Integrali comodi

$\displaystyle \int_{-\infty}^{+\infty}{x^{2n}\exp{(-\lambda x^2)}dx}=\sqrt{\frac{\pi}{\lambda^{2n+1}}}\frac{(2n-1)!!}{2^n}$

$\displaystyle \int_{-\infty}^{+\infty}{x^{2n+1}\exp{(-\lambda x^2)}dx}=\frac{n!}{2^{\lambda{n+1}}}$

$\displaystyle \int_{0}^{+\infty}{x^{n}\exp{(-\lambda x)}dx}=\frac{n!}{\lambda^{n+1}}$



Carlo 2008-03-02