1. Gaussian integral

$\displaystyle \int_{-\infty}^{\infty}{e}^{{-\alpha}{b-{b}_{0}}^{2}}+{e}^{ibx}db$
Where i is complex.

I need to integrate with respect to b. If the complex part wasn't there i'd know how to go about it, but with it being there i am unsure,
any help is much appreciated!

2. Originally Posted by imagemania
$\displaystyle \int_{-\infty}^{\infty}{e}^{{-\alpha}{b-{b}_{0}}^{2}}+{e}^{ibx}db$
Where i is complex.

I need to integrate with respect to b. If the complex part wasn't there i'd know how to go about it, but with it being there i am unsure,
any help is much appreciated!
Is the integral $\displaystyle \int_{-\infty}^{+\infty}{e}^{{-\alpha}{b-b}^{2}}\times {e}^{ibx}db$ ?
We have $\displaystyle -\alpha b-b^2 = -(b-\frac{\alpha}2)^2+\frac{\alpha^2}4$ hence put $\displaystyle t :=b-\frac{\alpha}2$. Let $\displaystyle f(x) = \int_{-\infty}^{+\infty}e^{-t^2}e^{itx}dt$. Try to find a differential equation satisfied by $\displaystyle f$.

3. ugh my bad my latex skills need polishing,
$\displaystyle \int_{-\infty}^{\infty}{e}^{{-\alpha}({b-{b}_{0}})^{2}}{e}^{ibx}db$
I also missed a braket off...

4. $\displaystyle e^{-\alpha(b-b_0)^2}e^{ibx}=e^{-\alpha\left((b-b_0)^2-\frac{ibx}{\alpha} \right)}=e^{-\alpha\left( \left(b-b_0-\frac{ix}{2\alpha}\right)^2-\left(b_0+\frac{ix}{2\alpha}\right)^2+b_0^2\right) }=e^{-\alpha\left( -\left(b_0+\frac{ix}{2\alpha}\right)^2+b_0^2\right) }e^{-\alpha\left(b-b_0-\frac{ix}{2\alpha}\right)^2}$

Let $\displaystyle t=\sqrt{\alpha}\left(b-b_0-\frac{ix}{2\alpha}\right)\implies dt=\sqrt{\alpha}db$

Now we're left with a standard Gaussian integral:
$\displaystyle \frac{1}{\sqrt{\alpha}}e^{-\alpha\left( -\left(b_0+\frac{ix}{2\alpha}\right)^2+b_0^2\right) }\int_{-\infty}^{+\infty}e^{-t^2}dt$