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Web AppAs I proved in my article on integration by parts (at the end), the factorial of a natural number $n$ can be calculated using the following integral:
$$ n! = ∫_0^∞ x^ne^{-x}\,dx $$The interesting thing about this integral is that it also works for $n$ that is not a natural number (and the result is a nice smooth function of $n$). The factorial of one half ($0.5$) is thus defined as
$$ (1/2)! = ∫_0^∞ x^{1/2}e^{-x}\,dx $$We will show that:
How to go about calculating the integral? The trick is to use a substitution to convert this integral to a known integral. First, we use integration by parts once, which will give us a form that is easier to work with:
$$ ∫_0^∞ \overbrace{x^{1/2}}^u\overbrace{e^{-x}}^{v’}\,dx = \l[\overbrace{x^{1/2}}^u\overbrace{(-e^{-x})}^v\r]_0^∞ - ∫_0^∞ \overbrace{\frac12 x^{-1/2}}^{u’}\overbrace{(-e^{-x})}^v\,dx $$The term in the brackets is $0$. Let’s apply the substitution $y = x^{1/2}$ (i.e. $dy = \frac12 x^{-1/2}dx$) to the second integral (with the negative sign taken out):
$$ ∫_0^∞ \frac12 x^{-1/2}e^{-x}\,dx = ∫_0^∞ e^{-y^2}\,dy\,. $$Since $e^{-y^2}$ is an even function, it is easy to see that
$$ ∫_0^∞ e^{-y^2}\,dy = ∫_{-∞}^0 e^{-y^2}\,dy = \frac12 ∫_{-∞}^∞ e^{-y^2}\,dy $$(the function $e^{-y^2}$ looks the same “to the right of $0$” as “to the left of $0$”). The rightmost integral is known as the Gaussian integral, and I have shown a way to solve it in a separate article. The value turns out to be $√\pi$, so we get:
$$ (1/2)! = \frac12 ∫_{-∞}^∞ e^{-y^2}\,dy = \frac{√\pi}{2}\,. $$
