integrating this infinite gaussian integral How does one integrate 
$\int_{-\infty}^{+\infty}x e^{-\lambda ( x-a )^2 }dx $
where $\lambda$ is a positive constant.
My integral tables are not returning anything useable. The best it return is non-definite gaussian integrals. Useless!
Help please
 A: $$
\begin{align}
\int_{-\infty}^{+\infty}x e^{-\lambda(x-a)^2}\,\mathrm{d}x
&=\int_{-\infty}^{+\infty}(x+a) e^{-\lambda x^2}\,\mathrm{d}x\tag{1}\\
&=\frac{a}{\sqrt{\lambda}}\int_{-\infty}^{+\infty}e^{-x^2}\,\mathrm{d}x\tag{2}\\
&=a\sqrt{\frac\pi\lambda}\tag{3}
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto x+a$
$(2)$: $\int_{-\infty}^{+\infty}xe^{-\lambda x^2}\,\mathrm{d}x=0\because$ odd integrand, then substitute $x\mapsto\frac x{\sqrt\lambda}$
$(3)$: $\int_{-\infty}^{+\infty}e^{-x^2}\,\mathrm{d}x=\sqrt\pi$
A: If the integral follows the comments above here is a hint.
$$
\int_{-\infty}^{\infty} x\mathrm{e}^{-\lambda(x-a)^2}dx
$$
let $\lambda(x-a)^2 = \frac{z^2}{2}\to z=\sqrt{2\lambda}(x-a)$ 
we get
$$
\frac{1}{\sqrt{2\lambda}}\int_{-\infty}^{\infty}\left(\frac{z}{\sqrt{2\lambda}}+a\right)\mathrm{e}^{-\frac{z^2}{2}}dz
$$
A: If the integral is the one you wrote, then of course no table gives you any hint: that integral does not converge. Indeed:
$$\int_{-\infty}^{+\infty}xe^{-\lambda}(x-a)^2\ \text{d}x = e^{-\lambda}\int_{-\infty}^{+\infty}(x^3 + a^2 x - 2ax^2)\ \text{d}x$$
And you easily can integrate it, but it will diverge.
So I think you wrote wrong, and the real integral is
$$\int_{-\infty}^{+\infty}x e^{-\lambda(x-a)^2}\ \text{d}x$$
At this point you can use
$$\sqrt{\lambda}(x-a) = y ~~~~~~~ \text{d}y = \sqrt{\lambda}\ \text{d}x ~~~~~~~ x = \frac{y}{\sqrt{\lambda}}+a$$
Can you proceed?
A: Another approach
\begin{equation*}
\int_{-\infty }^{+\infty }dxx\exp [-\lambda (x-a)^{2}]=\int_{-\infty
}^{+\infty }dx(x+a)\exp [-\lambda x^{2}]=a\int_{-\infty }^{+\infty }dx\exp
[-\lambda x^{2}]
\end{equation*}
