How can I compute the Fourier transform of the following function:

$$f\left(x \right) = \frac{1}{|x|+a}$$

I have attempted to solve this problem. Here is where I am so far:

\begin{equation} \mathscr{F} \left[ f \left(x \right) \right] = \int\limits_{-\infty}^{\infty} \frac{e^{iqx} dx}{|x|+a} = 2\int\limits_{0}^{\infty} \frac{\cos(qx) dx}{\left( \sqrt{x} \right)^2+a} = 2\int\limits_{0}^{\infty} \frac{\cos(qx) d\sqrt{x}}{2\sqrt{x}\left[ \left( \sqrt{x} \right)^2+a\right]}= \end{equation}

$$ 2\int\limits_{0}^{\infty} d\sqrt{x} \left( \frac{1}{\sqrt{x} + i\sqrt{a}} + \frac{1}{\sqrt{x} - i\sqrt{a}} \right) \cos(qx) $$

If I had $\sqrt{x}$ in the cosine in the last line then I would easily find the Fourier transforms...


1 Answer 1


I have found the answer with the help of "Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables" edited by Abramowitz and Stegun:

$$ \mathscr{F} \left[ f \left(x \right) \right] = \int\limits_{-\infty}^{\infty} \frac{e^{iqx} dx}{|x|+a} = 2\int\limits_{0}^{\infty} \frac{\cos(qx) dx}{x+a} = $$ $$ - 2\mathrm{Ci}(qa) \cos (qa)-\left[2\mathrm{Si} (qa) -\pi\right]\sin(qa), \text{ for } \mathrm{Re}(a) >0,$$ where $\mathrm{Ci}(x)$ and $\mathrm{Si}(x)$ are cosine and sine integrals: $$\mathrm{Ci}(x) = \gamma + \ln x +\int\limits_0^{x} \frac{\cos t - 1}{t} dt,$$ $$\mathrm{Si}(x) = \int\limits_0^{x} \frac{\sin t}{t} dt.$$

However, I am still struggling with the derivation of this expression.


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