How to differentiate $a(t-1)+bt+(1-t)\int_{0}^{t}\frac{dB_s}{1-s}$

Question

someone can help me to differentiate $$a(t-1)+bt+(1-t)\int_{0}^{t}\dfrac{dB_s}{1-s}?$$

I've tried but I really don't know how to do with the last part..

Thank you somuch for your help

Answer 1

Consider $$X_t=a(t-1)+bt+(1-t)Y_t,\qquad Y_t=\int_{0}^{t}\frac{\mathrm dB_s}{1-s},$$ then $$\mathrm dY_t=\frac{dB_t}{1-t},\qquad \mathrm dX_t=a\mathrm dt+b\mathrm dt+(1-t)\mathrm dY_t-Y_t\mathrm dt,$$ hence $$\mathrm dX_t=\mathrm dB_t+\left(a+b-\int_{0}^{t}\frac{\mathrm dB_s}{1-s}\right)\mathrm dt,$$ and $$\mathrm d\langle X\rangle_t=\mathrm dt.$$ Edit: In full generality, if $Z_t=U_tV_t$, Itô's formula reads $$\mathrm dZ_t=U_t\mathrm dV_t+V_t\mathrm dU_t+\mathrm d\langle U,V\rangle_t.$$ Using this for $U_t=1-t$ and $V_t=Y_t$, one gets $\mathrm d\langle U,V\rangle_t=0$ because the quadratic variation of $U$ is zero, hence one is left with the sum $$U_t\mathrm dV_t+V_t\mathrm dU_t=(1-t)\mathrm dY_t-Y_t\mathrm dt.$$

Answer 2

Hint: use the Ito formula in the form:

$$ d\left[\left(\alpha(t) +\int_0^t a_s dB_s\right) \left(\beta(t) + \int_0^t b_s dB_s\right)\right] \\= \left(\alpha(t)+\int_0^t a_s dB_s\right)\left(d\beta(t) + b_t dB_t\right) + \left( d\alpha(t) + a_tdB_t \right)\left(\beta(t) + \int_0^t b_s dB_s\right) + a_sb_s ds $$ where $\alpha,\beta:\Bbb R^+\to \Bbb R$.