You're Just So Integreat

Aim: To prove that $\displaystyle\Gamma(n)=\int_{0}^{1} \left(\ln \frac{1}{y}\right)^{n-1} \ dy$ .

Proof:

We know by definition , $\Gamma(n)=\int_{0}^{\infty} x^{n-1}e^{-x} \ dx$ .

Now substitute $x=\ln \frac{1}{y} \Rightarrow y=e^{-x} \Rightarrow dy=-e^{-x} \ dx$ . Also note that as $x=0$ , $y=1$ and as $x\rightarrow \infty$ , $y=0$ and we change the limits accordingly. Hence we have

$\Gamma(n)=-\int_{1}^{0} \left(\ln \frac{1}{y}\right)^{n-1}=\int_{0}^{1} \left(\ln \frac{1}{y}\right)^{n-1} \ dy$

which was to be proved :)

Now we use $\displaystyle\Gamma(n)=\int_{0}^{1} \left(\ln \frac{1}{y}\right)^{n-1} \ dy$ .In this problem , $n=\frac{1}{2}$ , hence ,

$\int_{0}^{1} \dfrac{1}{\sqrt{\ln \frac{1}{y}}} \ dy =-\int_{1}^{0} \dfrac{1}{\sqrt{\ln \frac{1}{y}}} \ dy= -\Gamma\left(\frac{1}{2}\right)=-\sqrt{\pi}$

Since we are asked to find the square , we have the answer as $\pi\approx \boxed{3.1415}$ .

lets just take the integral and rewrite it

$\int_{1}^{0} (-ln(y))^{-1/2}$

Using u sub where $u=-ln(y)$ ; $du=-dy/y$ ; $y=e^{-u}$ ; $dy=-ydu$ ; and $dy=-e^{-u}du$

We can rewrite the integral as

$-\int_{0}^{\infty} \frac{1}{\sqrt{u}} e^{-u}du$

Make a new substitution $u=v^{2}$ ; $du=2vdv$

and we get

$-\int_{0}^{\infty} \frac{1}{v} e^{-v^{2}}2vdv$

simplify

$-2\int_{0}^{\infty} e^{-v^{2}}dv$

This is a known integral, the indefinite integral of $e^{-x^2}$ is defined as the $erf(x)$ and the integral from 0 to infinity is known to be $\sqrt{\pi}/2$

so plug that in

$-2 \times \sqrt{\pi}/2$

$-\sqrt{\pi}$

square this, and you get $\pi$ for the answer.