I am required to find the Fourier transform of the following function:
$$f(x) = \dfrac{e^{-ax}}{x}$$
Where $a$ is an arbitrary constant.
This is what I've tried, but I'm not quite sure:
The Fourier transform (denoted by $\mathcal{F}(s)$) is given by:
$$\mathcal{F}(s) = \sqrt\frac{2}{\pi}.\int_{0}^{\infty}f(x)\sin(sx) dx \tag{1}$$
Where $f(x) = \frac{e^{-ax}}{x}$.
Differentiating with respect to $s$ on both sides and simplifying:
$$\frac{d}{ds}(\mathcal{F}(s)) = \sqrt\frac{2}{\pi}.\int_{0}^{\infty}\frac{\partial}{\partial s}\left(\dfrac{e^{-ax}}{x}.\sin(sx)\right) dx$$
$$=\sqrt\frac{2}{\pi}.\int_{0}^{\infty}e^{-ax}\cos(sx)dx$$
This is a standard integral; we get:
$$\frac{d}{ds}(\mathcal{F}(s)) = \sqrt\frac{2}{\pi}.\dfrac{a}{a^2+s^2}$$
Now,
$$\mathcal{F}(s) = \sqrt\frac{2}{\pi}.\int \dfrac{a}{a^2+s^2} ds$$
Which is also a standard integral. The result is
$$\mathcal{F}(s) = \sqrt\frac{2}{\pi}.\tan^{-1}\left(\dfrac{s}{a}\right) + C \tag{2}$$
Also, from $(1)$, we know that $\mathcal{F}(0) = 0$. Using this in $(2)$ we get $$C = 0$$
Therefore, the Fourier transform $\mathcal{F}(s)$ of the given function $f(x)$ is
$$\mathcal{F}(s) = \sqrt\frac{2}{\pi}.\tan^{-1}\left(\dfrac{s}{a}\right) $$
Is my working correct?
I have a strong feeling that I'm missing something; I concede that I do not have a good understanding of Fourier transforms. Any and all corrections and/or tips are highly appreciated. Thank you.