Evaluate $\int_{-\infty}^{\infty}x^2 e^{-\alpha x^2+\beta x}dx$

Question

As a consequence of this Q, I need some help evaluating the following integral: $$\int_{-\infty}^{\infty}x^2 e^{-\alpha x^2+\beta x}dx$$

Integration by parts wouldn't simplify things and I guess that a subtle manipulation on the integrand is needed.

Answer 1

Start with evaluating the integral

$$I=-\int_{R} e^{-\alpha x^2+\beta x} dx $$

Which is Gausian type then differentiate the answer with respect to alpha.

Note: you can use the new formula I derived a while ago which handles a more general integral than yours. See here.

Answer 2

$$-\alpha x^2+\beta x=-\alpha\left(x^2-\frac\beta\alpha x\right)=-\alpha\left(x-\frac\beta{2\alpha}\right)^2+\frac{\beta^2}{4\alpha}\implies$$

$$I(\alpha):=\int_{\Bbb R}e^{-\alpha\left(x-\frac\beta{2\alpha}\right)^2}e^{\frac{\beta^2}{4\alpha}}\;dx=e^{\frac{\beta^2}{4\alpha}}\sqrt{\frac\pi{\alpha}}\implies$$

$$\implies I'(\alpha)=\sqrt\pi\; e^{\beta^2/4\alpha}\left(-\frac{\beta^2}{4\alpha^{5/2}}-\frac1{2\alpha^{3/2}}\right)=-\sqrt\pi\;e^{b^2/4\alpha}\frac{\beta^2+2\alpha}{4\alpha^{5/2}}$$

But also

$$\frac d{d\alpha}\int_{\Bbb R}e^{-\alpha x^2+\beta x}dx=-\int_{\Bbb R}x^2e^{-\alpha x^2+\beta x}dx$$

because differentiation under the integral sign is allowed here since

$$f(x,\alpha):=e^{-\alpha x^2+\beta x}\;\;\text{and}\;\;f'_\alpha$$

are continuous functions everywhere in the $\;x\alpha-$plane