Evaluating $\Gamma$-function at $x=1/2$ I was following an explanation of the gamma function and everything made sense until the author evaluated the function at 1/2. The $\Gamma$-function is defined as the following integral: $$\Gamma(x)=\int_0^\infty x^{n-1}e^{-x}\,dx$$And now evaluating $\Gamma$($1/2$) we get: $$\Gamma(1/2)=\int_0^\infty x^{-1/2} e^{-x} \, dx$$This is further simplified by setting $x=y^2$ and therefore $dx=2y\,dy$. From this, we can get the following integral: $$2\int_0^\infty e^{-y^2}\,dy$$This next step is where I am confused. The author proceeds to make the following equality: $$2\int_0^\infty e^{-y^2}\,dy = 2\int_0^\infty e^{-x^2} \, dx$$ I don't understand how this equality is reached. Doesn't this mean $\int_0^\infty x^{-1/2}e^{-x} \, dx$ is equal to $2\int_0^\infty e^{-x^2} \, dx$? Does it have something to do with the bounds?
Thanks for the help in advance!
 A: In a definite integral, the variable used is a just a placeholder. Therefore, in this statement:
$$2\int_0^\infty e^{-y^2}\,dy = 2\int_0^\infty e^{-x^2} \, dx$$
The $y$ can be changed to $x$ because this is a definite integral, the meaning of the integral has not changed. It's like assigning the name "Bob" to a kid, then naming him "Tom" tomorrow. Both are different names and they might have different definitions, but they both refer to that kid.
Anyways this is just the Gaussian integral, so I hope from here you can understand.
Oh, and is this step necessary? No. If you want, you can just keep the y and it will still be a Gaussian integral.
A: $$
2\int_0^\infty e^{-y^2}\,dy = 2\int_0^\infty e^{-x^2} \, dx
$$
The way in which this equality is justified is that $y$ and $x$ are bound variables and thus can be freely renamed.
$$
\sum_{j=1}^3 j^2 = \underbrace{1^2 + 2^2 + 3^2}_\text{No $j$ or $k$ appears here.} = \sum_{k=1}^3 k^2.
$$
The value of the expression $1^2 + 2^2 + 3^2$ does not depend on the value of any quantity called $j$ or $k.$ The symbols $j$ and $k$ are bound variables. One can freely change the name of a bound variable from $j$ to $k$ or from $x$ to $y.$
In your example, this does imply that $\displaystyle\int_0^\infty x^{-1/2} e^{-x}\,dx = 2\int_0^\infty e^{-x^2}\, dx.$
