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One set of eigenfunction for the following fractional integral operator is $f(z)=e^{-bz}$ for any constant Re$b>0$, with eigenvalue $\lambda=\frac{\Gamma(\alpha)}{b^\alpha}$, $$\int_z^\infty (\zeta-z)^{\alpha-1}f(\zeta) d\zeta = \lambda f(z) $$ where $\alpha$ is a constant with Re$(\alpha)>0$.

I found it by looking first at positive integer $\alpha$.

My question is: How does one solve for this eigenfunction problem in general and are there more eigenfunctions and eigenvalues?

I have tried Laplace and Fourier transforms. The integration limits and function growth rate block their respective applicability in the most straightforward manner --- that does preclude clever ways of application that circumvent these obstacles.

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  • $\begingroup$ Apply Laplace transform and see what happens. $\endgroup$ Jan 8, 2014 at 15:34
  • $\begingroup$ @MichaelRenardy: That was the first thing I tried, together with Fourier. The integral limits and function growth rate block their applicability respectively. $\endgroup$
    – Hans
    Jan 8, 2014 at 15:49
  • $\begingroup$ You can interpret the Fourier transform of $x^{\alpha-1}_+$ as a distribution. Then apply Fourier transform to your equation. The eigenfunctions of the transformed problem are delta functions. so your original eigenfunctions are exponentials as you found. Convergence of the integral restricts you to Re(b)>0. $\endgroup$ Jan 8, 2014 at 16:51
  • $\begingroup$ @MichaelRenardy: You are right. I suppose I can limit $f(x)$ to $x>0$ and let it vanish for negative $x$. When I was doing the Fourier before, I worried too much about convergence of $e^{-bx}$ over the whole axis of $x$. Still, how do I reconcile the convergence issue if I does not force $f$ to vanish on the negative axis when solving it? $\endgroup$
    – Hans
    Jan 8, 2014 at 17:59
  • $\begingroup$ The Fourier transforms of exponential functions belong to a class called analytic functionals. Gelfand and Shilov (Generalized Functions) discuss the Fourier transform in this context. $\endgroup$ Jan 8, 2014 at 18:30

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