Show that $F_x(x,y) = P(x,y), F_y(x,y) = Q(x,y)$.












1












$begingroup$


In Complex Variables and Applications by Brown and Churchill it comes:




When the point $(x_0,y_0)$ is kept fixed and $(x , y)$ is allowed to vary throughout a simply connected domain $D$, the integral represents a single-valued function $$F(x,y) = int_{(x_0,y_0)}^{(x,y)} P(s,t) ds+ Q(s,t) dt$$ of $x$ and $y$ whose first-order partial derivatives are given by the equations $$F_x(x,y) = P(x,y), F_y(x,y) = Q(x,y).$$




My question is how $F_x(x,y) = P(x,y), F_y(x,y) = Q(x,y)$ hold?



I searched MSE for similar question, but no much similarities with other questions and Leibniz integral rule couldn't be much help: For, there are variables in upper bound (actually two), partial derivative with respect to one variable makes the other function zero!! and confusion of different variables 'inside' the integral.






real-analysis integration complex-analysis partial-derivative greens-theorem







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    In Complex Variables and Applications by Brown and Churchill it comes:




    When the point $(x_0,y_0)$ is kept fixed and $(x , y)$ is allowed to vary throughout a simply connected domain $D$, the integral represents a single-valued function $$F(x,y) = int_{(x_0,y_0)}^{(x,y)} P(s,t) ds+ Q(s,t) dt$$ of $x$ and $y$ whose first-order partial derivatives are given by the equations $$F_x(x,y) = P(x,y), F_y(x,y) = Q(x,y).$$




    My question is how $F_x(x,y) = P(x,y), F_y(x,y) = Q(x,y)$ hold?



    I searched MSE for similar question, but no much similarities with other questions and Leibniz integral rule couldn't be much help: For, there are variables in upper bound (actually two), partial derivative with respect to one variable makes the other function zero!! and confusion of different variables 'inside' the integral.






    real-analysis integration complex-analysis partial-derivative greens-theorem







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      In Complex Variables and Applications by Brown and Churchill it comes:




      When the point $(x_0,y_0)$ is kept fixed and $(x , y)$ is allowed to vary throughout a simply connected domain $D$, the integral represents a single-valued function $$F(x,y) = int_{(x_0,y_0)}^{(x,y)} P(s,t) ds+ Q(s,t) dt$$ of $x$ and $y$ whose first-order partial derivatives are given by the equations $$F_x(x,y) = P(x,y), F_y(x,y) = Q(x,y).$$




      My question is how $F_x(x,y) = P(x,y), F_y(x,y) = Q(x,y)$ hold?



      I searched MSE for similar question, but no much similarities with other questions and Leibniz integral rule couldn't be much help: For, there are variables in upper bound (actually two), partial derivative with respect to one variable makes the other function zero!! and confusion of different variables 'inside' the integral.






      real-analysis integration complex-analysis partial-derivative greens-theorem







      share|cite|improve this question











      $endgroup$




      In Complex Variables and Applications by Brown and Churchill it comes:




      When the point $(x_0,y_0)$ is kept fixed and $(x , y)$ is allowed to vary throughout a simply connected domain $D$, the integral represents a single-valued function $$F(x,y) = int_{(x_0,y_0)}^{(x,y)} P(s,t) ds+ Q(s,t) dt$$ of $x$ and $y$ whose first-order partial derivatives are given by the equations $$F_x(x,y) = P(x,y), F_y(x,y) = Q(x,y).$$




      My question is how $F_x(x,y) = P(x,y), F_y(x,y) = Q(x,y)$ hold?



      I searched MSE for similar question, but no much similarities with other questions and Leibniz integral rule couldn't be much help: For, there are variables in upper bound (actually two), partial derivative with respect to one variable makes the other function zero!! and confusion of different variables 'inside' the integral.






      real-analysis integration complex-analysis partial-derivative greens-theorem




      real-analysis integration complex-analysis partial-derivative greens-theorem






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 15 '18 at 7:18







      72D

















      asked Dec 14 '18 at 2:25









      72D72D

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      514116






















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          $begingroup$

          Fix a point $(x,y)$. Since the domain $D$ is open, there is some small disk $Delta$ centered at $(x,y)$ which is entirely contained in $D$. The line segment $(a,y) to (x,y)$ is thus contained in $D$ for some constant $a$ sufficiently close to $x$. Since we have independence of the integration path, choose a path which starts by going from $(x_0,y_0)$ to $(a,y)$, followed by the line segment $(a,y) to (x,y)$. We then have
          $$F(x,y)=int_{(x_0,y_0)}^{(a,y)} P(s,t) ,mathrm{d} s+ Q(s,t), mathrm{d}t+int_{(a,y)}^{(x,y)} P(s,t) ,mathrm{d} s+ Q(s,t), mathrm{d}t . $$
          Taking the partial derivative w.r.t. $x$ we notice that the first term vanishes, meaning
          $$F_x(x,y)=frac{partial}{partial x} int_{(a,y)}^{(x,y)} P(s,t) ,mathrm{d} s+ Q(s,t), mathrm{d}t. $$

          This integral can be rewritten using the parameterization $s=u,t=y$, for $a leqslant u leqslant x$, giving
          $$F_x(x,y)=frac{partial}{partial x} int_a^x P(u,y) mathrm{d} u. $$
          The Fundamental Theorem of Calculus then gives $F_x=P$. A similar approach gives $F_y=Q$.






          share|cite|improve this answer









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            1












            $begingroup$

            Fix a point $(x,y)$. Since the domain $D$ is open, there is some small disk $Delta$ centered at $(x,y)$ which is entirely contained in $D$. The line segment $(a,y) to (x,y)$ is thus contained in $D$ for some constant $a$ sufficiently close to $x$. Since we have independence of the integration path, choose a path which starts by going from $(x_0,y_0)$ to $(a,y)$, followed by the line segment $(a,y) to (x,y)$. We then have
            $$F(x,y)=int_{(x_0,y_0)}^{(a,y)} P(s,t) ,mathrm{d} s+ Q(s,t), mathrm{d}t+int_{(a,y)}^{(x,y)} P(s,t) ,mathrm{d} s+ Q(s,t), mathrm{d}t . $$
            Taking the partial derivative w.r.t. $x$ we notice that the first term vanishes, meaning
            $$F_x(x,y)=frac{partial}{partial x} int_{(a,y)}^{(x,y)} P(s,t) ,mathrm{d} s+ Q(s,t), mathrm{d}t. $$

            This integral can be rewritten using the parameterization $s=u,t=y$, for $a leqslant u leqslant x$, giving
            $$F_x(x,y)=frac{partial}{partial x} int_a^x P(u,y) mathrm{d} u. $$
            The Fundamental Theorem of Calculus then gives $F_x=P$. A similar approach gives $F_y=Q$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Fix a point $(x,y)$. Since the domain $D$ is open, there is some small disk $Delta$ centered at $(x,y)$ which is entirely contained in $D$. The line segment $(a,y) to (x,y)$ is thus contained in $D$ for some constant $a$ sufficiently close to $x$. Since we have independence of the integration path, choose a path which starts by going from $(x_0,y_0)$ to $(a,y)$, followed by the line segment $(a,y) to (x,y)$. We then have
              $$F(x,y)=int_{(x_0,y_0)}^{(a,y)} P(s,t) ,mathrm{d} s+ Q(s,t), mathrm{d}t+int_{(a,y)}^{(x,y)} P(s,t) ,mathrm{d} s+ Q(s,t), mathrm{d}t . $$
              Taking the partial derivative w.r.t. $x$ we notice that the first term vanishes, meaning
              $$F_x(x,y)=frac{partial}{partial x} int_{(a,y)}^{(x,y)} P(s,t) ,mathrm{d} s+ Q(s,t), mathrm{d}t. $$

              This integral can be rewritten using the parameterization $s=u,t=y$, for $a leqslant u leqslant x$, giving
              $$F_x(x,y)=frac{partial}{partial x} int_a^x P(u,y) mathrm{d} u. $$
              The Fundamental Theorem of Calculus then gives $F_x=P$. A similar approach gives $F_y=Q$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Fix a point $(x,y)$. Since the domain $D$ is open, there is some small disk $Delta$ centered at $(x,y)$ which is entirely contained in $D$. The line segment $(a,y) to (x,y)$ is thus contained in $D$ for some constant $a$ sufficiently close to $x$. Since we have independence of the integration path, choose a path which starts by going from $(x_0,y_0)$ to $(a,y)$, followed by the line segment $(a,y) to (x,y)$. We then have
                $$F(x,y)=int_{(x_0,y_0)}^{(a,y)} P(s,t) ,mathrm{d} s+ Q(s,t), mathrm{d}t+int_{(a,y)}^{(x,y)} P(s,t) ,mathrm{d} s+ Q(s,t), mathrm{d}t . $$
                Taking the partial derivative w.r.t. $x$ we notice that the first term vanishes, meaning
                $$F_x(x,y)=frac{partial}{partial x} int_{(a,y)}^{(x,y)} P(s,t) ,mathrm{d} s+ Q(s,t), mathrm{d}t. $$

                This integral can be rewritten using the parameterization $s=u,t=y$, for $a leqslant u leqslant x$, giving
                $$F_x(x,y)=frac{partial}{partial x} int_a^x P(u,y) mathrm{d} u. $$
                The Fundamental Theorem of Calculus then gives $F_x=P$. A similar approach gives $F_y=Q$.






                share|cite|improve this answer









                $endgroup$



                Fix a point $(x,y)$. Since the domain $D$ is open, there is some small disk $Delta$ centered at $(x,y)$ which is entirely contained in $D$. The line segment $(a,y) to (x,y)$ is thus contained in $D$ for some constant $a$ sufficiently close to $x$. Since we have independence of the integration path, choose a path which starts by going from $(x_0,y_0)$ to $(a,y)$, followed by the line segment $(a,y) to (x,y)$. We then have
                $$F(x,y)=int_{(x_0,y_0)}^{(a,y)} P(s,t) ,mathrm{d} s+ Q(s,t), mathrm{d}t+int_{(a,y)}^{(x,y)} P(s,t) ,mathrm{d} s+ Q(s,t), mathrm{d}t . $$
                Taking the partial derivative w.r.t. $x$ we notice that the first term vanishes, meaning
                $$F_x(x,y)=frac{partial}{partial x} int_{(a,y)}^{(x,y)} P(s,t) ,mathrm{d} s+ Q(s,t), mathrm{d}t. $$

                This integral can be rewritten using the parameterization $s=u,t=y$, for $a leqslant u leqslant x$, giving
                $$F_x(x,y)=frac{partial}{partial x} int_a^x P(u,y) mathrm{d} u. $$
                The Fundamental Theorem of Calculus then gives $F_x=P$. A similar approach gives $F_y=Q$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 18 '18 at 9:02









                user1337user1337

                16.7k43391




                16.7k43391






























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