In relation with the question I posed on MathSE here, I want to ask how can Mathematica give an answer to my problem.
The context
I am trying to get rid of the integral over $y$ in
$$begin{align}
int_0^{infty}dy ; psileft(frac{y}{q}right)phi(y) int_{-infty}^{infty}frac{dp}{sqrt{2pi}} p^{gamma} e^{-ip(y-x)};,
end{align}$$
and to obtain an analytical expression in terms of $phi(x)$, $psi(x/q)$ and their derivatives. I assume that $gamma=1-delta$, with $0<delta<1$. I tried to break down the problem (though I don’t know if there is a better way) so that I am left with
$$ begin{align}
int_0^{infty} dy ; psileft(frac{y}{q}right)phi(y) times left(frac{delta(y-x)}{(y-x)^{1-delta}} – (1-delta)frac{H(y-x)}{(y-x)^{2-delta}}right);,
end{align}$$
where $delta$ is the Dirac delta and $H$ the Heaviside function.
The problem
Since the parameter $delta$ (not to be confused with the Dirac) is constrained as $0<delta<1$, I expect troubles when evaluating the first term in the brackets. However, when I define $psi$ and $phi$, MMA returns an expression for this integral, which leaves me confused. Basically, my code is
(Psi)(x_) := 9*6^{-1/2} x^{3/2} Exp(-3 x/2);
(Phi)(x_) := x; (* Or x^2, or E^(x)...*)
fourpfrac(x_, y_, (Delta)_) := (2*Pi)^(1/2)/Gamma((Delta))*HeavisideTheta(y - x)*(y - x)^((Delta) - 1); (* Fourier Transform of p^(1-(Delta)) *)
x1frac(x_, (Nu)_, (Delta)_, (Alpha)_, q_) := Integrate(Derivative(0, (Nu), 0)(fourpfrac)(x, y, (Delta))*(Psi)(y/q)*(Phi)(y)*y^(-(Alpha) - 1), {y, 0, Infinity}); (* The whole integral *)
x1frac(x, 1, (Delta), -1, q)
which returns in traditional form
$$
left{text{ConditionalExpression}left(frac{45 sqrt{3} x ((pi delta ) sin ) Gamma left(-delta -frac{3}{2}right) (-x)^{delta +frac{1}{2}} , _1F_1left(frac{7}{2};delta +frac{5}{2};-frac{3 x}{2 q}right)}{8 q^{3/2}}+frac{sqrt{pi } 2^{delta +frac{3}{2}} 3^{-delta } Gamma left(delta +frac{3}{2}right) q^{delta } , _1F_1left(2-delta ;-delta -frac{1}{2};-frac{3 x}{2 q}right)}{Gamma (delta -1)},x<0right)right}
$$
and in input form
{ConditionalExpression((2^(3/2 + (Delta)) 3^-(Delta) Sqrt((Pi)) q^(Delta) Gamma(3/2 + (Delta)) Hypergeometric1F1(2 - (Delta), -(1/2) - (Delta), -((3 x)/(2 q))))/Gamma(-1 + (Delta))
+ (45 Sqrt(3) (-x)^(1/2 + (Delta))x Gamma(-(3/2) - (Delta)) Hypergeometric1F1(7/2, 5/2 + (Delta), -((3 x)/(2 q))) Sin((Pi) (Delta)))/(8 q^(3/2)), x < 0)}
The thing is that in no way the form I have given to $phi$ and $psi$ could have canceled the singularity.
- Is the result I have obtained with MMA correct?
- What did MMA do to evaluate the integral?
- Is there a better way to resolve the problem?
Many thanks!