Some gamma function properties

Question

So my Professor used two identities of the Gamma function but I found no proofs in books or the internet.

The first part is:

(i) $ n \in \mathbb {N} $ holds for all $$\int_ {0}^{\infty} e^{- t ^ {2}} t ^ {2 n} d t = \sqrt {\pi} \frac {(2 n-1)! !} {2 ^ {n + 1}} $$ where $ (2 n-1)! ! = 1 \times 3 \times 5 \times \ldots \times (2 n-1) $.

(ii) For all $ a> -1 $ holds $$ \int_ {0} ^ {1} \left (\log \frac {1} {t} \right) ^ {a} d t = \Gamma (a + 1). $$

Maybe one can prove (i) with the fact that $ \Gamma \left (\frac {1} {2} \right) = \sqrt {\pi} $.

If anyone knew passages in books with a proof for them or is able to prove it by himself or herself, I would be very happy to see such results or even hints.

Answer 1

(i) can be found here, the method is a classic example of "differentiating under the integral sign" (very common in physics for example). There is no need to use any gamma function.

(ii) Hint: just do a substitution with $z=\log(1/t)$.

Answer 2

For (ii), using the @Simon clue, $t = e^{-z}\rightarrow dt = -e^{-z} dz$. As $t\rightarrow 0$, $z\rightarrow \infty$, and as $t\rightarrow 1$, $z\rightarrow 0$. Hence, $$ \int_ {0} ^ {1} \left (\log \frac {1} {t} \right) ^ {a} d t = \int^{\infty}_{0} z^a e^{-z}dz = \Gamma (a + 1). $$ You should just notice changing the limit of integration which cancels the minus sign that will arise.

Answer 3

For the first part, there are multiple proofs if all you want is the Legendre Duplication Formula for the $n\in\mathbb{N}$, then here's an outline.

Let $I=\displaystyle\int_ {0}^{\infty} e^{- t ^ {2}} t ^ {2 n} d t $

$I=\displaystyle\frac{1}{2}\int_{0}^{\infty}x^{n-\frac{1}{2}}e^{-x}\mathrm{d}x$, $t^2=x$

Now using the functional form of the Gamma function we have,

$I=\displaystyle\frac{\Gamma(n+\frac{1}{2})}{2}$

All that is left to do is as sketched below, we know that $\displaystyle\int_{0}^{\infty}e^{-x^2}\mathrm{d}x=\frac{\sqrt{\pi}}{2}$,

$\displaystyle\Gamma\left(n+\frac{1}{2}\right)=\left(n-\frac{1}{2}\right)\Gamma\left(n-\frac{1}{2}\right)=\left(n-\frac{1}{2}\right)\left(n-\frac{3}{2}\right)\Gamma\left(n-\frac{3}{2}\right)=\cdots=\sqrt {\pi} \frac {(2 n-1)! !} {2 ^ {n + 1}}$

Note that the last term in the product would be $\displaystyle\Gamma\left(\frac{1}{2}\right)=\frac{\sqrt{\pi}}{2}$.And as for whether this formula implies $\displaystyle\Gamma\left(\frac{1}{2}\right)=\frac{\sqrt{\pi}}{2}$, the answer is as you can see $\textbf{NO}$, you need to know that fact beforehand in order to derive this result, the proof of the result is quite readily available. Hope this helps, if not just check out the link below, though the derivation is easy, it will only be good for University proofs if you know the rigorous derivation of Differential Under the Integral Sign.