$''$ Prove that the following integral operator

$ Ku(x) = \int_{0}^{ \infty } \ e ^{-xy} u(y) dy $

has as eigenfunction the

$ φ_α(x) = \sqrt {Γ(α)} x^{-α} + \sqrt {Γ(1-α)} x^{α-1} $

for $ α\in (0,1) $ and $Γ$ the gamma function. $''$

I replaced $u(x)$ with $φ_α(x)$ so

$ Kφ_α(x) = \int_{0}^{ \infty } \ e ^{-xy} ( \sqrt {Γ(α)} y^{-α} + \sqrt {Γ(1-α)} y^{α-1} ) dy $

and after some calculations I concluded to this:

$ Kφ_α(x) = \sqrt {Γ(α){Γ^{2}(1-α)}} x^{α-1} + \sqrt {Γ(α){Γ^{2}(1-α)}} x^{-α}$ =

$ \sqrt {Γ(α)Γ(1-α)} \sqrt {Γ(1-α)} x^{α-1} + \sqrt {Γ(α)Γ(1-α)}\sqrt {Γ(α)} x^{-α} $=

$ \sqrt {Γ(α)Γ(1-α)} (\sqrt {Γ(1-α)} x^{α-1} +\sqrt {Γ(α)} x^{-α} ) $

My questions are :

  1. Is the above calculation enough in order to say someone that the eigenvalue of $K$ operator is the $\sqrt {Γ(α)Γ(1-α)} $ which is equal to $ \sqrt {\frac {π}{\sin πα}} $ and so the eigenfunction is $φ_α(x)$ ?

  2. If the above isn't enough, how can i prove this?

  • $\begingroup$ You seem to have stated the problem incorrectly. Since the given $\phi(x)$ is a function rather than a number, you want to prove it is an Eigenfunction or eigenvalue, not an "eigenvalue". Yes, what you have done, showing that $K\phi(x)= \lambda \phi(x)$, with $\lambda= \sqrt{\Gamma(\alpha)\Gamma(1- \alph)}$ is sufficient to show that $\phi(x)$ is an eigenfunction. $\endgroup$ – user247327 May 4 '16 at 13:35
  • $\begingroup$ I haven't seen this mistake . Thank you for pointing it out!! $\endgroup$ – kaithkolesidou May 4 '16 at 15:17

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