# 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

• I suspect it has to do with taking the limit in the second integral. But it's hard (for me) to say anything more without knowing the original problem and context. Commented May 30, 2023 at 17:59
• It's from chapter 3 of that book. He's proving the Gil-Pelaez inversion theorem. The book can be found online but I'm not sure if it's against SE rules to share those kind of links
– KT8
Commented May 30, 2023 at 18:11
• Well, yes, so the inner integral in 3.32 is an indefinite integral and hence you need to take the limit which can be evaluated using the residue theorem. Commented May 30, 2023 at 18:21
• Why would you use the res.thm in the first as the integration interval does not include the pole u= -i? To evaluate the second integral I think some kind of contour is used with radius R and then the limit is taken. So this contour will have the pole inside it. This is how I see it (without actually doing the calculations) Commented May 30, 2023 at 18:52
• I'll try to look into this more when I have more time and give a proper answer hopefully, but I suspect the reason is as explained in my previous comment. Commented May 30, 2023 at 19:03

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.

• Thanks a lot Hans!
– KT8
Commented Jun 26, 2023 at 8:55

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.

• Thanks for your answer p.s.! But, couldn't you close the contour for that case ($l<0$) using the upper half-plane?
– KT8
Commented Jun 12, 2023 at 13:00
• It depends on $\varphi_2$ whether the integral on the semi-circle would be zero. Also, you wouldn't get the residue at $-i$ if you close in the upper half plane.
– p.s.
Commented Jun 12, 2023 at 16:55
• I agree on the $\varphi_2$ argument. But what I meant about the upper half plane is closing it at the $-i$ straight. Wouldn't that be possible? Sorry but I'm a bit rusty in complex calculus at the moment...
– KT8
Commented Jun 13, 2023 at 6:45
• These are a lot of iffy statements. See my direct and simple proof quant.stackexchange.com/a/75949/6686.
– Hans
Commented Jun 26, 2023 at 14:47