Finding integral of $2 \lambda\int_0^{\infty} x^{2n}xe^{-\lambda x^2} \ dx$ Finding integral of $$2 \lambda\int_0^{\infty} x^{2n}xe^{-\lambda x^2} \ dx$$
using integral by parts I get that Finding integral of $$2 \lambda\int_0^{\infty} x^{2n}xe^{-\lambda x^2} \ dx = \dfrac{n}{\lambda} \cdot \dfrac{n-1}{\lambda} \int_0^\infty x^{2n-4} \cdot x e^{-\lambda x^2} \ dx$$
so I assume that $$2 \lambda\int_0^{\infty} x^{2n}xe^{-\lambda x^2} \ dx = \dfrac{n!}{\lambda^n}$$. How can I formally prove this?

Hint: Evaluate $I(\lambda)=\displaystyle\int_0^\infty xe^{-\lambda x^2}~dx$, and then differentiate it n times with regard to $\lambda$.
