I assumed that
$$\Gamma\left(k+\frac{1}{2}\right)=2\int^\infty_0 e^{-x^2}x^{2k}\,dx=\frac{\sqrt{\pi}(2k)!}{4^k k!} \,,\space k>-\frac{1}{2}$$
and that
$$\Gamma\left(k+\frac{3}{2}\right)=2\int^\infty_0 e^{-x^2}x^{2(k+1)}\,dx$$
and my goal is to solve the integral and get a function in terms of $k$ for $\Gamma\left(k+\frac{3}{2}\right)$
I use partial integration and differentiate $x^2$ and integrate the rest:
$$=\left[x^2.2\int^\infty_0 e^{-x^2}x^{2k}\,dx \right]^\infty_0 – \int^\infty_02x\left(2\int^\infty_0 e^{-x^2}x^{2k}\,dx\right)\,dx$$
and then I substitute the above function in terms of k and get:
$$=\left[x^2\frac{\sqrt{\pi}(2k)!}{4^k k!}\right]^\infty_0 – \int^\infty_02x\frac{\sqrt{\pi}(2k)!}{4^k k!}\,dx$$
$$=\left[x^2\frac{\sqrt{\pi}(2k)!}{4^k k!}\right]^\infty_0 – \left[x^2\frac{\sqrt{\pi}(2k)!}{4^k k!}\right]^\infty_0 =0$$
I know for sure that the final answer is wrong. I think my problem has to do with the substitution of the definite integral in the penultimate step. How can I make the math work out?
EDIT: Sorry for not mentioning previously but this is part of a proof by induction. The first statement is only assumed to be true.
Best Answer
Let us assume that
$$\Gamma\left(k+\frac{1}{2}\right)=2\int^\infty_0 e^{-x^2}x^{2k}\,dx=\frac{\sqrt{\pi}(2k)!}{4^k k!}$$
1- for $k=0$ we have
$$\Gamma\left(\frac{1}{2}\right)=2\int^\infty_0 e^{-x^2}\,dx=\sqrt{\pi}$$
which holds true since
$$\int^\infty_{-\infty} e^{-x^2}\,dx=\sqrt{\pi}$$
2- We need to prove the case $P(k)\to P(k+1)$
$$\Gamma\left(k+1+\frac{1}{2}\right) = \left( k+\frac{1}{2}\right)\Gamma\left( k+\frac{1}{2}\right)$$
From the inductive step we have
$$\left( k+\frac{1}{2}\right)\Gamma\left( k+\frac{1}{2}\right) =\left( k+\frac{1}{2}\right) \frac{\sqrt{\pi}(2k)!}{4^k k!} = \sqrt{\pi}\frac{(2k+1)(2k)!}{2\times4^kk!} $$
Mutliply and divide by $(2k+2)$
$$ \frac{\sqrt{\pi}}{4}\frac{(2k+2)(2k+1)(2k)!}{ 4^k (k+1)k!} =\frac{\sqrt{\pi}(2(k+1))!}{4^{(k+1)}(k+1)!}\blacksquare$$