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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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What have you tried? –  Austin Mohr Sep 4 '12 at 21:25
    
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
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2 Answers 2

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
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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
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