This is related to my previous question. I assume that the two wires have the same charge density with the same sign ($\lambda_1 = \lambda_2 = \lambda$)
If I want to take the same approach, the potential is still the same,
$V = k\lambda \log(1/r)$.
The energy is, I think, the integral over the (second) wire, since this is the direction where I add charges. Since the wires are paralleled, $r$ is constant in the integral
$U = \int_{-\infty}^{\infty} l\lambda^2 \log(1/r) dx$
Which explodes.
I actually expect this, since I keep adding charges at a finite distance from the other wire.
On the other hand though, I think there are well defined solutions for the energy of infinite area capacitors which are worse. Or is it only the fields?