How to attack the gamma function manually

Question

Thoughts before Friday.

If $\operatorname{\Gamma(1.5)}=\displaystyle \int_0^\infty e^{-x}\sqrt{x}=\dfrac{\sqrt{\pi}}{2}$, how was this calculated? I proved the relevance of the formula as an analytic continuation of the factorial function but I've been stuck on calculating the value of $0.5!$ from the integral itself.

I'm not exactly sure which pattern to produce and I also have no idea where the π came from. Was it guessed or is there a step-by-step solution? Any help would be appreciated.

Answer 1

This definite integral is difficult, and you need "tricks".

For example, you start from $$\int_0^\infty\frac{e^{-x}}{2\sqrt x}dx=\int_0^\infty e^{-x^2}dx$$

and switch to $2D$ using Fubini's theorem:

$$I^2=\int_0^\infty e^{-x^2}dx\int_0^\infty e^{-y^2}dy=\int_0^\infty\int_0^\infty e^{-(x^2+y^2)}dx\,dy.$$

Next, you convert to polar coordinates, using the element of area $dx\,dy=r\,dr\,d\theta$, and

$$\int_0^\infty\int_0^\infty e^{-(x^2+y^2)}dx\,dy=\int_0^{\pi/2}\int_0^\infty re^{-r^2}\,dr\,d\theta=\int_0^{\pi/2}d\theta\int_0^\infty re^{-r^2}\,dr=\frac\pi4.$$

This shows you how $\pi$ can appear.

Answer 2

This is equivalent to $\Gamma(\tfrac12)=\int_0^\infty\frac{1}{\sqrt{x}}e^{-x}dx=\sqrt{\pi}$. With $y=\sqrt{x}$, this is equivalent to normalizing a Gaussian integral, usually done this way. Alternatively, you may work with the Beta function:$$\Gamma^2(\tfrac12)=\operatorname{B}(\tfrac12,\,\tfrac12)=2\int_0^{\pi/2}d\theta=\pi.$$

Answer 3

Here is a possible solution: $$ \Gamma ^2 \left( {\frac{3}{2}} \right) = \frac{1}{4}\Gamma ^2 \left( {\frac{1}{2}} \right) = \frac{1}{4}\left( {\int_0^{ + \infty } {e^{ - t} \frac{1}{{\sqrt t }}dt} } \right)^2 \\ = \left( {\int_0^{ + \infty } {e^{ - x^2 } dx} } \right)^2 = \int_0^{ + \infty } {\int_0^{ + \infty } {e^{ - x^2 - y^2 } dy} dx} \\ = \int_0^{ + \infty } {\int_0^{ + \infty } {xe^{ - x^2 (1 + s^2 )} ds} dx} = \int_0^{ + \infty } {\int_0^{ + \infty } {xe^{ - x^2 (1 + s^2 )} dx} ds} \\ = \int_0^{ + \infty } {\left[ {\frac{1}{{ - 2(1 + s^2 )}}e^{ - x^2 (1 + s^2 )} } \right]_{x = 0}^{x = + \infty } ds} \\ = \frac{1}{2}\int_0^{ + \infty } {\frac{{ds}}{{1 + s^2 }}} = \frac{1}{2}\left[ {\arctan s} \right]_{s=0}^{s= + \infty } = \frac{\pi }{4} . $$

Answer 4

Another possible solution to this may be via the normal probability distribution.

$$p(x) = \frac{e^{-(x-\mu)^2/\left(2\sigma^2\right)}}{\sqrt{2\pi\sigma^2}}$$

Take the case where the mean ($\mu$) is $0$ and the standard deviation($\sigma$) is $\frac{1}{\sqrt2}$

So $p(x)$ here becomes: $p(x) = \dfrac{e^{-x^2}}{\sqrt\pi}$

The area under probability dist. curve is $1$

$$\int_{-\infty}^{\infty} p(x)\,dx = 1$$ So

$$\int_{0}^{\infty} p(x)\,dx = \frac{1}{2}$$

$$\int_{0}^{\infty} \frac{e^{-x^2}}{\sqrt\pi}\,dx = \frac{1}{2} ........(1)$$

$\Gamma\left(\frac{1}{2}\right) = \int_{0}^{\infty} x^{-\frac{1}{2}}e^{-x}\,dx$

Put $u = \sqrt{x}$

$$\Gamma\left(\frac{1}{2}\right) = \int_{0}^{\infty}2e^{-u^2}\,du = 2\left(\frac{\sqrt{π}}{2}\right) = \sqrt{\pi}$$

So from here we can apply the Gamma function's property,

$\Gamma(z+1) = z \Gamma(z)$

$$\Gamma(1.5) = \frac{1}{2}\Gamma(0.5) = \frac{\sqrt{\pi}}{2}$$