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I want to evaluate

$$ \int_{-\infty}^\infty a e^{-bt^2} \cos(\omega t) dt $$

With Euler's identity

$$ \int_{-\infty}^\infty a e^{-bt^2} \cos(\omega t) dt = \int_{-\infty}^\infty a e^{-bt^2} \big\{ {{e^{i \omega t} + e^{-i \omega t}} \over 2}\big\} dt $$

$$ = \int_{-\infty}^\infty \bigg({a e^{-bt^2} e^{i \omega t} \over 2} + {a e^{-bt^2} e^{-i \omega t} \over 2}\bigg)\; dt $$

$$ = {a \over 2} \int_{-\infty}^\infty (e^{-bt^2+i \omega t} + e^{-bt^2-i \omega t})\; dt $$

$$ = {a \over 2} \bigg({e^{-bt^2+i \omega t} \over -2bt+i \omega}\bigg|_{-\infty}^\infty + {e^{-bt^2-i \omega t} \over {-2bt-i \omega}}\bigg|_{-\infty}^\infty\bigg) $$

This leads to problems with infinities.

Veak
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  • Try using Euler's identity. – Jair Taylor Apr 15 '22 at 22:29
  • I don't think there is an elementary antiderivative. If you want to compute the definite integral, then replace $\cos(\omega t)$ with $e^{i\omega t}$ to reduce it to a well known integral. – Mason Apr 15 '22 at 22:31
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Shaun Apr 15 '22 at 22:36
  • You are on the right track. However the last equation is not correct. As Mason says, there's no elementary derivative. Have you seen $\int_{-\infty}^\infty e^{-u^2} , du= \sqrt{\pi}$? Try completing the square and and then reduce it to a form like $\int_{-\infty}^\infty e^{-u^2},du$. – Jair Taylor Apr 16 '22 at 19:32

1 Answers1

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Let's define the function $I(\xi) = \int_{-\infty}^{\infty} e^{- x^2}\cos(2 \xi x)\, \mathrm{d}x $. Using Feynman's trick we see that \begin{align} I'(\xi) &=\int_{-\infty}^{\infty} \left[-2xe^{- x^2}\right]\sin(2 \xi x)\, \mathrm{d}x\overset{\text{I.B.P.}}{=} -2\xi \underbrace{\int_{-\infty}^{\infty} e^{- x^2}\cos(2 \xi x)\, \mathrm{d}x}_{\color{blue}{I(\xi)}} \end{align} So we get $$ \frac{\mathrm{d} I}{\mathrm{d}\xi} = -2\xi I \mathbin{\color{purple}{\implies}} \int_0^{\xi}\frac{1}{I}\frac{\mathrm{d} I}{\mathrm{d}\widetilde{\xi}} \mathrm{d} \widetilde{\xi} = \int_0^{\xi}-2 \widetilde{\xi}\, \mathrm{d}\widetilde{\xi} \mathbin{\color{purple}{\implies}} \ln\Bigg|\frac{I(\xi)}{I(0)}\Bigg| = -\xi^2 \mathbin{\color{purple}{\implies}} I(\xi) = I(0)e^{-\xi^2} $$ But since from our original defintion we know $I(0) =$ $\int_{-\infty}^{\infty} e^{- x^2}\, \mathrm{d}x = \sqrt{\pi}$ we can conclude $$ \int_{-\infty}^{\infty} e^{- x^2}\cos(2 \xi x)\, \mathrm{d}x = \sqrt{\pi} e^{-\xi^2} $$

Using the previous equation we can conclude your problem as follows: \begin{align} \int_{-\infty}^{\infty} a e^{-bt^2}\cos(\omega t) \,\mathrm{d}t & \overset{\color{blue}{\sqrt{b}t = x}}{=}\frac{a}{\sqrt{b}} \int_{-\infty}^{\infty} e^{-x^2}\cos\left(2 \frac{\omega}{2\sqrt{b}} x\right)\mathrm{d}x = \boxed{a \sqrt{\frac{\pi}{b}} e^{-\frac{\omega^2}{4b}}} \end{align}

Robert Lee
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