Leibniz' Rule: Prove $\int_0^y u_{tt}(x,t) dt = u_y (x,y) - u_y (x,0)$ A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Thm 6.8
Statement of Thm 6.8:

Suppose $u$ is harmonic on $\mathbb C$. Then $$v(x,y) := \int_0^y u_x(x,t) dt - \int_0^x u_y(t,0) dt$$  is a harmonic conjugate for $u$.

Pf of Thm 6.8:



Question:

How do we show $$\int_0^y u_{tt}(x,t) dt = u_y (x,y) - u_y (x,0)?$$

It's been awhile since I've done PDEs or multivariable calc, but here goes:
First approach:
$$\int_0^y u_{tt}(x,t) dt = u_t(x,t)|_{\color{green}{0}}^{\color{blue}{y}} = u_t(x,t)|_{t=\color{blue}{y}} - u_t(x,t)|_{t=\color{green}{0}} = u_\color{blue}{y}(x,\color{blue}{y}) - u_\color{green}{0?}(x,\color{green}{0})$$
I mainly don't get why we might have the $y$ in the last term $$u_\color{green}{y}(x,\color{green}{0}).$$
Second approach:
$$\int_0^y u_{tt}(x,t) dt = u_t(x,t)|_{\color{green}{0}}^{\color{blue}{y}} = u_t(x,\color{blue}{y}) - u_t(x,\color{green}{0}) \stackrel{(*)}{=} u_\color{red}{y}(x,\color{blue}{y}) - u_\color{red}{y}(x,\color{green}{0})$$
(*) Here, I guess I get how for the second term we have $u_t(x,\color{green}{0}) = u_\color{red}{y}(x,\color{green}{0}),$ but not how for the first term we have $$u_t(x,\color{blue}{y})=u_\color{red}{y}(x,\color{blue}{y}).$$
 A: If I asked you why
$$\tag 1 \int_a^b f''(t)\,dt = f'(b)-f'(a),$$
you would have an answer: It follows from the FTC, since $f'$ is an antiderivative of $f''.$ That's all that is going on in your situation.
Things might be clearer if we use $D_1, D_2$ for the partial derivatives with respect the first and second variables. With $x$ fixed, define $f(t) = u(x,t).$ Then by the definition of partial derivatives, we have $f'(t) = D_2u(x,t)$ and $f''(t) = D_2(D_2 u)(x,t).$ Thus by $(1),$
$$\int_a^b D_2(D_2 u)(x,t) dt = \int_a^b f''(t)\,dt$$ $$ = f'(b)-f'(a) = (D_2 u)(x,b) - (D_2 u)(x,a).$$
If you think of $a=0,b=y,$ we obtain exactly the expression you asked about.
A: Based on JessicaK's comment:
$$\int_{0}^{y} u_{tt}(x,t)\operatorname{d}t = u_{t}(x,t)\big|_{0}^{y} = [\lim_{h\rightarrow 0} \frac{u(x,t+h)- u(x,t)}{h}]\big|_{0}^{y} \stackrel{(**)}{=} \lim_{h\rightarrow 0} [\frac{u(x,t+h)- u(x,t)}{h}\big|_{0}^{y}] = \lim_{h\rightarrow 0} [[\frac{u(x,y+h)- u(x,y)}{h}] - [\frac{u(x,0+h)- u(x,0)}{h}]]$$
$$ = \lim_{h\rightarrow 0} [\frac{u(x,y+h)- u(x,y)}{h}] - \lim_{h\rightarrow 0} [\frac{u(x,0+h)- u(x,0)}{h}]$$
$$ = u_{y}(x,y)-u_{y}(x,0)$$
Indeed, $t$ is a dummy so we could instead have
$$ = u_{y}(x,y)-u_{t}(x,0)$$
Therefore, the first approach is wrong, and we should actually have
$$\int_0^y u_{tt}(x,t) dt = u_t(x,t)|_{\color{green}{0}}^{\color{blue}{y}} = u_t(x,t)|_{t=\color{blue}{y}} - u_t(x,t)|_{t=\color{green}{0}} = u_t(x,\color{blue}{y}) - u_t(x,\color{green}{0}) = u_y(x,\color{blue}{y}) - u_y(x,\color{green}{0})$$
Note that this is what is done for the second approach and that we do not plug in $t=y$ into the partial derivative, i.e. we say not that $$u_t(x,t)|_{t=\color{blue}{y}} = u_\color{blue}{y}(x,\color{blue}{y})$$ but that $$u_t(x,t)|_{t=\color{blue}{y}} = u_t(x,\color{blue}{y}) = u_y(x,\color{blue}{y}).$$

(**) This relies on convincing one's self that
$$[\lim_{h\rightarrow 0} f(x+h)]\big|_{a}^{b} = \lim_{h\rightarrow 0} [f(x+h)\big|_{a}^{b}],$$
where
$$LHS = [\lim_{h\rightarrow 0} f(x+h)]\big|_{b} - [\lim_{h\rightarrow 0} f(x+h)]\big|_{a}$$
and
$$RHS = \lim_{h\rightarrow 0} [f(b+h) - f(a+h)] = \lim_{h\rightarrow 0} f(b+h) - \lim_{h\rightarrow 0} f(a+h),$$
i.e. convincing one's self that $$[\lim_{h\rightarrow 0} f(x+h)]\big|_{b} = \lim_{h\rightarrow 0} f(b+h),$$
if that isn't already true by definition. I believe we could define $$g(x):=\lim_{h\rightarrow 0} f(x+h)$$ s.t. $$g(b):=\lim_{h\rightarrow 0} f(b+h)=:[\lim_{h\rightarrow 0} f(x+h)]\big|_{b}.$$
