Chebyshev function: variational formulation The Wikipedia article on the Chebyshev function $\psi(x)$ states that $\psi(e^t)$ minimizes the functional 
$$J[f] = \int_0^\infty \dfrac{f(s)\zeta'(s+c)}{\zeta(s+c)(s+c)}ds - \int_0^\infty \int_0^\infty e^{-st}f(s)f(t)ds dt,$$
so that $f(t) = \psi(e^t)e^{-ct}$ for $c>0$.
Wikipedia gives no source for this. Can someone please point me to a text proving this (and possibly more information on the connection between analytic number theory and the calculus of variations)?
 A: As it is it is a nonsense just because $\psi(e^u) \sim e^u$ and $\frac{\zeta'}{\zeta}$ has a pole at $1$ and it is negative on $(1,\infty)$. But one may try correcting the typos.

Think to the Laplace transform as a linear map $L^2(\Bbb{R}_{>0}) \to L^2(\Bbb{R}_{>0})$, for $c>1$ and $t>0$
$$ \mathcal{L}[ \psi(e^u)e^{-cu}](t) = \int_0^\infty  \psi(e^u)e^{-cu} e^{-tu}du = \frac1{t+c}\frac{-\zeta'(t+c)}{\zeta(t+c)}$$
With $\langle .,.\rangle$ the standard inner product on $L^2(\Bbb{R}_{>0})$ and $f\in L^2(\Bbb{R}_{>0})$ let
$$\mathcal{J}[f]=\langle f- \psi(e^u)e^{-cu}, \mathcal{L}[f- \psi(e^u)e^{-cu}]\rangle$$
(wikipedia & OP's functional is $J[f]= \frac{-1}4\mathcal{J}[-2f] + A$)
That $\mathcal{J}$ is unbounded above is obvious (since the Laplace transform sends non-negative functions to non-negative functions).

Due to its symmetric kernel $ \mathcal{L}$ is self-adjoint, if it was positive definite we would have $\mathcal{J}[f] =0 \iff f- \psi(e^u)e^{-cu}=0$.

But wolfram & parigp intnum(u=0,50,(1-u^2)*exp(-u)*(1/(u+1)-2/(u+1)^3))
both tell me
$$\langle (1-u^2) e^{-u}, \mathcal{L}[(1-u^2) e^{-u}]\rangle =\langle  (1-u^2) e^{-u}, \frac1{u+1}-\frac2{(u+1)^3} \rangle \approx -0.42$$
Whence $ \mathcal{L}$ is not positive semi-definite, $\mathcal{J}$ is unbounded below and wikipedia's claim is uncorrect.
