# calculate $\int_{-\infty}^{+\infty} \cos(at) e^{-bt^2} dt$

Could someone please help me to calculate the integral of:

$$\int_{-\infty}^{+\infty} \cos (at) e^{-bt^2} dt.$$

a and b both real, b>0.

I have tried integration by parts, but I can't seem to simplify it to anything useful. Essentially, I would like to arrive at something that looks like: 7.4.6 here: textbook result

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yep! But... I can't seem to simplify it to anything useful. Essentially, I would like to arrive at something that looks like: 7.4.6 here people.math.sfu.ca/~cbm/aands/page_302.htm – confused Sep 4 '12 at 21:26
You need to specify the domain. – Tunococ Sep 4 '12 at 21:30
Sorry. done now – confused Sep 4 '12 at 21:32

Hint:

Use the fact that $$\int_{-\infty}^\infty e^{iat- bt^2}\,dt = \sqrt{\frac{\pi}{b}} e^{-a^2/4b}$$ which is valid for $b>0$.

To derive this formula, complete the square in the exponent and then shift the integration contour a bit.

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Hey @fabian , what does it mean to 'shift the integration contour'? I am not familiar with such a thing/method? – confused Sep 4 '12 at 22:47
Thanks! I think I see what you mean now. – confused Sep 5 '12 at 0:11
I think it must be $\,a^2/4b\,$ in the exponent of the exponential, because of that $\,i\,$ there... – DonAntonio Sep 5 '12 at 3:38
@DonAntonio: I checked again. I believe it is correct. Note that $i a t - b t^2 = -a^2/4b -b(t-ia/2b)^2$. – Fabian Sep 5 '12 at 7:16
You're right, of course. I oversaw that minus sign before the quadratic term. Thanks. – DonAntonio Sep 5 '12 at 9:27

Do you have restrictions on 'a' and 'b'?

For example, they are real and > 0.

Otherwise, things are messy!

See here for details.

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a and b are both real; and b>0. Hmm.. I should have written this question up a bit better. I apologise. – confused Sep 4 '12 at 21:40