$\displaystyle$\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$$then they claim changing the order or integration should result in \displaystyle \zeta_T(v) = \int_{-\infty}^0 ds e^{-rT} q_T(s) \int_{s}^{\infty *} (e^{(1+iv)k} - e^s e^{ivk})dk + \int_{0}^{\infty} e^{-rT} q_T(s) \int_{0}^{s} ( e^s e^{ivk}- e^{(1+iv)k})dk$$
What I dont get is why the first inner integral marked by * goes from s to $\displaystyle$\infty$$and not from s to 0. Can someone explain this to me? thx in advance. 2. I agree with you. I think it should be s to 0. 3. Originally Posted by Ackbeet I agree with you. I think it should be s to 0. thanks a lot. ... so there seems to some hope ;-) so in the follow up they go on that \displaystyle \zeta_T(v) = \int_{-\infty}^0 ds e^{-rT} q_T(s) \int_{s}^{0} (e^{(1+iv)k} - e^s e^{ivk})dk + \int_{0}^{\infty} e^{-rT} q_T(s) \int_{0}^{s} ( e^s e^{ivk}- e^{(1+iv)k})dk$$,
$\displaystyle$\zeta_T(v) =e^{-rT}[\frac{1}{1 + iv} - \frac{e^{rT}}{iv} - \frac{\psi_T(v-i)}{v^2 - iv}]$$where \displaystyle \psi_T(u) = \int_{-\infty}^{\infty}e^{ius}q_T(s) ds$$
$\displaystyle \frac{(i (i + v) (exp(s + i s v) - 1))}{(v^2 + 1)} - \frac{(i exp(s) (exp(i s v) - 1))}{v}$$so the entire integral then is \displaystyle \zeta_T(v) = \int_{-\infty}^{\infty} e^{-rT} q_T(s)\left[\frac{(i (i + v) (exp(s + i s v) - 1))}{(v^2 + 1)} - \frac{(i exp(s) (exp(i s v) - 1))}{v}\right] ds$$ from there i can't see how the solution is derived. There are some things which just seem strange, e.g. in the solution,$\displaystyle $\zeta_T(v) =e^{-rT}[\frac{1}{1 + iv} - \frac{e^{rT}}{iv} - \frac{\psi_T(v-i)}{v^2 - iv}]$$the second term in the braces is not multiplied by \displaystyle e^{-rT}$$. How is this possible if the entire integral is basically multiplied by$\displaystyle $e^{-rT}$\$? any help or hints are greatly appreciated ;-) thx in advance