Integral of cos(Ax)e^(-Bx²) from 0 to infinity

[quasar987]

Now I have to evaluate

$$\int_{-\infty}^{\infty} e^{-Bx^2} e^{-iAx} dx$$

Splitting it in two using Euler's identity show that the imaginary part is 0 (cuz integrand is odd). Remains the real part

$$2 \int_0^{\infty} cos(-Ax) e^{-Bx^2} dx$$

for which integration by parts leads nowhere.

 

[Tide]

I recommend completing the square in the exponential then suitably choosing a loop around which to integrate in the complex plane.

 

[quasar987]

I should have mentionned that I have no knowledge of complex analysis whatesoever. I don't know anything about residues, integration in the complex plane, and stuff like that.

 

[Tide]

Since a complex number appears in the exponential you are at least familiar with some of the basics. I still recommend completing the square in the exponential. It should provide some illumination. :)

 

[quasar987]

I love ilumination. Let me try just that.

 

[quasar987]

Is it just $\sqrt{\pi / B}$?

After completing the square, I'm left with

$$\mbox{exp}(-A^2 / 4B) \int \mbox{exp}(-B(x+Ai/2B)^2) dx$$

And so with substitution y = x+Ai/2B, I get the integral

$$\mbox{exp}(-A^2 / 4B) \int_{-\infty}^{\infty} \mbox{exp}(-By^2) dy$$

which is $\sqrt{\pi / B}$ as pointed out by another thread by Tom Mattson.

Is this valid with complex too?

 

[Tide]

quasar,

Way to go!