Toronto Math Forum
APM346-2018S => APM346--Tests => Term Test 2 => Topic started by: Victor Ivrii on March 23, 2018, 06:18:55 AM
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Find Fourier transforms of the function
\begin{equation}
f(x)=\cos^2(x)e^{-|x|}
\end{equation}
and write this function as a Fourier integral.
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We first note that $ \cos^2 x = (\frac{(e^{ix} + e^{-ix})}{2})^2 = \frac{2 + e^{2ix} + e^{-2ix}}{4} $
From there, we take the fourier transform: $$\hat{f(}k) = \frac{1}{2\pi} \int_{-\infty }^{\infty } e^{-ikx} e^{-|x|} ( \frac{2 + e^{2ix} + e^{-2ix}}{4}) = \frac{1}{2\pi }\left ( \int_{0}^{\infty } \frac{2e^{-x(1+ki)} + e^{-x(1+ki - 2i)} + e^{-x(1+ki + 2i))}}{4} dx +\int_{-\infty }^{0} \frac{2e^{x(1- ki)} + e^{x(1- ki - 2i) }+ e^{x(1- ki + 2i))}}{4} dx\right )$$
We integrate that thing, and note that by evaluating at the bounds of $0,\pm \infty $, only the terms evaluated at 0 can be kept, netting us:
$$\hat{f(}k) = \frac{1}{2\pi}( \frac{1}{2(1+ki)} + \frac{1}{4(1+ki - 2i)} + \frac{1}{4(1+ki+2i)} + \frac{1}{2(1-ki)} + \frac{1}{4(1-ki-2i)} + \frac{1}{4(1-ki+2i)}) $$
While this expression may be ugly, what is useful about it is the fact that these are fractions of complex numbers and their conjugates (noteably first and fourth terms, 2nd and 6th, 3rd and 5th) so we'd get the following fraction
$$\hat{f(}k) = \frac{1}{2\pi}( \frac{2}{4(1+ k^2)} + \frac{2}{16(1+ (k-2)^2)} + \frac{2}{16(1+ (k+2)^2)} ) $$
Thus, we can plug this function into the IFT:
$$ f(x) = \int_{-\infty}^{\infty} \hat{f}(k) e^{ikx} dk = \int_{-\infty}^{\infty} \frac{1}{4\pi}( \frac{1}{(1+ k^2)} + \frac{1}{4(1+ (k-2)^2)} + \frac{1}{4(1+ (k+2)^2)}) e^{ikx} dk $$
Leaving the function as a Fourier integral.
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I suggest somewhat different numerical factor. Not much different.
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I suggest somewhat different numerical factor. Not much different.
Please correct.
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\begin{equation}\begin{split}
\hat{f}(\omega)&=\frac{1}{2\pi}\int_{-\infty}^\infty (\frac{1}{4}e^{2ix}+\frac{1}{4}e^{-2ix}+\frac{1}{2})e^{-|x|}e^{-i\omega x}\,dx\\
&=\frac{1}{2\pi}\int_{-\infty}^\infty \frac{1}{4}e^{-|x|}e^{-i(\omega-2) x}\,dx+\frac{1}{2\pi}\int_{-\infty}^\infty \frac{1}{4}e^{-|x|}e^{-i(\omega+2) x}\,dx+\frac{1}{2\pi}\int_{-\infty}^\infty \frac{1}{2}e^{-|x|}e^{-i\omega x}\,dx\\
&=\frac{1}{4\pi(1+(x-2)^2)}+\frac{1}{4\pi(1+(x+2)^2)}+\frac{1}{2\pi(1+x^2)}
\end{split}\end{equation}