# Complex integral (1 Viewer)

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#### quasar987

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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

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

#### quasar987

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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

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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

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I love ilumination. Let me try just that. :tongue2:

#### quasar987

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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?

Last edited:

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quasar,

Way to go!

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