integration - Integrating the gamma function

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.

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 $(2k+2)$

$$\frac{\sqrt{\pi}}{4}\frac{(2k+2)(2k+1)(2k)!}{ 4^k (k+1)k!} =\frac{\sqrt{\pi}(2(k+1))!}{4^{(k+1)}(k+1)!}\blacksquare$$