Set $z=x_0+iy$. Since $|M(z)|\leq\mathbb Ee^{x_0\xi}$ and $|z(z-1)|^{-1}$ is integrable over $y\in\mathbb R$, the proposed contour integral is absolutely convergent. The same bound justifies the <Fubini's theorem> when substituting $M(z)=\mathbb Ee^{z\xi}$:
$$
\begin{aligned}
\frac1{2\pi i}\int_{x_0-i\infty}^{x_0+i\infty}\frac{M(z)}{z(z-1)e^{(z-1)k}}\,dz
&=e^k\mathbb E\!\left[\frac1{2\pi i}\int_{x_0-i\infty}^{x_0+i\infty}\frac{e^{z(\xi-k)}}{z(z-1)}\,dz\right]\\
&=e^k\mathbb E(e^{\xi-k}-1)^+=C(k).
\end{aligned}
$$
The last equality uses the contour identity provided in the original PDF. Consequently
$$
\boxed{C(k)=\frac1{2\pi i}\int_{x_0-i\infty}^{x_0+i\infty}\frac{M(z)}{f(z,k)}\,dz\qquad(x_0>1).}
$$
This is <contour inversion for call prices>. The supplied TeX corrupts the permitted identity into a reciprocal and drops the imaginary units from its endpoints; the PDF has $(e^a-1)^+$ and the vertical contour used above.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 211 / 5 / b / Solution

The <Mellin transform of call prices> can be inverted along a vertical contour with real part $x_0>1$: $C(k)=\frac1{2\pi i}\int_{x_0-i\infty}^{x_0+i\infty}\mathbb Ee^{z\xi}/(z(z-1)e^{(z-1)k})\,dz$. A finite <exponential moment> at $x_0$ and the quadratic denominator give absolute convergence.

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