On trouve

I_n = \left|\frac{x^{n+1}}{n+1} e^{-ax^2}\right|_{0}^{\infty} + \frac{2a}{n+1}\int_{0}^{\infty}x^{n+2} e^{-ax^2} \ dx

Soit

I_n = \frac{2a}{n+1} I_{n+2}