Expected value $\mathbb{E}\left[e^{-kt/m}\int_0^t e^{ks/m} dB_s\right]$

Question

I want to show expected value of below Ito integral goes to zero $$\mathbb{E}\left[e^{-kt/m}\int_0^t e^{ks/m} dB_s\right]$$ where $k, m $ are constant .

Yesterday I found a question like this in math stack exchange ,But I can't find it now . I have to solve it , and I don't have clue . I am thankful if someone guide me to prove this . Thanks in advanced .( a clue or address of a page that contain like this question )

Answer 1

Isn't it a property of the Itô integral that $$\mathbb{E} \left[ \int_0^t f(s,\omega) \, dB(s) \right] = 0,$$ for $f$ progressively measurable and such that $\mathbb{E}[\int_0^t f(t,\omega)^2 dt] < +\infty$?

Perhaps this is the question you are looking for, but the integral to evaluate in this question has a random variable (i.e., $X$) as a limit of the integral, while in your it seems to be a real number (i.e., $t$) (or better it is not specified).

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Edit

I found also this question that could really be what you are looking for!