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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 $x=a$...2012-10-19

2 Answers 2

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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. $