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$$f(x) = \frac{\cos(bx)}{a^2 +x^2} $$ for $a, b$ $>0$

$$\hat f(\omega) = \int\limits^{\infty}_{-\infty} \dfrac{1}{2{\pi}}\cdot\dfrac{\cos\left(bx\right)}{a^2+x^2}\mathrm{e}^{-\mathrm{i}wx}\,\mathrm{d}x = \int\limits^{\infty}_{-\infty} \dfrac{\mathrm{e}^{\mathrm{i}bx}+\mathrm{e}^{-\mathrm{i}bx}}{4\pi\left(a^2+x^2\right)}\mathrm{e}^{-\mathrm{i}wx}\,\mathrm{d}x=$$

$$ \int\limits^{\infty}_{-\infty} \dfrac{\mathrm{e}^{\mathrm{i}x\left(b-w\right)}}{2\left(a^2+x^2\right)}\,\mathrm{d}x + \int\limits^{\infty}_{-\infty} \dfrac{\mathrm{e}^{-\mathrm{i}x\left(b+w\right)}}{2\left(a^2+x^2\right)}\,\mathrm{d}x$$

How should I continue from here?

I checked on wolfram alpha but the answer should not expressed with dirac function since we didn't learn it.

OmnipotentEntity
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Sagigever
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  • The Fourier transform of a cosine will always involve a pairs of dirac deltas. In this particular case; however, because it is being multiplied by 1/(a^2 + x^2) in frequency space the dirac deltas are being convolved with the fourier transform of 1/(a^2 + x^2) which results in the dirac deltas being removed by the convolution integral. So the answer should not contain a dirac delta. (And does not when I ask WA to tell me the transform.) – OmnipotentEntity Jan 07 '21 at 22:10
  • For clarity, are you simply having difficulties computing the integrals? Because as far as I can tell, thus far you've done everything correct (other than accidentally dropping a factor of $2\pi$) – OmnipotentEntity Jan 07 '21 at 22:12
  • Yes I am having trouble to calculate the integral. and also wanted to know if my approach is fine – Sagigever Jan 07 '21 at 22:18
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    Yes, you are correct. Next step, observe that the imaginary part of the integral is zero because the product of an odd function (sin) with an even function ($1/(a^2+x^2)$) is odd, and therefore the integral vanishes. In general, the Fourier transform of a real and even function is still real and even. – Kosh Jan 07 '21 at 22:24
  • now I have left with integral from the kind $\displaystyle\int\limits^{\cssId{upper-bound-mathjax}{\infty}}_{\cssId{lower-bound-mathjax}{0}} \dfrac{\cos\left(wx\right)}{a^2+x^2},\cssId{int-var-mathjax}{\mathrm{d}x}$ anyone can help me simplfy it? – Sagigever Jan 07 '21 at 22:35
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    @Sagigever See for example: https://math.stackexchange.com/questions/272622/integral-evaluation-int-infty-infty-frac-cos-ax-pi-1x2dx – S.H.W Jan 07 '21 at 22:37

1 Answers1

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Note that the factor $\cos(bx)$ is just a modulation, i.e., a translation of the spectrum. So we can try to determine the Fourier transform $G(\omega)$ of

$$g(x)=\frac{1}{a^2+x^2}\tag{1}$$

and from $G(\omega)$ determine the Fourier transform of $f(x)$:

$$F(\omega)=\frac12\big[G(\omega-b)+G(\omega+b)\big]\tag{2}$$

Rewrite $g(x)$ as

$$g(x)=\frac{1}{2a}\left[\frac{1}{a+jt}+\frac{1}{a-jt}\right]\tag{3}$$

It's straightforward to see that

$$\mathcal{F}\left\{\frac{1}{a+jt}\right\}=2\pi e^{a\omega}u(-\omega),\qquad a>0\tag{4}$$

and

$$\mathcal{F}\left\{\frac{1}{a-jt}\right\}=2\pi e^{-a\omega}u(\omega),\qquad a>0\tag{5}$$

where $u(\omega)$ is the Heaviside unit step function.

Using $(4)$ and $(5)$, the Fourier transform of $g(x)$ is obtained as

$$G(\omega)=\frac{\pi}{a}e^{-a|\omega|}\tag{6}$$

The Fourier transform of $f(x)$ can be obtained by plugging $(6)$ into $(2)$.

Matt L.
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