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How do you compute the following integral:

$$\int_0^t e^{\mu s + \sigma W_s} ds$$

or

$$\int_0^t e^{\mu s + \sigma W_s} dW_s$$

?

Are those integrals stochastic processes of some well-know type (log-normal, for instance)?

Since they are interrelated through Ito's integral, if we compute one of those we will immediately get the other one, so It doesn't matter which integral to consider:

Ito's integral for $e^{\mu t + \sigma W_t}$:

$e^{\mu t + \sigma W_t} = e^{W_0} + \int\limits_0^t (\mu + \frac{\sigma^2}{2})e^{\mu s + \sigma W_s} ds + \int\limits_0^t \sigma e^{\mu s + \sigma W_s} d W_s $

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  • $\begingroup$ There is Ito's symmetry about the variance ... but what do you mean be it does not matter which one to consider? $\endgroup$ – Ric Oct 29 '15 at 8:04
  • $\begingroup$ It does not appear to have an expression in finite form. But it will be very interesting that someone can provide one. $\endgroup$ – Gordon Oct 29 '15 at 14:59
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    $\begingroup$ It will not be log-normal, as a weighted sum of log-normal random variables is no longer log-normal. $\endgroup$ – Gordon Oct 29 '15 at 17:31
  • $\begingroup$ @Richard I meant that if we get the analytical form of one of them, we will be able to get the form for another throught Ito's integral. I've added the formula to clarify my point. $\endgroup$ – Oleg Oct 29 '15 at 19:04

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