# Fourier transform of a European put

In book The concepts and practice of mathematical finance, in the context of illustrating the stochastic volatility model, the Fourier transform $$\hat{P}(\xi, V, T)$$ of a European put $$P(x, V, T)$$ is listed as

$$\hat{P}(\xi, V, T) = - \frac{K^{i \xi + 1}}{\xi^2 - i \xi}$$

where $$x = \log S$$ is the logarithm of real world underlying stock values $$S$$, $$\xi$$ the respective Fourier variable, $$V$$ is square vol, $$T$$ the time maturity, $$i$$ the imaginary unit. My development of the Fourier integral, using $$K - e^x$$ as payout for the put, leads to a different result:

$$\hat{P}(\xi, V, T) = \int e^{i \xi x} ( K - e^x ) dx =\\ \frac{K}{i \xi} e^{i \xi x} - \frac{1}{1 + i \xi} e^{(1 + i \xi) x} = e^{i \xi x} \frac{K + i \xi (K - e^x)}{i \xi - \xi^2}$$

Where have I gone wrong? How to reach the book result?

The generalised Fourier transform $$\hat{P}(z)$$ of the payoff of a put option with $$P(x)=\max\{e^k-e^x,0\}$$ is \begin{align*} \hat{P}(z) &= \int_{-\infty}^\infty e^{izx} \left( e^k - e^x \right)^+ \mathrm{d}x \\ &= \int_{-\infty}^k \left( e^ke^{izx}-e^{i(z-i)x} \right)\mathrm{d}x \\ &= \left[ e^k\frac{e^{izx}}{iz} -\frac{e^{i(z-i)x}}{i(z-i)} \right]_{-\infty}^k \\ &= \left( e^k\frac{e^{izk}}{iz} - \frac{e^{i(z-i)k}}{i(z-i)} \right)-0\\ &= \frac{e^{i(z-i)k}}{iz} - \frac{e^{i(z-i)k}}{i(z-i)} \\ &= -\frac{e^{ik(z-i)}}{z(z-i)}. \end{align*} The computation above is only valid if the summand for $$x=-\infty$$ indeed equals zero. In general, if $$z\in\mathbb{R}$$, $$\lim\limits_{x\to-\infty}e^{ixz}$$ does not make sense since $$e^{izx}$$ merely describes points on the unit circle around the origin. However, if $$z=a+ib$$ is complex, $$e^{izx}=e^{-bx} e^{iax}$$ which at least converges to zero as $$x\to-\infty$$ if $$b=\text{Im}(z)<0$$ since it contracts the unit circle into the origin. Equivalently, $$\lim\limits_{x\to-\infty} |e^{izx}|=\lim\limits_{x\to-\infty} e^{-bx}=0$$ if $$b=\text{Im}(z)<0$$.
Thus, we require $$\text{Im}(z)<0$$ for the first summand and $$\text{Im}(z-i)<0$$ for the latter. Both conditions together lead to $$\text{Im}(z)<0$$. Consequently, the generalised Fourier transform $$\hat{P}(z)$$ is only well-defined in the open strip $$\mathcal{S}_P=\{z\in\mathbb{C}:\text{Im}(z)<0\}$$.
The strip of regularity for a call option is $$\mathcal{S}_C=\{z\in\mathbb{C}:\text{Im}(z)>1\}$$.
Note: it is no coincidence to have $$i$$ and $$0$$ as poles of the payoff transform. You can use inversion theorems to see how they relate to $$N(d_1)$$ and $$N(d_2)$$ (or more general exercise probabilities).
• Thank you for specifying the range where the integral does not diverge to $\infty$. So substitute $K = e^{k}$ to obtain the book result. Jul 18, 2021 at 16:30
• @Giogre Precisely. You only need to be careful with the integral boundaries. The calculation itself is easy. And you’re right. I wrote $k=\ln(K)$ for the log-strike price. Jul 18, 2021 at 16:31