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The relevant boundary values are $a=\pi$ and $\sigma =0$, but the original complex integral defining $g(0)$ requires $\sigma >0$. Substituting zero into it is invalid: its real part has a nonintegrable $dx/(x\log x)$ singularity near zero. Its imaginary part is different, since for positive $\sigma$This has a finite limit as $\sigma \downarrow 0$. On $(0,1)$, $x^{\sigma -1}\le x^{-1}$, whose displayed logarithmic denominator makes the integral finite; at infinity, a fixed upper bound on $\sigma$ and exponential damping provide an integrable dominator. Thus dominated convergence gives $N(t)={\lim }_{\sigma \downarrow 0}F(t,\pi ,\sigma )$.

For the asymptotics, take the imaginary-part parameter identity before passing to that limit, with a fixed $0<s<1/2$. Its endpoint factor converges toby the Gamma reflection formula. Near zero, $\sin (\pi r)\Gamma (\sigma +r)$ is bounded uniformly for small $\sigma \ge 0$, because the sine zero compensates the gamma pole. At $s>0$, the remainder integral itself remains absolutely convergent at $\sigma =0$ and is $O(t^{-s})$. ConsequentlyThe endpoint function is analytic with value one, so Watson's lemma proves the logarithmic expansion of Ramanujan's functionIn particular, with $\gamma$ Euler's constant,This use of imaginary-part regularization at a zero Mellin exponent matters: the positive-$\sigma$ coefficient ${\beta }_0$ in part (b) is always zero, whereas the boundary endpoint function has $h_0(0)=1$. The fixed-positive-parameter expansion is not uniform as $\sigma \downarrow 0$, so its coefficients cannot simply be sent to that boundary term by term.