integrating this infinite gaussian integral

Question

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

Answer 1

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 $$ \frac{1}{\sqrt{2\lambda}}\int_{-\infty}^{\infty}\left(\frac{z}{\sqrt{2\lambda}}+a\right)\mathrm{e}^{-\frac{z^2}{2}}dz $$

Answer 2

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?

Answer 3

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*}

Answer 4

$$ \begin{align} \int_{-\infty}^{+\infty}x e^{-\lambda(x-a)^2}\,\mathrm{d}x &=\int_{-\infty}^{+\infty}(x+a) e^{-\lambda x^2}\,\mathrm{d}x\tag{1}\\ &=\frac{a}{\sqrt{\lambda}}\int_{-\infty}^{+\infty}e^{-x^2}\,\mathrm{d}x\tag{2}\\ &=a\sqrt{\frac\pi\lambda}\tag{3} \end{align} $$ Explanation:
$(1)$: substitute $x\mapsto x+a$
$(2)$: $\int_{-\infty}^{+\infty}xe^{-\lambda x^2}\,\mathrm{d}x=0\because$ odd integrand, then substitute $x\mapsto\frac x{\sqrt\lambda}$
$(3)$: $\int_{-\infty}^{+\infty}e^{-x^2}\,\mathrm{d}x=\sqrt\pi$