Contour Closure and Fourier Integration in the Heston Model

# Complex Integral in Rouah's Heston book


# Complex Integral in Rouah's Heston book












I have a silly question regarding complex calculus, in which I'm a bit rusty at the moment. In F. Rouah's book The Heston Model and Its Extensions in Matlab and C# the following appears:

> Now evaluate the inner integral in Equation (3.32), as was done in (3.14). This produces $$\begin{align} \Pi_1 & = \dfrac{1}{2\pi}\int_{-\infty}^{\infty} \varphi_2(u) \dfrac{e^{−i(u+i)l}}{ i(u + i)} du − \dfrac{1}{2\pi} \text{lim}_{R\to\infty}\int_{-\infty}^{\infty} \varphi_2(u) \dfrac{e^{−i(u+i)R}}{ i(u + i)} du\\ & = I_1 − I_2. \end{align}$$ The second integral is a complex integral with a pole at $u = −i$. The residue there is, therefore, $\varphi_2(−i)/i$. Applying the Residue Theorem, we obtain $$ I_2 = \text{lim}_{R\to\infty} \dfrac{1}{2\pi} \Bigg[ -2 \pi i \times \dfrac{ \varphi_2 (−i)}{i} \Bigg] = -\varphi_2(-i).$$

I think I see the point on solving the integral using the residue theorem and how it works. However, the question that rises is: Why can't I just do the same for the first integral?

Thanks

Edit: a screenshot of that page, for completion

## Answer by Hans (score 2, accepted)

https://quant.stackexchange.com/a/75949

The derivation in the book is unnecessarily complicated, and plain erroneous in many places, such as taking limit of $R$. Here is a simple and direct proof.

Proof: Suppose $q\in L_1(-\infty,\infty)\cap L_2(-\infty,\infty)$. This implies the Fourier transform $\hat q\in L_2(-\infty,\infty)$. This is not but needed to be stated by the book.

\begin{align} \Pi_1&=\int_l^\infty e^xq(x)\,dx \\ &=\bigg(\int_{-\infty}^\infty-\int_{-\infty}^l\bigg)e^xq(x)\,dx \\ &= 1-\int_{-\infty}^l dx\,e^x \frac1{2\pi}\int_{-\infty}^\infty du e^{-iux}\hat q(u) \\ &= 1- \frac1{2\pi}\int_{-\infty}^\infty du\, \hat q(u)\int_{-\infty}^l dx\,e^{(1-iu)x} \tag1\label{eq:Fb}\\ &= 1+\frac{e^l}{2\pi i}\int_{-\infty}^\infty du\,\hat q(u)\frac{e^{-iul}}{u+i}. \end{align} The interchanging of the order of integration resulting in Equation \eqref{eq:Fb} holds because $$\bigg(\int_{-\infty}^l dx\int_{-\infty}^\infty du\,e^x |e^{-iux}\hat q(u)|\bigg)^2\le \int_{-\infty}^l dx\,e^{2x}\int_{-\infty}^\infty du\,|\hat q(u)|^2 <\infty$$ by the Cauchy-Schwardtz inequality and $\hat q\in L_2(-\infty,\infty)$ satisfying the premise of the Fubini's theorem. $\quad\blacksquare$

Indeed, the same methodology in effect proves the Plancherel's theorem of which this problem is an example.

## Answer by p.s. (score 2)

https://quant.stackexchange.com/a/75836

More context is needed, but presumably the difference is that $\ell <0$. Since $R>0$, you can close the integral in the bottom half-plane as the expression $e^{-i(u+i)R}$ goes to zero exponentially in the bottom great semi-circle. (I assume $\varphi_2$ doesn't increase too quickly to cause a problem.) But if $\ell<0$ then the expression $e^{-i(u+i)\ell}$ increases exponentially in the bottom half-plane, so if you were to try to apply the residue theorem there, you'd get a term for the semi-circle which doesn't go to zero.