I'm working on a problem I have been dealing with unsuccessfully for months now, so any help is greatly appreciated!
The context
I have to solve an integral of the form
$$\begin{align}
\int_0^{\infty}dy \; \psi\left(\frac{y}{q}\right)\phi(y) \int_{-\infty}^{\infty}\frac{dp}{\sqrt{2\pi}} p^{\gamma} e^{-ip(y-x)}\;,
\end{align}$$
where $\psi$ is square-integrable and $\phi$ is undefined. When $\gamma \in \mathbb{N}$, we can write
$$\begin{align}
\int_0^{\infty}dy \; \psi\left(\frac{y}{q}\right)\phi(y) \int_{-\infty}^{\infty}\frac{dp}{\sqrt{2\pi}} (-i)^{-\gamma}\partial_{y}^{\gamma} \left[ e^{-ip(y-x)} \right]\;,
\end{align}$$
and we obtain a Dirac delta which, upon integration by parts and discarding the surface terms thanks to $\psi$, helps us getting rid of the integral over $y$ and obtaining an analytic expression in terms of $\phi(x)$, $\psi(x/q)$ and their derivatives. I now want to do the same for $\gamma$ real. For this, I suppose that $\gamma = \nu-\rho$, with $\nu \in \mathbb{N}$ and $0<\rho<1$, because I can express the Fourier transform as
$$\begin{align}
\int_{-\infty}^{\infty}\frac{dp}{\sqrt{2\pi}} p^{\nu-\rho} e^{-ip(y-x)} &= \int_{-\infty}^{\infty}\frac{dp}{\sqrt{2\pi}} p^{-\rho} (-i)^{-\nu} \partial_{y}^{\nu}\left[e^{-ip(y-x)}\right] = i^{\nu}\partial_{y}^{\nu} \left[\int_{-\infty}^{\infty}\frac{dp}{\sqrt{2\pi}} p^{-\rho} e^{-ip(y-x)}\right] \\
&= i^{\nu-\rho} \partial_{y}^{\nu} \left[ \frac{\sqrt{2\pi}}{\Gamma(\rho)} \frac{H(y-x)}{(y-x)^{1-\rho}} \right] \;,
\end{align}$$
where $H$ is the Heaviside function.
The problem
The trouble starts now, which is why I consider $\nu=1$ for simplicity, so that I have to evaluate an integral of the form
$$\begin{align}
\int_0^{\infty}dy \; \psi\left(\frac{y}{q}\right)\phi(y) \times \partial_y \left[\frac{H(y-x)}{(y-x)^{1-\rho}}\right]\;.
\end{align}$$
Deriving, I obtain
$$ \begin{align}
\int_0^{\infty} dy \; \psi\left(\frac{y}{q}\right)\phi(y) \times \left[\frac{\delta(y-x)}{(y-x)^{1-\rho}} – (1-\rho)\frac{H(y-x)}{(y-x)^{2-\rho}}\right]\;,
\end{align}$$
in which the first term is singular and the second is another Heaviside function. Amazingly, if I define $\psi$ and $\phi$, Mathematica is able to return a function of $x$, even in the presence of this singular term (I have tried with $\phi$ being a polynomial or an exponential), which convinced me that this integral should be solvable somehow (though I have asked a similar question on MathematicaSE here).
Does anybody know how I could get rid of the integral? Also, if somebody knows of an alternative way to treat this problem, that would be awesome!
Many thanks!
Best Answer
Hint:
I would try to replace the Heaviside function with one of its analytic approximations $$ \eqalign{ & H(x) = \mathop {\lim }\limits_{\varepsilon \to 0} {1 \over 2}\left( {1 + {x \over {\sqrt {x^{\,2} + \varepsilon ^{\,2} } }}} \right) \cr & H(x) = \mathop {\lim }\limits_{\varepsilon \to 0} {1 \over {1 + e^{\, - \,2\,x/\varepsilon } }} \cr} $$
Putting $$ {\partial \over {\partial y}}\left( {{{H(y - x)} \over {\left( {y - x} \right)^{\,1 - r} }}} \right) = \left. {{\partial \over {\partial z}}\left( {{{H\left( z \right)} \over {z^{\,1 - r} }}} \right)} \right|_{\,z = y - x} $$ the first expression would give $$ \eqalign{ & \mathop {\lim }\limits_{\varepsilon \to 0} {\partial \over {\partial z}}\left( {{{H\left( z \right)} \over {z^{\,1 - r} }}} \right) = \cr & = {{z^{\,r - 1} } \over {2\left( {\sqrt {z^{\,2} } } \right)^3 }}\left( {z + \sqrt {z^{\,2} } } \right) \left( {\left( {r - 2} \right)z + \sqrt {z^{\,2} } } \right) = \cr & = {1 \over 2}\left( {\sqrt {z^{\,2} } } \right)^{\,r - 4} \left( {z + \sqrt {z^{\,2} } } \right) \left( {\left( {r - 2} \right)z + \sqrt {z^{\,2} } } \right) = \cr & = \left( {r - 1} \right)z^{\,r - 2} H\left( z \right) \cr} $$
Alternative hint:
Consider that $$ \eqalign{ & \int_{y = 0}^\infty {f(y){{\delta \left( {y - x} \right)} \over {\left( {y - x} \right)^{\,1 - \rho } }}dy} \quad \Rightarrow \quad \mathop {\lim }\limits_{y \to x} {{f(y)} \over {\left( {y - x} \right)^{\,1 - \rho } }} \;\left| {\,0 < x} \right. \cr & \int_{y = 0}^\infty {f(y){{H\left( {y - x} \right)} \over {\left( {y - x} \right)^{\,2 - \rho } }}dy} \quad \Rightarrow \quad \int_{y = x}^\infty {{{f(y)} \over {\left( {y - x} \right)^{\,2 - \rho } }}} \;\left| {\,0 < x} \right. \cr} $$ with a lot of cautions about formal aspects , obligatory when dealing with distributions.
In particular in the first line there might be the introduction of a $1/2$ factor, depending on the definition of the Heaviside ...
The result of the limit depends on $\rho$ : most probably you are having that both terms $ \to \infty$ and whether their subtraction leads to a finite result will depend on $f$.