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Year 4 of the mathematics course of the University of Oxford

Perturbation methods and applied complex variables course of the University of Oxford

Past exam of the perturbation methods and applied complex variables course of the University of Oxford

2009

{c}

On the selected <steepest descent contour> put $t=e^{i\pi/6}r$, $r\geq0$. Then $it^3=-r^3$ and $dt=e^{i\pi/6}\,dr$. Its exact contribution is
$$
I_0(x)=e^{i\pi/6}\int_0^\infty e^{-xr^3}\,dr
=\frac{e^{i\pi/6}}{3x^{1/3}}\int_0^\infty e^{-s}s^{-2/3}\,ds.
$$
Use the <Gamma function> integral and its recurrence $\Gamma(4/3)=\Gamma(1/3)/3$ to obtain
$$
\boxed{I_0(x)=e^{i\pi/6}\Gamma(4/3)x^{-1/3}.}
$$
There is no remainder here: the complete infinite descent ray was integrated exactly.

Solution

Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2009 / A / 1 / iii / Solution

Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2009 / A / 1 / iii / Solution

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Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2009 / A / 1 / iii / Solution

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