Complex Gaussian Integral - Cauchy Integral Theorem

[VVS]

Homework Statement

I have to prove that I(a,b)=\int_{-\infty}^{+\infty} exp(-ax^2+bx)dx=\sqrt{\frac{\pi}{a}}exp(b^2/4a) where a,b\in\mathbb{C}.
I have already shown that I(a,0)=\sqrt{\frac{\pi}{a}}.
Now I am supposed to find a relation between I(a,0) and \int_{-\infty}^{+\infty} exp(-a(x-c)^2)dx where c\in\mathbb{C} using the Cauchy integral theorem and prove using this the result above.

Homework Equations

The Cauchy integral theorem states that \oint_\gamma f(z) dz=2\pi i Res(f(z)).

The Attempt at a Solution

This is what I got, but I am pretty sure it doesn't lead me anywhere.
Now I am not sure whether \int_{-\infty}^{+\infty} exp(-a(x-c)^2)dx is analytic in c. But I think what I can do is integrate over a closed contour \gamma and use Cauchy's integral theorem, change the order of integration and thus equate it to the Residue.
\int_{-\infty}^{+\infty} \oint_\gamma exp(-a(x-c)^2)dcdx=2\pi i Res(f(z))
Then I can expand to yield:
\int_{-\infty}^{+\infty} exp(-ax^2) \oint_\gamma exp(-ac^2+2axc)dcdx=2\pi i Res(f(z))
I can sort of see the relation between the integrals now but I am kind of stuck.

 

[Svein]

I think you are complicating this a bit too much. After all, -ax^{2}+bx=-a(x^{2}-\frac{b}{a}x)=-a(x-\frac{b}{2a})^{2}+\frac{b^{2}}{4a}.

 

[Ray Vickson]

Homework Statement

I have to prove that I(a,b)=\int_{-\infty}^{+\infty} exp(-ax^2+bx)dx=\sqrt{\frac{\pi}{a}}exp(b^2/4a) where a,b\in\mathbb{C}.
I have already shown that I(a,0)=\sqrt{\frac{\pi}{a}}.
Now I am supposed to find a relation between I(a,0) and \int_{-\infty}^{+\infty} exp(-a(x-c)^2)dx where c\in\mathbb{C} using the Cauchy integral theorem and prove using this the result above.

Homework Equations

The Cauchy integral theorem states that \oint_\gamma f(z) dz=2\pi i Res(f(z)).

The Attempt at a Solution

This is what I got, but I am pretty sure it doesn't lead me anywhere.
Now I am not sure whether \int_{-\infty}^{+\infty} exp(-a(x-c)^2)dx is analytic in c. But I think what I can do is integrate over a closed contour \gamma and use Cauchy's integral theorem, change the order of integration and thus equate it to the Residue.
\int_{-\infty}^{+\infty} \oint_\gamma exp(-a(x-c)^2)dcdx=2\pi i Res(f(z))
Then I can expand to yield:
\int_{-\infty}^{+\infty} exp(-ax^2) \oint_\gamma exp(-ac^2+2axc)dcdx=2\pi i Res(f(z))
I can sort of see the relation between the integrals now but I am kind of stuck.

Are you being forced to use Cauchy's integral theorem here? I really don't see how it helps or why anyone would think it is needed. To me, it just gets in the way. A simple change of variables does the trick.