Write $w=z-1=X+iY$. The phase and its derivative are
$$
\phi=w^3+3w,
\qquad
\phi'=3(w^2+1),
$$
so the <saddle points> are
$$
\boxed{z_+=1+i,qquad z_-=1-i,}
$$
with $\phi(z_+)=2i$ and $\phi(z_-)=-2i$. The contour geometry can be drawn from
$$
\operatorname{Re}\phi=X(X^2-3Y^2+3),
\qquad
\operatorname{Im}\phi=Y(3X^2-Y^2+3).
$$
The stationary-phase level $\operatorname{Re}\phi=0$ consists of $X=0$ and $X^2-3Y^2+3=0$, meeting at the saddles. The steepest curves through $z_\pm$ are the levels $\operatorname{Im}\phi=\pm2$. Far away, sectors with $\cos(3\arg w)>0$ are exponential hills and those with $\cos(3\arg w)<0$ are valleys.

The stated contour deforms through the upper saddle. If $z=z_++s$, then
$$
\phi(z)=2i+3is^2+s^3.
$$
The descent tangent has $s=e^{i\pi/4}t$, because $3is^2=-3t^2$. The <simple-saddle contribution in steepest descent> is therefore
$$
f(\lambda)\sim e^{2i\lambda}e^{i\pi/4}
\int_{-\infty}^{\infty}e^{-3\lambda t^2}dt
=\boxed{\sqrt{\frac\pi{3\lambda}}e^{i\pi/4+2i\lambda}.}
$$

When the contour begins at $z=1$, deform it first from the endpoint into the decaying negative-real direction and then onto the same upper-saddle descent path. Near the endpoint, $s=z-1$ and $\phi=3s+O(s^3)$, so the <endpoint contribution in steepest descent> is
$$
\int_0^{-\infty}e^{3\lambda s}ds=-\frac1{3\lambda}.
$$
Adding the saddle and endpoint pieces gives
$$
\boxed{f(\lambda)\sim
\sqrt{\frac\pi{3\lambda}}e^{i\pi/4+2i\lambda}
-\frac1{3\lambda}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 336 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-336/1/b/solution