \pi s}$

We now input $s=\frac{1}{2}$ and then we just have to the square root !!!!

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.

Gamma (1/ 2) = Sqrt (Pi) = 1.7724538509055160272981674833411