Leibniz's integral rule :
Suppose $p(x)$ and $q(x)$ are differentiable on $\Bbb R$. Let $f:\Bbb R^2\to \Bbb R$ and $\frac{\partial f}{\partial y} $ are continuous on $\Bbb R$. Define $F(y)=\int _{p(y)} ^{q(y)} f(x,y) dx$ , $y\in \Bbb R$. Then $F$ is differentiable on $\Bbb R$ and $F'(y)=f(q(y),y)q'(y)-f(p(y),y)p'(y)+\int_{p(y)} ^{q(y)}\frac{\partial f}{\partial y}(x,y)dx$.
My attempt:
Let $G(x_1,x_2,x_3)=\int_{x_1}^{x_2}f(x,x_3)dx$. I want to show that $G$ is a $C^1$ mapping.
$\frac{\partial G}{\partial x_1}(x_1,x_2,x_3)=\frac{\partial}{\partial x_1} \int_{x_1}^{x_2}f(x,x_3)dx=-\frac{\partial}{\partial x_1}\int_{x_2}^{x_1}f(x,x_3)dx$ $\;\;\;(\star)$
My question is can I use the FTC ( Fundamental theorem of calculus ) to $(\star)$. I am confused about that the FTC is one variable, but $f$ is two variables function. Should I think $x_3$ in $(\star)$ to be an fixed number then it is one variable function?
If I use the FTC to $(\star)$, then what is the indefinite integral of $f$ ? Is $H(x_1)=\int _a ^{x_1}f(x,x_3)dx$ or $H(x_1,x_3)=\int _a ^{x_1}f(x,x_3)dx$ ? ( because $f$ is continuous on $\Bbb R^2$, so I can write $\int_{x_2}^{x_1}f(x,x_3)dx$ to be $\int_{a}^{x_1}f(x,x_3)dx-\int_{a}^{x_2}f(x,x_3)dx$ for some $a\in \Bbb R$. )
Appreciate your help!
By the way, I have read the answer in Proof for the "Fundamental Calculus Theorem" for two variables.
Best Answer
Yes, you can apply the Fundamental Theorem of Calculus to compute$$\frac\partial{\partial x_1}\int_{x_1}^{x_2}f(x,x_3)\,\mathrm dx,\tag1$$since here you can treat $x_2$ and $x_3$ as if they where constants; you will get that $(1)$ is equal to $-f(x_1,x_3)$. And you will also get that$$\frac\partial{\partial x_2}\int_{x_1}^{x_2}f(x,x_3)\,\mathrm dx=f(x_2,x_3).$$