Question 1,
Can this Gamma expression be further simplified to its lowest form? $$\frac{a^2\Gamma\left(1+\frac{1}{a}\right)\Gamma\left(a-\frac{1}{a}\right)}{\Gamma^2\left(\frac{1}{a}\right)\Gamma\left(-\frac{1}{a}\right)}$$
An example would be, $\frac{4}{6}=\frac{2}{3}$
Question 2,
Show that,
$$\frac{a^2\Gamma\left(1+\frac{1}{a}\right)\Gamma\left(a-\frac{1}{a}\right)}{\Gamma^2\left(\frac{1}{a}\right)\Gamma\left(-\frac{1}{a}\right)}=\prod_{n=1}^{a-1}\left(\frac{an-1}{an}\right)^{(-1)^{n-1}}$$
ignore what I did below of how I got the answer, I would like to see your our way of showing this formula
Question 3,
As a tends to infinity does this product converges?
We found out that this product converges to 0.5
$$\lim_{a \to \infty}\frac{a^2\Gamma\left(1+\frac{1}{a}\right)\Gamma\left(a-\frac{1}{a}\right)}{\Gamma^2\left(\frac{1}{a}\right)\Gamma\left(-\frac{1}{a}\right)}=\prod_{n=1}^{a-1}\left(\frac{an-1}{an}\right)^{(-1)^{n-1}}=\frac{1}{2}$$
I don't know how to use limits to verify this, can anyone help me?
(1) $$\int_0^\infty\frac{1}{(1+x^a)^a}dx=\frac{\Gamma\left(1+\frac{1}{a}\right)\Gamma\left(a-\frac{1}{a}\right)}{\Gamma(a)} $$ (2) $$\int_0^\infty\frac{1}{(1+x^a)^a}dx=\frac{\pi}{a}\cdot\frac{1}{\sin\left(\frac{\pi}{a}\right)}\prod_{n=1}^{a-1}\left(\frac{an-1}{an} \right)^{(-1)^{n-1}}$$
Reflection formula
(3) $$\frac{\Gamma\left(\frac{1}{a}\right)\Gamma\left(-\frac{1}{a}\right)}{a}=\frac{\pi}{\sin\left(\frac{\pi}{a}\right)}$$
Substitute (3) into (2) and it is equal to (1), simplify and we have:
$$\frac{a^2\Gamma\left(1+\frac{1}{a}\right)\Gamma\left(a-\frac{1}{a}\right)}{\Gamma^2\left(\frac{1}{a}\right)\Gamma\left(-\frac{1}{a}\right)}=\prod_{n=1}^{a-1}\left(\frac{an-1}{an}\right)^{(-1)^{n-1}}$$
I need help on how to further simplify the left hand side of the product, can anyone help?
Once simplify the LHS,if possible.
Can anybody prove this product in a different way better and more professional? Just imagine you never saw how I arrived at this product.
Let a tends to infinity does this product converges to a closed form?