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Do you have any idea how can we calculate the integral of Fourier cosine transform of the so-called function? Thanks.

Christian Blatter
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MAh2014
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  • Is that a multiplication or a convolution? Also, what are your own thoughts about this? – mickep Oct 15 '15 at 14:08
  • a multiplication.as we know a relation between the cosine transform of a derivative of a function and itself,I thought about $x*exp(-x^2)$ that it's derivative contains our function and also $exp(-x^2)$ so now we should calculate cosine(sine)transform of the function I've suggested and $exp(-x^2)$,then we could find our answer .but finding these new transforms is not much easier than the main one! – MAh2014 Oct 15 '15 at 14:23
  • You should include such thoughts in your question. – Silvia Ghinassi Oct 15 '15 at 14:37
  • And something else I just think about is that $x exp(-x^2)$ is also the derivative of the function $exp(-x^2)$(indeed a multiple of it),so now we just need the transform of $exp(-x^2)$. – MAh2014 Oct 15 '15 at 14:40
  • A possible duplicate of http://math.stackexchange.com/q/14540/191425 – Hamed Oct 15 '15 at 14:47

1 Answers1

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I'm not sure if what I'm about to say is what you are looking for. There is in general a neat trick in doing such integrations. Note that $$ x^2 e^{-x^2} = \left.\frac{\partial}{\partial\lambda} e^{-\lambda x^2}\right|_{\lambda = 1}$$ So if you are looking for $$ \int dx x^2 e^{-x^2}\cos(kx) = \lim_{\lambda \to 1}\frac{d}{d\lambda}\int dx e^{-\lambda x^2}\cos(kx) $$ the second integral is easy to calculate by writing $\cos (kx) = (e^{ikx}+e^{-ikx})/2$ and completing the square.

Hamed
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  • Now what is the integral of $exp(-x^2)$? – MAh2014 Oct 15 '15 at 14:58
  • You mean integral of $e^{-x^2+ikx}$? Well first of all $\int dx e^{-x^2} =\sqrt{\pi}$ (the simplest way to prove it is by letting $I=\int dx e^{-x^2}$, and then $I^2 = \int dx dy e^{-x^2-y^2}$. Now go to polar coordinates.) Once you know this integral by completing the square in $e^{-x^2+ikx}$ the integral is easy. – Hamed Oct 15 '15 at 15:01
  • I meant$ \int d x e^{-x^2} $it self because the solution you suggested ends in finding this integral.Thank you. – MAh2014 Oct 15 '15 at 15:09