Erroneous reasoning over equivalence of hankel function with logarithm We have the following equation/solution pairs:
$$(\nabla^2+k^2)G(\mathbf{x},\mathbf{x'}) = \delta(\mathbf{x}-\mathbf{x'})  \to G(\mathbf{x},\mathbf{x'}) = \frac{i}{4}H_0^{(1)}(k|\mathbf{x}-\mathbf{x'}|)$$
$$\nabla^2G(\mathbf{x},\mathbf{x'}) = \delta(\mathbf{x}-\mathbf{x'})  \to G(\mathbf{x},\mathbf{x'}) = -\frac{1}{2\pi}\ln(|\mathbf{x}-\mathbf{x'}|)$$
If these two equations are to be consistent with each other as we let $k \to 0$, I draw the conclusion:
$$\lim_{k \to 0} H_0^{(1)}(kz) = -\frac{2}{\pi i}\ln(z).$$
However from  I find the definition 
$$H_n^{(1)}(kz) = \frac{1}{i\pi}\int_0^{\infty} \frac{e^{kz/2(t-1/t)}}{t^{n+1}} dt$$
and so 
$$\lim_{k \to 0} H_0^{(1)}(kz) = \frac{1}{i\pi}\int_0^{\infty} \frac{1}{t} dt = \frac{1}{i\pi}[\ln t]_0^{\infty}.$$
Which doesnt seem to mathc with what I got above. I must have made some faulty assumption somewhere but cant quite figure out where.

The very first conclusion is, strictly speaking, incorrect. The asymptotics of the Hankel function is
$$H_0^{(1)}(z\to 0)\sim \frac{2i}{\pi}\ln z+O(1).$$
This looks similar to your third formula, but in fact that equation should contain an additional term proportional to $\ln k$ which diverges as $k\to 0$. This reflects the logarithmic divergence you get in the end from the integral representation. 
On the other hand, adding a constant to any solution of the second equation we still obtain a solution. Deriving its fundamental solution can therefore be heuristically understood as the limit of the fundamental solution of the first equation supplemented by its regularization by a divergent (but constant!) $\ln k$-piece.
