Determining a constant in a probability density function I know there is an easier way to solve this, but can someone point me to my error. I am losing a constant of 4 somewhere.
There is p.d.f $Ax^2 e^{-\lambda x^2}$ for $x>0$ and $0$ for $x\leq 0$.
I have to show: $A=4\sqrt{\frac{\lambda^3}{\pi}}$
This condition has to be met, if it is a p.d.f.: $\int_0^\infty Ax^2 e^{-\lambda x^2} dx = 1$, or $A = \frac{1}{\int_0^\infty x^2 e^{-\lambda x^2}}$.
First integrating by parts I obtain this result:
$\int_0^\infty Ax^2 e^{-\lambda x^2} dx \equiv \frac{1}{\lambda} \int_0^\infty e^{-\lambda x^2} dx$
Now let $I = \frac{1}{\lambda} \int_0^\infty e^{-\lambda x^2} dx $, then $I^2 = \frac{1}{\lambda^2} \int_0^\infty e^{-\lambda x^2}\int_0^\infty e^{-\lambda y^2} dx$
Using the substitution $x=r\cos \theta$ and $y = r \sin \theta$
I obtain:
$-\frac{1}{2\lambda ^3} \int_0^{2\pi}\left[ e^{-\lambda r^2}\right]_0^\infty d\theta$, which evaluates to $\frac{\pi}{\lambda^3}$ or $A=\sqrt{\frac{\lambda^3}{\pi}}$...
Not what I was meant to show.

When you made your substitution you were wrong, you better remarked that : 
$$\int_0^{\infty} \int_0^{\infty} e^{-\lambda (x^2+y^2)}dxdy = \frac{1}{4} \int_{\mathbb{R}^{2}}e^{-\lambda (x^2+y^2)} dxdy$$ the integral : 
$$\int_{\mathbb{R}^{2}}e^{-\lambda (x^2+y^2)} dxdy$$ is precisely what you calculated with your substitution (draw the area $\left \{ \theta \in [0,2 \pi], r \ge 0\right \}$)
Still here for more details. 
