Integrate $\int_{-\infty}^\infty xe^{-\alpha x^2+\beta x}dx$ I am familiar with the gauusian integral
$$\int_{-\infty}^\infty e^{-\alpha x^2+\beta x}dx=\sqrt{\frac{\pi}{\alpha}}e^{\beta^2/(4\alpha)}$$
Could anyone help me to find out the value of the following?
$$\int_{-\infty}^\infty xe^{-\alpha x^2+\beta x}dx$$
Integration by parts turns out to contain error function which I am not familiar with.

$$\int_{-\infty}^{\infty}xe^{-\alpha x^2+\beta x}\ dx=-\frac{1}{2\alpha}\int_{-\infty}^{\infty}(-2\alpha x+\beta-\beta)e^{-\alpha x^2+\beta x}\ dx$$
$$=-\frac{1}{2\alpha}\int_{-\infty}^{\infty}(-2\alpha x+\beta)e^{-\alpha x^2+\beta x}\ dx+\frac{\beta}{2\alpha}\int_{-\infty}^{\infty}e^{-\alpha x^2+\beta x}\ dx$$
$$=-\frac{1}{2\alpha}\int_{-\infty}^{\infty}(-2\alpha x+\beta)e^{-\alpha x^2+\beta x}\ dx+\frac{\beta}{2\alpha}\sqrt{\frac{\pi}{\alpha}}e^{\beta^2/4\alpha}$$
since first integral diminishes, 
$$=0+\frac{\beta}{2\alpha}\sqrt{\frac{\pi}{\alpha}}e^{\beta^2/4\alpha}$$
$$=\frac{\beta}{2\alpha}\sqrt{\frac{\pi}{\alpha}}e^{\beta^2/4\alpha}$$
