Semi-integral of $f(x) = \frac{1}{x}$ I wish to know what the semi-integral (fractional integral of order one-half) of the function $f(x) = \frac{1}{x}$ is.
From the definition of the Riemann–Liouville fractional integral one has
$$D^{-\nu} f(t) = \frac{1}{\Gamma (\nu)} \int_0^t (t - u)^{\nu - 1} f(u) \, du.$$
Here $t > 0$ and $\nu$ is a positive real number.
Setting $\nu = \frac{1}{2}$ and $f(t) = \frac{1}{t}$, after enforcing a substitution of $u \mapsto ut$, one has
$$D^{-\frac{1}{2}} \left (\frac{1}{t} \right ) = \frac{1}{\sqrt{\pi t}} \int_0^1 \frac{du}{u\sqrt{1 - u}} = -2 \operatorname{arctanh} (\sqrt{1 - u}) \Big{|}_0^1 \to \infty,$$
which clearly diverges.
However, when I use this semi-integral operation found here one finds
$$D^{-\frac{1}{2}} \left (\frac{1}{t} \right ) = -i \sqrt{\frac{\pi}{t}}.$$
Here $i$ is the imaginary unit.
So can anyone tell me what they think is going on here? Perhaps a different definition for the semi-integral has been used?
 A: Each and every one of these steps is going to be a heuristic for something made much more rigorous by the theory of distributions - but the point of distributions is to make the heuristics true almost by definition, thus these calculations should suffice. What was done to calculate the half antiderivative was the Fourier Transform. It does not matter which convention we take, so I will use the computationally easy convention
$$\mathcal{F}\{f(t)\}(\omega) = \int_{-\infty}^\infty f(t)e^{-i\omega t}\:dt$$
$$\mathcal{F}^{-1}\{f(\omega)\}(t) = \frac{1}{2\pi}\int_{-\infty}^\infty f(\omega)e^{i\omega t}\:d\omega$$
Using the property
$$D^{a}f(t) = \mathcal{F}^{-1}\left\{(i\omega)^a\mathcal{F}\{f(t)\}(\omega)\right\}(t)$$
we can find the half integral. First
$$\mathcal{F}\left\{\frac{1}{t}\right\}(\omega) = -2\pi i H(\omega)$$
where $H$ is the Heaviside function. Next what we want is to multiply by $(i\omega)^{-\frac{1}{2}}$ and then take the inverse Fourier Transform of the result.
$$D^{-\frac{1}{2}}f(t) = -\sqrt{i}\int_0^\infty \frac{e^{i\omega t}}{\sqrt{\omega}}\:d\omega = -\sqrt{i}\int_{-\infty}^{\infty} e^{-(-it)\alpha^2}\:d\alpha "=" -i\sqrt{\frac{\pi}{t}}$$
by an analytic continuation of the result
$$\int_{-\infty}^\infty e^{-ax^2}\:dx = \sqrt{\frac{\pi}{a}}$$
to the imaginary axis by an application of the Paley-Wiener theorem.
