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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

enter image description here

KT8
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    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. – Frido May 30 '23 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 May 30 '23 at 18:11
  • I don't recommend sharing links of questionable legality. You could however post a screenshot or even better write down the main points. – Frido May 30 '23 at 18:13
  • Edited the question including that page from the book – KT8 May 30 '23 at 18:19
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    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. – Frido May 30 '23 at 18:21
  • Sure, my question is another though. Why can't I just use that same argument to evaluate the first integral? – KT8 May 30 '23 at 18:24
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    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) – Frido May 30 '23 at 18:52
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    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. – Frido May 30 '23 at 19:03
  • Thanks Frido. My question comes from the fact that I think that you can also do that same contour integral in the first term. It seems like you can't (otherwise why isn't it done), but I can't come out with a mathematical argument for that. – KT8 May 30 '23 at 19:29
  • Your doubt is completely legitimate. There is no ground for applying the residue theorem. This derivation of the book is woefully unsubstantiated and wrong. I will give you a rigorous derivation, later. – Hans Jun 20 '23 at 04:26
  • @Frido: See my proof below. The route in the book is illegitimate and unnecessarily convoluted. – Hans Jul 01 '23 at 16:52
  • @Hans Thanks Hans, +1 for your derivation, agree that Rouah's text seems to complicate matters only. – Frido Jul 02 '23 at 19:36

2 Answers2

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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.

p.s.
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  • Thanks for your answer p.s.! But, couldn't you close the contour for that case ($l<0$) using the upper half-plane? – KT8 Jun 12 '23 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. Jun 12 '23 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 Jun 13 '23 at 06:45
  • These are a lot of iffy statements. See my direct and simple proof https://quant.stackexchange.com/a/75949/6686. – Hans Jun 26 '23 at 14:47
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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.

Hans
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