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.
 A: We write (assuming $\alpha>0$)
 \begin{align}
\int_{-\infty}^{\infty}xe^{-\alpha x^2+\beta x}&=\frac 1{-2\alpha}\int_{-\infty}^{\infty}-2\alpha xe^{-\alpha x^2+\beta x}+\frac{\beta}{-2\alpha}\int_{-\infty}^{\infty}e^{-\alpha x^2+\beta x}-\frac{\beta}{-2\alpha}\int_{-\infty}^{\infty}e^{-\alpha x^2+\beta x}\\
&=\frac 1{-2\alpha}\int_{-\infty}^{\infty}(-2\alpha+\beta) xe^{-\alpha x^2+\beta x}+\frac{\beta}{2\alpha}\sqrt{\frac{\pi}{\alpha}}e^{\beta^2/(4\alpha)}\\
&=\frac{1}{-2\alpha}\left[e^{-\alpha x^2+\beta x}\right]_{-\infty}^{\infty}+\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)}
\end{align}
A: You should complete $ax^2-\beta x$ to quadratic form $a (x-\beta/a)^2-\beta^2/a^2$. After this apply the change of variable $u=ax-\beta$ and integrate $\int u \exp(-u^2/2)du=\exp(-u^2/2)$.
A: Hint:  $xe^{-\alpha x^2 + \beta x} = \frac{d}{d\beta}e^{-\alpha x^2 + \beta x}$
A: Hint: Write the integrand as $(e^{\beta x})(-\frac1{2\alpha}e^{-\alpha x^2})'$ and use integration by parts. You get
$$\left[-\tfrac1{2\alpha}e^{-\alpha x^2+\beta x}\right]_{-\infty}^{\infty} + \tfrac{\beta}{2\alpha}\int_{-\infty}^{\infty}e^{-\alpha x^2 + \beta x}\; dx$$
The first term vanishes and the second term is $\tfrac{\beta}{2\alpha}$ times your familiar integral.
A: Hint: try with integration by parts:
$$\int_{-\infty}^{+\infty}\ x e^{F(x)}\ \text{d}x = x\cdot \int_{-\infty}^{+\infty}\ e^{F(x)}\ \text{d}x - \int_{-\infty}^{+\infty} e^{F(x)}\ \text{d}x$$
Where $F(x) = -ax^2 + \beta x$
Or use the Feynman trick: differentiation under the integral sign. Since you know what the result of $\int_{-\infty}^{+\infty}\ e^{-ax^2 + \beta x}\ \text{d}x$ just notice that
$$x e^{-ax^2 + \beta x} = \frac{\partial}{\partial \beta} e^{-ax^2 + \beta x}$$
thence, just differentiate the result of the known integral with respect upon $\beta$:
$$\int_{-\infty}^{+\infty}\ x e^{-ax^2 + \beta x}\ \text{d}x = \int_{-\infty}^{+\infty}\  \text{d}x \frac{\partial}{\partial \beta} e^{-ax^2 + \beta x}\ = \frac{\beta}{2a}\sqrt{\frac{\pi}{a}}\ e^{\beta^2/4a}$$
A: $$\int_{-\infty}^{\infty}xe^{-\alpha x^2+\beta x}\ dx=-\frac{1}{2\alpha}\int_{-\infty}^{\infty}(-2\alpha x+\beta-\beta)e^{-\alpha x^2+\beta x}\ dx$$
$$=-\frac{1}{2\alpha}\int_{-\infty}^{\infty}(-2\alpha x+\beta)e^{-\alpha x^2+\beta x}\ dx+\frac{\beta}{2\alpha}\int_{-\infty}^{\infty}e^{-\alpha x^2+\beta x}\ dx$$
$$=-\frac{1}{2\alpha}\int_{-\infty}^{\infty}(-2\alpha x+\beta)e^{-\alpha x^2+\beta x}\ dx+\frac{\beta}{2\alpha}\sqrt{\frac{\pi}{\alpha}}e^{\beta^2/4\alpha}$$
since first integral diminishes, 
$$=0+\frac{\beta}{2\alpha}\sqrt{\frac{\pi}{\alpha}}e^{\beta^2/4\alpha}$$
$$=\frac{\beta}{2\alpha}\sqrt{\frac{\pi}{\alpha}}e^{\beta^2/4\alpha}$$
