My Problem in my Thesis

Question

Good day! Does anyone have any idea on how to evaluate this integral? $$ \frac{d}{du}\int_{0}^{u}\int_{0}^{u} f(x,y)dxdy $$ I only know some of the Fundamental Theorem of Calculus, but how would I evaluate the derivative of this with respect to $u$ when it is a double integral with $u$ as the upper limits. I would appreciate your responses. I hope you can help me to work this out.

Answer 1

Suppose that $f:[a,b]\times[c,d]\rightarrow\mathbb{R}$ is continuous. Define $F:[a,b]\times[c,d]\rightarrow\mathbb{R}$ by $F(x,y)=\int_{a}^{x}\left(\int_{c}^{y}f(u,v)dv\right)du$. Note that $\frac{\partial F}{\partial x}$ and $\frac{\partial F}{\partial y}$ exist and $\frac{\partial F}{\partial x}(x,y)=\int_{c}^{y}f(x,v)dv$, $\frac{\partial F}{\partial y}(x,y)=\int_{a}^{x}f(u,y)du$. Clearly, $\frac{\partial F}{\partial x}$ and $\frac{\partial F}{\partial y}$ are continuous. Finally, if $a=c$ and $b=d$, we may define $G:[a,b]\rightarrow\mathbb{R}$ by $G(t)=F(t,t)$. Then, by chain rule, \begin{eqnarray*} G'(t) & = & F_{x}(t,t)\cdot1+F_{y}(t,t)\cdot1\\ & = & \int_{a}^{t}f(t,v)dv+\int_{a}^{t}f(u,t)du. \end{eqnarray*}

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Justification: Since $f$ is continuous, $F$ is well-defined and the order of integration can be interchanged (Fubini Theorem). Fix $(x_{0},y_{0})\in[a,b]\times[c,d].$ Let $\phi(u)=\int_{c}^{y_{0}}f(u,v)dv.$ Observe that $\phi$ is continuous on $[a,b]$. For, since $f$ is continuous on the compact set $[a,b]\times[c,d]$, $f$ is uniformly continuous (Cantor Theorem). Let $\varepsilon>0$ be given. Choose $\delta>0$ such that $|f(u,v)-f(u',v')|<\varepsilon$ whenever $(u-u')^{2}+(v-v')^{2}<\delta^{2}$. Let $u,u'\in[a,b]$ with $|u-u'|<\delta$, then for each $v\in[c,d]$, we have $d((u,v),(u',v))=|u-u'|<\delta$, so $|f(u,v)-f(u',v)|<\varepsilon$. It follows that $|\phi(u)-\phi(u')|\leq\int_{c}^{y_{0}}|f(u,v)-f(u',v)|dv\leq\varepsilon(d-c)$. This shows that $\phi$ is continuous. Now, $F(x,y_{0})=\int_{a}^{x}\phi(u)du$. By Fundamental Theorem of Calculus, $\frac{\partial F}{\partial x}(x_{0},y_{0})$ exists and \begin{eqnarray*} \frac{\partial F}{\partial x}(x_{0},y_{0}) & = & \phi(x_{0})\\ & = & \int_{c}^{y_{0}}f(x_{0},v)dv. \end{eqnarray*} Similarly, by writing $F(x,y)=\int_{c}^{y}\left(\int_{a}^{x}f(u,v)du\right)dv$, we can prove that $\frac{\partial F}{\partial y}(x_{0},y_{0})=\int_{a}^{x_{0}}f(u,y_{0})du.$