Solving SDE example Let's assume $f_{B_t}(t,B_t) = t*B_t$ ($B_t$ is Brownian motion)

*

*$f(t,B_t) = \frac{1}{2}t*B_t^2$

*$f_t(t,B_t) = \frac{1}{2}*B_t^2$

*$∫_0^Tf_t(t,B_t)dt = ∫_0^T\frac{1}{2}*B_t^2dt$
Above those are solution.
First, what I confused is that How Brownian motion can remain uninfluenced when we calculate derivative about time '$t$'?? (Question about number 2 equation)
Second, if number 2 equation true, it seems that we treat $B_t$ as constant when we do some calculus about '$t$'
then Why $∫_0^T\frac{1}{2}*B_t^2dt$ isn't simplified? I mean, why integral terms are not going away?
I think it would be simplified like this : $\frac{1}{2}*B_t^2*T$
 A: It is somehow missleading with those indices $t,B_t$ at the bottom of $f$.
You have some function $f:\mathbb R_+ \times \mathbb R \to \mathbb R$ and you evaluate it at point $(t,B_t)$ (where the latter is understood in pointwise sense $B_t(\omega)$.
By $f_t$ they mean derivative with respect to the first argument and by $f_{B_t}$ they mean derivative with respect to second argument (you can think of value $x$ instead of $B_t$, it would make it more clear).
(To give you example, if you have some function $g(t,x) = tx$, then they call $g_t$ a derivative with respect to first argument, that is $g_t(t,x) = x$, right?  So that $g_t$ is just a projection on the second argument. Even if now, you change names of your variables, $g_t$ is still the same function, it's just a denotation. Having said that, for some functions $\psi,\varphi$ of variable $t$, you still have $g_t(\varphi(t),\psi(t)) = \psi(t)$, because $g_t$ just leaves us with second argument, which in the latter case is just $\psi(t)$ and it does not matter that $\psi$ depends on $t$, because that letter $t$ in $g_t$ is just for convenience, it indicates we're differentiating with respect to first variable)
If you want to be more clear with yourself, you can change $f_t,f_{B_t}$ with $f_1,f_2$ meaning derivatives with respect to first and the second variable.
In that langueage, your equations means $f_2(t,B_t) = t B_t$ (or $f_2(t,x) = tx$ if you evaluate it pointwise for some $B_t(\omega) = x$).
And I think it's not that hard to understand it with $f_2(t,x) = tx$, since then $f(t,x) = \frac{1}{2}tx^2 $ (one of possibilities, we could add any function of variable $t$, because it's constant with respect to $x$) so that $f_t(t,x) := f_1(t,x)= \frac{1}{2}x^2 $ and as we said, $f_t(t,B_t) = f_1(t,B_t) = \frac{1}{2}B_t^2$, since it's just function $f_1$ evaluated at point $(t,B_t)$.
As for the last equality with integral, it's a little bit different. $B_t$ depends on $t$ on it's own.
$$ \int_0^T f_1(t,B_t)dt = \int_0^T \frac{1}{2}B_t^2 dt $$ You see, last "dt" affects whole term under integral, which is $f_1(t,B_t)$.
It's the same as with function $g$ above. Having $\int_0^T g_t(t,x)dt = \int_0^T x dt = Tx$, but if you want to evaluate $g$ at those points $(\varphi(t),\psi(t))$ then you indeed have $\int_0^T g_t(\varphi(t),\psi(t))dt = \int_0^T \psi(t) dt$ and you cannot just leave it as $T\psi(t)$
To cuy long story short, subsript near function (even if they use some variable for it) indicates which variable (meaning which one of arguments) we're integrating with respect to, but under integral, changes $dx,dt,dy$ or so on, affects whole term under integral, every one depending on the variable which we're integrating with respect to.
