Trying to solve infinite integral involving error function

Question

I am trying to evaluate

$$I = \int^{\infty}_0 (y^{\frac{1}{2}}e^{-\frac{1}{2}y})dy$$

using the fact that

$$\int^{\infty}_0 (e^{-x^2})dx = \frac{\sqrt{\pi}}{2}$$

Not sure how to do this. I tried transforming with $y=2x^2$ substitution, but found this suggests

$$I = \sqrt{2} \int^{\infty}_0 (xe^{-x^2})dx$$

By this point Wolfram tells me I've made a mistake as evaluating the above definite integral yields

$$I=\frac{\sqrt{2}}{2}$$

which does not involve $\sqrt{\pi}$ and is surely incorrect. Any help?

Answer 1

When I substitute $y=2x^2$ I get $$I=4\sqrt2\int_0^\infty x^2e^{-x^2}\,dx.$$ By integration by parts, this can be reduced to the usual "probability integral" $$\int_0^\infty e^{-x^2}\,dx.$$

Answer 2

$$I=\int_{0}^{\infty} y^{1/2} e^{-y/2} dy$$ Let $y-2x^2, dy=4x dx$, then $$I=4 \sqrt{2} \int_{0}^{\infty} x^2 e^{-x^2} dx= 4 \sqrt{2} \int_{0}^{\infty} x. x e^{-x^2} dx$$ Integration by parts leads to $$I=4\sqrt{2}~ [-\frac{x}{2} e^{-x^2}|_{0}^{\infty}+\frac{1}{2} \int_{0}^{\infty}e^{-x^2} dx~]$$ $$I=2\sqrt{2}\sqrt{\frac{\pi}{2}}=\sqrt{2\pi}$$