I am trying to get the cosine Fourier transform of $e^{- x^2}$ where I am stuck in simplifying the integration further. The integration is between the limits $0$ to $\infty$!
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The integrand is even so $\int_{-\infty}^\infty$ is the same. Click on edit in other discussion to see how we use latex – reuns Oct 10 '18 at 18:13
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Hint: Try calling the integral , say I, and differentiate under the integral sign with respect to s (or $\omega$). It should turn the problem into a differential equation you can solve. – Vegeta the Prince of Saiyans Oct 10 '18 at 18:18
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1The other option is to write the cosine function in terms of complex exponentials. – Andrei Oct 10 '18 at 18:20
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https://math.stackexchange.com/q/317249/321264 – StubbornAtom Jan 14 '22 at 18:27
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To reiterate, call $$I = \int_{0}^{\infty} e^{-x^2}cos(sx)dx$$ Now, $$\frac{dI}{ds}= \int_{0}^{\infty} -xe^{-x^2}sin(sx)dx= \frac{1}{2}\int_{0}^{\infty} sin(sx)(-2xe^{-x^2})dx$$ Use integration by parts to get:$$\frac{1}{2} [sin(sx)e^{-x^2}\Big|_0^\infty- s\int_{0}^{\infty}cos(sx)e^{-x^2}dx]= -\frac{s}{2}\int_{0}^{\infty} e^{-x^2}cos(sx)dx= -\frac{s}{2} I$$ So we have a differential equation: $$\frac{dI}{I} = -\int \frac{s}{2}ds + log(c).$$ Hence, $$log(I) =-\frac{s^2}{4} + log(c)= log(ce^{\frac{-s^2}{4}}).$$ Solving for $I$, we get that $$I =ce^{\frac{-s^2}{4}}=\int_{0}^{\infty} e^{-x^2}cos(sx)dx$$. I'll leave you to solve for $c$, just plug in $s=0$ and then you get $c=...$
Abhilekh Gautam
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