- #1
homer
- 46
- 0
Homework Statement
Let [itex]a,b[/itex] be real with [itex]a > 0[/itex]. Compute the integral
[tex]
I = \int_{-\infty}^{\infty} e^{-ax^2 + ibx}\,dx.
[/tex]
Homework Equations
Equation (1):
[tex]\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}[/tex]
Equation (2):
[tex]-ax^2 + ibx = -a\Big(x - \frac{ib}{2a}\Big)^2 - \frac{b^2}{4a}[/tex]
The Attempt at a Solution
Completing the square in [itex]-ax^2 + ibx[/itex] gives me Equation (2), so that my integral is now
[tex]
I = e^{-b^2/4a}\int_{-\infty}^{\infty} e^{-a(x-ib/2a)^2}\,dx.
[/tex]
Making the substitution [itex]u = \sqrt{a}(x-ib/2a)[/itex] I get [itex]du = \sqrt{a}\,dx[/itex] so tht my integral becomes
[tex]
I = \frac{1}{\sqrt{a}}\,e^{-b^2/4a}\int_{-\infty - ib/2\sqrt{a}}^{\infty - ib/2\sqrt{a}} e^{-u^2}\,du.
[/tex]
But this doesn't seem right.