= Solution
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.
Back to article page
