Changing order or integration

**hofi** · May 27th 2011, 06:43 AM

Hi all, reading a paper i struggled over the following formula:
$\zeta_T(v) = \int_{-\infty}^0 dk e^{ivk} e^{-rT}\int_{-\infty}^k (e^k - e^s)q_T(s)ds + \int_{0}^{\infty} dk e^{ivk} e^{-rT} \int_{k}^{\infty} ( e^s- e^k)q_T(s)ds$
then they claim changing the order or integration should result in
$\zeta_T(v) = \int_{-\infty}^0 ds e^{-rT} q_T(s) \int_{s}^{\infty *} (e^{(1+iv)k} - e^s e^{ivk})dk + \int_{0}^{\infty} e^{-rT} q_T(s) \int_{0}^{s} ( e^s e^{ivk}- e^{(1+iv)k})dk$
What I dont get is why the first inner integral marked by * goes from s to $\infty$ and not from s to 0. Can someone explain this to me? thx in advance.

**Ackbeet** · May 27th 2011, 08:31 AM

I agree with you. I think it should be s to 0.

**hofi** · May 27th 2011, 03:33 PM

Originally Posted by Ackbeet

I agree with you. I think it should be s to 0.

thanks a lot. ... so there seems to some hope ;-)
so in the follow up they go on that
$\zeta_T(v) = \int_{-\infty}^0 ds e^{-rT} q_T(s) \int_{s}^{0} (e^{(1+iv)k} - e^s e^{ivk})dk + \int_{0}^{\infty} e^{-rT} q_T(s) \int_{0}^{s} ( e^s e^{ivk}- e^{(1+iv)k})dk$ ,
i.e. the correct formula (i assume it was a typo) can be rewritten as
$\zeta_T(v) =e^{-rT}[\frac{1}{1 + iv} - \frac{e^{rT}}{iv} - \frac{\psi_T(v-i)}{v^2 - iv}]$
where
$\psi_T(u) = \int_{-\infty}^{\infty}e^{ius}q_T(s) ds$
I struggle with deriving this formula too. Doing the inner integrations, both read
$\frac{(i (i + v) (exp(s + i s v) - 1))}{(v^2 + 1)} - \frac{(i exp(s) (exp(i s v) - 1))}{v}$$ so the entire integral then is
$\zeta_T(v) = \int_{-\infty}^{\infty} e^{-rT} q_T(s)\left[\frac{(i (i + v) (exp(s + i s v) - 1))}{(v^2 + 1)} - \frac{(i exp(s) (exp(i s v) - 1))}{v}\right] ds$
from there i can't see how the solution is derived. There are some things which just seem strange, e.g. in the solution,
$\zeta_T(v) =e^{-rT}[\frac{1}{1 + iv} - \frac{e^{rT}}{iv} - \frac{\psi_T(v-i)}{v^2 - iv}]$
the second term in the braces is not multiplied by $e^{-rT}$ . How is this possible if the entire integral is basically multiplied by $e^{-rT}$ ? any help or hints are greatly appreciated ;-) thx in advance

btw. the paper is can be downloaded here: http://citeseerx.ist.psu.edu/viewdoc...=rep1&type=pdf, relevant equations: 13,14 and 15

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