Answer to Question #97506 in Calculus for Rachel

Question #97506
Let f be one of the following functions on R. Compute f-hat. (a) f(x)=cosx, xER. (b) f(x)=sinx, xER. (c) f(x)=e^(iax), xER, where aER.
1
Expert's answer
2019-11-12T13:00:53-0500

Fourier transform of function is:

f^(k)=12πdxeikxf(x)\hat{f}(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} dx\, e^{ikx}f(x)

and integral representation of Dirac delta-function is

δ(x)=12πeikxdk\delta(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ikx}\, dk

a)

f(x)=cos(x)f^(k)=12πdxeikxcos(x)=12πdxeikxeix+eix2=π2δ(k1)+π2δ(k+1)f(x)= \cos(x)\\ \hat{f}(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} dx\, e^{ikx}\cos(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}dx\, e^{ikx}\frac{e^{ix}+e^{-ix}}{2}=\\ \sqrt{\frac{\pi}{2}}\delta(k-1)+\sqrt{\frac{\pi}{2}}\delta(k+1)

b)

f(x)=sin(x)f^(k)=12πdxeikxsin(x)=12πdxeikxeixeix2i=iπ2δ(k1)iπ2δ(k+1)f(x)= \sin(x)\\ \hat{f}(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} dx\, e^{ikx}\sin(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}dx\, e^{ikx}\frac{e^{ix}-e^{-ix}}{2i}=\\ i\sqrt{\frac{\pi}{2}}\delta(k-1)-i\sqrt{\frac{\pi}{2}}\delta(k+1)

c)

f(x)=eiaxf^(k)=12πdxeikxeiax==12πdxei(k+a)x=2πδ(k+a)f(x)= e^{iax}\\ \hat{f}(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} dx\, e^{ikx}e^{iax} =\\ =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} dx\, e^{i(k+a)x} = \sqrt{2\pi}\delta(k+a)


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