How to solve the following Fourier Transform, knowing the note below?

Question

Find $$\mathcal F \left( \frac{ b x \cos(\omega x)}{\left( x^2 + a^2 \right)^2} \right) $$ Note that $$\frac{{\rm d}}{{\rm d}x} \left( \frac{1}{x^2 + a^2} \right) = \frac{-2 x}{\left( x^2 + a^2 \right)^2}$$

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I took a calculus test last week, and one of the questions I couldn't solve. I wrote it down so I could get a solution later. I thought of using some table properties like "transformed from the derivative of a function" or even the "product of a function with cosine". But I couldn't put the ideas together clearly.

Answer 1

You could use several properties of the Fourier transform to reduce it to just a few standard transforms. If will assume

$$\mathcal F(f)(\xi) = \int_{-\infty}^\infty f(x) e^{-i \xi x}\, dx.$$

Alternative definitions add a normalization constant or scale the angular frequency by $2\pi$ to get the spatial frequency.

We will use the following properties:

• The convolution theorem: $\mathcal F(fg) = \frac1{2\pi}\mathcal Ff\ast\mathcal Fg$
• The behaviour of the Fourier transform under derivatives: $\mathcal F(f')(\xi) = i\xi\mathcal F(f)(\xi)$
• $\mathcal F(\cos\omega x)(\xi) = \pi\left(\delta(\xi - \omega) + \delta(\xi + \omega)\right)$
• $\mathcal F\left(\frac1{x^2 + a^2}\right)(\xi) = \pi\frac{e^{-a|\xi|}}a$
• $\delta(x - a)\ast f(x) = f(x - a)$

From the hint it follows that you have to compute

$$\mathcal F\left(-\frac b2\cos\omega x\left(\frac d{dx}\frac1{x^2 + a^2}\right)\right) = -\frac b2\mathcal F\left(\cos\omega x\right) \ast \mathcal F\left(\frac d{dx}\frac1{x^2 + a^2}\right).$$

The second factor is

$$\mathcal F\left(\frac d{dx}\frac1{x^2 + a^2}\right) = i\xi\mathcal F\left(\frac1{x^2 + a^2}\right) = i\xi\pi\frac{e^{-a|\xi|}}a.$$

Putting it all together gives us (if I didn't make any mistakes):

$$-\pi i b\xi\frac{e^{-a|\xi - \omega|} + e^{-a|\xi + \omega|}}{2a}.$$

If you take good care of the signs, you could write this in terms of hyperbolic functions.