Homework Help: Integral of gaussian and more

1. Dec 4, 2012

aaaa202

I have an integral of the form (from -inf to inf):

∫exp((a+ib)x2+icx)dx

How do I solve that?

I have tried setting y = √(a+ib)(x+ic/√(a+ib))

And you then get an integral of exp(-y2) times a constant. But I don't even know if this is legal since part of the exponentials are complex. Is it possible to obtain the result with my method?

2. Dec 4, 2012

Fightfish

Have you tried completing the square on the exponent?

3. Dec 4, 2012

aaaa202

what do you mean? Isn't that what I do by the introduction of my y?

4. Dec 4, 2012

Fightfish

Oh okay that didnt look clear to me as I expected you to leave it in the square form directly. I can't prove it off-hand, but the gaussian integration formula holds for complex a,b and c (where the exponent is in the form -ax^2 + bx + c) as long as the real part of a is positive

5. Dec 4, 2012

Ray Vickson

What you want is not too hard to prove if you have a bit of complex function theory. If you take
$$I = \int_{-\infty}^{\infty} e^{-(a + ib)(x + iy)^2} \, dx, \: a > 0$$
you can write it in the form of a complex contour integral:
$$I = \int_{\Gamma} e^{- (a + ib) z^2} \, dz,$$
where $\Gamma$ is the line from -∞ + iy to +∞ + iy. This is the limit as M → ∞ of I_M, the integral from -M + iy to +M + iy. Since the integrand is an analytic function of z, get the same result I_M if we deform the contour into a path from -M+iy to -M, then from -M to +M, then from M to M+iy. The part from -M to +M goes along the real axis, and the parts from -M+iy to -M and from M to M+iy go to zero as M → ∞, so we are left with
$$I = \int_{-\infty}^{\infty} e^{-(a + ib) x^2} \, dx.$$
Now the usual method of squaring I and converting the double integral to polar coordinates applies essentially unchanged.