Inverse Fourier transorm of $f(\omega)=\frac{1}{\omega \tau}$ I am stuck at calculating the inverse Fourier transform of the function $f(\omega)=\dfrac{1}{\omega \tau}$.
Working by definition leads me to
$$f(t)=\dfrac{1}{2\pi}\int_{-\infty}^{\infty}\dfrac{1}{\tau \omega}\cdot e^{i\omega t}d\omega$$ which is not an analytical inntegral.
I can't think of any other ways to compute $f(t)$.
This integral doesn't seem to work out because the exponent doesn't decay. It has a plus sign in its argument.
Any suggestions?
 A: If the Fourier Transform is interpreted in the Cauchy Principal Value sense, then we have
$$\begin{align}
f(t)&=\frac1{2\pi}\int_{-\infty}^\infty \frac{e^{i\omega t}}{\tau \omega}\,d\omega\\\\
&=\frac1{2\pi}\lim_{L\to \infty,\epsilon\to 0^+}\left(\int_{-L}^{-\epsilon} \frac{e^{i\omega t}}{\tau \omega}\,d\omega+\int_{\epsilon}^{L} \frac{e^{i\omega t}}{\tau \omega}\,d\omega\right) \tag 1\\\\
&=\frac{i}{2\pi}\lim_{L\to \infty,\epsilon\to 0^+}\left(\int_{-L}^{-\epsilon} \frac{\sin(\omega t)}{\tau \omega}\,d\omega+\int_{\epsilon}^{L} \frac{\sin(\omega t)}{\tau \omega}\,d\omega\right) \tag 2\\\\
&=\frac{i}{\pi}\int_{0}^\infty \frac{\sin(\omega t)}{\tau \omega}\,d\omega \tag 3\\\\
&=\frac{i}{\tau\pi}\,\text{sgn}(t)\int_{0}^\infty \frac{\sin(\omega )}{\omega}\,d\omega \tag 4\\\\
&=\frac{i}{2\tau} \,\text{sgn}(t)\tag 5
\end{align}$$
In going from $(1)$ to $(2)$, we exploited Euler's Identity $e^{i\omega t}=\cos (\omega t)+i\sin (\omega t)$ along with the fact that the integral of an odd function (i.e., $\frac{\cos \omega t}{\omega}$) about symmetric limits is zero.
In going from $(2)$ to $(3)$, we exploited the fact that the integral of an even function (i.e., $\frac{\sin \omega t}{\omega}$) about symmetric limits is twice the integral over one half interval.
In going from $(3)$ to $(4)$, we enforce the substitution $\omega \to \omega/t$.
And in going from $(4)$ to $(5)$, we use the well-known value of the Dirichlet Integral.
