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You can write $h(s)=\int_0^Tf(t)dtH(s)$ then your expectation could be written as $E[ \int_{o}^{T} h(s) dW(s)]$. This integral is $0$ if you can prove $(\int_{o}^{t} h(s) dW(s))$ to be a martingale, for example $E(\int_{o}^{T} h^2(s) ds)<+\infty$.

As you noticed in your comment, this procedure is correct if we can suppose that the process $h(s)$ is adapted, which is the case if $f$ is deterministic. Otherwise we can the decompose the integral as the in the following

$$\int_0^T(\int_0^Tf(t)dt)H(s)dW_s =\int_0^T(\int_0^sf(t)dt)H(s)dW_s+\int_0^T(\int_s^Tf(t)dt)H(s)dW_s$$

then with interverting of the order of integration in the second integral we can write $$\int_0^T(\int_s^Tf(t)dt)H(s)dW_s=\int_0^T(\int_0^tH(s)dW_s)f(t)dt$$

The process apperaing in the integrals are now adapted, and you can add the condition so that Fubini works and to justify the martangality of the integrals. Hope thsi help.

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You can write $h(s)=\int_0^Tf(t)dtH(s)$ then your expectation could be written as $E[ \int_{o}^{T} h(s) dW(s)]$. This integral is $0$ if you can prove $(\int_{o}^{t} h(s) dW(s))$ to be a martingale, for example $E(\int_{o}^{T} h^2(s) ds)<+\infty$.