Contour integral

[Jim Heckman]

Greetings all,

This is probably an easy question, but it's been decades since I've
done much complex analysis.

Pursuant to recent questions here and in other newsgroups, I've
been reviewing my understanding of basic quantum mechanics, and in
one of my texts I've come across this integral over the reals:

I(a,b) = Int_{-inf}^{+inf} exp[-a^2*(x+b)^2] dx

where a and b are complex and Re(a^2) > 0 (so the integral
converges).

The text claims without proof that:

The method of residues enables us to show that this integral
does not depend on b:

I(a,b) = I(a,0)

and that, when the condition -pi/4 < Arg(a) < +pi/4 is
fulfilled (which is always possible if Re(a^2) > 0), I(a,0)
is given by:

I(a,0) = I(1,0)/a = sqrt(pi)/a

Of course I know how to show that I(1,0) = sqrt(pi), but I'm having
trouble finding appropriate contours to confirm the other claims.

I have found a demonstration, in a textbook on engineering
mathematics, of how to integrate exp(i*z^2) around the pie-wedge
0 <= Arg(z) <= pi/4 to show that Int_0^{+inf} cos(x^2) dx =
Int_0^{+inf} sin(x^2) dx = sqrt(pi/8). I thought that that might be
useful, but alas so far I don't see how.

Any help would be much appreciated!

--
Jim Heckman

[Robin Chapman]

> I(a,b) = Int_{-inf}^{+inf} exp[-a^2*(x+b)^2] dx
>
> where a and b are complex and Re(a^2)> 0 (so the integral
> converges).
>
> The text claims without proof that:
>
> The method of residues enables us to show that this integral
> does not depend on b:
>
> I(a,b) = I(a,0)

Integrate exp(-a^2 z^2) over rectangular contour: vertices +-R,
+-R + i Im(b).

> and that, when the condition -pi/4< Arg(a)< +pi/4 is
> fulfilled (which is always possible if Re(a^2)> 0), I(a,0)
> is given by:
>
> I(a,0) = I(1,0)/a = sqrt(pi)/a

Integrate exp(-z^2) over "slice-of-pie" contour: centre 0
vertices at aR and |a|R.

[Jim Heckman]

On 25-May-2012, Robin Chapman <R.J.C...@ex.ac.uk>
wrote in message <jpnjar$lrp$1...@dont-email.me>:

Thanks, Robin. And may I say it's nice to see you here again in
<sci.math>. I've noticed you've poked your nose in once or twice
recently -- hope you stick around awhile.

--
Jim Heckman

[David C. Ullrich]

On Sat, 26 May 2012 04:53:18 GMT, "Jim Heckman"
<rot13(reply-to)@none.invalid> wrote:

>[...]

>
>Thanks, Robin. And may I say it's nice to see you here again in
><sci.math>. I've noticed you've poked your nose in once or twice
>recently -- hope you stick around awhile.

Second.