Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 48 / 2

For real $s$, the multiplier $e^{ias}$ has modulus one andThe triangle inequality applied to the integral therefore givesThe hypotheses $s+\sigma >0$ and $t>0$ ensure endpoint convergence. Differentiation in $s$ is justified on compact parameter intervals by integrable bounds with an additional factor $1+|\log x|$. Differentiating both the exponential multiplier and the power of $x$ makes the logarithmic denominator cancel:Here the scaled Gamma function integral is obtained by $u=tx$. Integrating this derivative from zero to $s$ yieldsThe same modulus bound also gives the useful parameter-shift remainder bound for a logarithmic Laplace integralbecause the denominator magnitude is at least $a$.

Fix a small $s>0$, for example $s<\sigma /2$, independently of $t$. Taking imaginary parts of the identity in part (a) givesThe parameter-shift remainder bound for a logarithmic Laplace integral makes the last term $O(t^{-\sigma -s})$, smaller than $t^{-\sigma }{L}^{-k}$ for every fixed $k$.

The endpoint function $h$ is analytic near zero. For any fixed $N$, its Taylor expansion isIntegrating this expansion against $e^{-Lr}$ useswhile the error is $O(L^{-N-2})$ and the extension from $s$ to infinity contributes an exponentially small tail. This is Watson's lemma, and givesEquivalently the coefficients have the stated exponential generating function $h(r)={\sum }_{n\ge 0}{\beta }_n{r}^n/n!$. In particular,Thus the apparent $t^{-\sigma }(\log t{)}^{-1}$ term vanishes: the first nonzero terms are $t^{-\sigma }[\Gamma (\sigma )(\log t{)}^{-2}+2{\Gamma }^{\prime }(\sigma )(\log t{)}^{-3}+\cdots ]$. The derivatives of the gamma function supply the successive logarithmic coefficients.

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.