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How can I calculate the integral: $$\int_{-\infty}^{+\infty}e^{-\frac{(x-a)^2}{0.01}}\cos(bx)dx$$ I can not find in the references. Excuse my bad English.

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You could expand the cosine in a Taylor series around $x=a$... –  Fabian Oct 19 '12 at 16:21
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up vote 5 down vote accepted

With

$$ \cos(bx)=\frac12\left(\mathrm e^{\mathrm ibx}+\mathrm e^{-\mathrm ibx}\right)\;, $$

the integrand becomes the sum of two Gaussians with complex exponents, whose integrals can be evaluated like this.

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A reference is Rudin, Principles of mathematical analysis, Example 9.43. Here the integral $$ \int_{-\infty}^{\infty}e^{-x^2}\cos(xt)\,dx $$ is calculated using the theory of ordinary differential equations. The integral is $$ \sqrt{\pi}\exp\left(-\frac{t^2}{4} \right). $$ (Hint) In your integral after introducing new variable you should calculate $$ \int_{-\infty}^{\infty}e^{-z^2}\cos(cz)\,dz $$ and $$ \int_{-\infty}^{\infty}e^{-z^2}\sin(cz)\,dz. $$

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