fourier transform of $\dfrac{f(x)}{x}$

Question

Question

If i have that the fourier transform of $f(x)$ is $\widehat{f(k)}$, then what is the fourier transform of $\dfrac{f(x)}{x}$? Problem 7.2.8 Chapter 7 Introduction to partial differential equation Peter Olver

I know that this can be done with covolution, but the text book that i'm following has not introduced yet how to work with covolution, so i wonder how to answer the question without covolution, could someone helpe me please

Thanks in advance

Answer 1

You get $$ \begin{gather} \int \frac1xf(x)e^{-i2\pi kx} dx = \\ \left[\frac 1x f(x) e^{-i2\pi kx} \frac1{-i2\pi k} \right] - \frac1{-i2\pi k}\int \frac{f'(x)x - f(x)}{x^2}e^{-i2 \pi kx} dx = \\ -\frac1{-i2\pi k} F(f'(x)/x) + \frac1{-i2\pi k}F(f(x)/x^2) + \left[\frac 1x f(x) e^{-i2\pi kx} \frac1{-i2\pi k} \right] \end{gather} $$ So you need more information.

Answer 2

Let $g(x)=f(x)/x$, and suppose that $f(0)=0$ and $f$ is of Schwartz class. Then $g$ is Schwartz, and $\widehat g$ is well defined. We want to write $\widehat g$ in terms of $\widehat f$. To do this, let us differentiate the equation $\widehat g(\xi)=\int e^{-i\xi x}\frac{f(x)}{x}dx$ by $\xi$:

$$ \widehat g'(\xi)=-i\int e^{-i\xi x}f(x)dx=-i\widehat f(\xi). $$

Note that differentiation is permissible since $g$ is Schwartz. So, by the fundamental theorem of calculus combined with $\widehat g(-\infty)=0$, we obtain:

$$ \widehat g(\xi)=\int_{-\infty}^\xi \widehat f(\eta)d\eta/i, $$

i.e. $\widehat g$ is the anti-derivative of $\widehat f$.