What is integral of a scalar field with respect to a variable?












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I was reading Tom Apostol calculus volume 2 and came across theorem 10.8 (page 350)




The theorem is:enter image description here




My doubt is, What is the definition of the right hand side of the underlined equation that is, What is the definition of $$int_{a}^b psi(vec x,t)dt $$, this integral formally? I know integral of real valued function formally to some extent? But what is the FORMAL DEFINITION OF THIS INTEGRAL?










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    0












    $begingroup$


    I was reading Tom Apostol calculus volume 2 and came across theorem 10.8 (page 350)




    The theorem is:enter image description here




    My doubt is, What is the definition of the right hand side of the underlined equation that is, What is the definition of $$int_{a}^b psi(vec x,t)dt $$, this integral formally? I know integral of real valued function formally to some extent? But what is the FORMAL DEFINITION OF THIS INTEGRAL?










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I was reading Tom Apostol calculus volume 2 and came across theorem 10.8 (page 350)




      The theorem is:enter image description here




      My doubt is, What is the definition of the right hand side of the underlined equation that is, What is the definition of $$int_{a}^b psi(vec x,t)dt $$, this integral formally? I know integral of real valued function formally to some extent? But what is the FORMAL DEFINITION OF THIS INTEGRAL?










      share|cite|improve this question









      $endgroup$




      I was reading Tom Apostol calculus volume 2 and came across theorem 10.8 (page 350)




      The theorem is:enter image description here




      My doubt is, What is the definition of the right hand side of the underlined equation that is, What is the definition of $$int_{a}^b psi(vec x,t)dt $$, this integral formally? I know integral of real valued function formally to some extent? But what is the FORMAL DEFINITION OF THIS INTEGRAL?







      real-analysis calculus functional-analysis multivariable-calculus vector-analysis






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      asked Dec 27 '18 at 17:46









      Bijayan RayBijayan Ray

      136112




      136112






















          1 Answer
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          $begingroup$

          You're confused because $psi(vec{x},,t)$ depends on multiple variables. But for each value of $vec{x}$, $psi(vec{x},,t)$ becomes just a function of $t$; you know how to integrate that. This recipe gives each $varphi(vec{x})$, thus defining the function $varphi$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I had thought of that about defining a different function g for each $vec x$, as a function of only t, then integrating g. But then how would you explain the next equation that is taking the partial derivative of $phi(vec x)$ ?
            $endgroup$
            – Bijayan Ray
            Dec 28 '18 at 5:09












          • $begingroup$
            @BijayanRag You have s function of $vec{x}$; differentiate it.
            $endgroup$
            – J.G.
            Dec 28 '18 at 7:18










          • $begingroup$
            your statement about function g (as per my last comment) is :- for all $vec x$ there exist a function g such that $psi (vec x,t)$=g(t)
            $endgroup$
            – Bijayan Ray
            Dec 28 '18 at 10:38












          • $begingroup$
            can you please tell me in formal words where are you getting a function of $vec x$ to differentiate?
            $endgroup$
            – Bijayan Ray
            Dec 28 '18 at 10:42










          • $begingroup$
            I think what I want to tell from the comments above is similar to the logical difference between uniform convergence and pointwise convergence & and also similar to logic behind existence of all directional derivative does not imply continuity
            $endgroup$
            – Bijayan Ray
            Dec 28 '18 at 10:52











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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

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          2












          $begingroup$

          You're confused because $psi(vec{x},,t)$ depends on multiple variables. But for each value of $vec{x}$, $psi(vec{x},,t)$ becomes just a function of $t$; you know how to integrate that. This recipe gives each $varphi(vec{x})$, thus defining the function $varphi$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I had thought of that about defining a different function g for each $vec x$, as a function of only t, then integrating g. But then how would you explain the next equation that is taking the partial derivative of $phi(vec x)$ ?
            $endgroup$
            – Bijayan Ray
            Dec 28 '18 at 5:09












          • $begingroup$
            @BijayanRag You have s function of $vec{x}$; differentiate it.
            $endgroup$
            – J.G.
            Dec 28 '18 at 7:18










          • $begingroup$
            your statement about function g (as per my last comment) is :- for all $vec x$ there exist a function g such that $psi (vec x,t)$=g(t)
            $endgroup$
            – Bijayan Ray
            Dec 28 '18 at 10:38












          • $begingroup$
            can you please tell me in formal words where are you getting a function of $vec x$ to differentiate?
            $endgroup$
            – Bijayan Ray
            Dec 28 '18 at 10:42










          • $begingroup$
            I think what I want to tell from the comments above is similar to the logical difference between uniform convergence and pointwise convergence & and also similar to logic behind existence of all directional derivative does not imply continuity
            $endgroup$
            – Bijayan Ray
            Dec 28 '18 at 10:52
















          2












          $begingroup$

          You're confused because $psi(vec{x},,t)$ depends on multiple variables. But for each value of $vec{x}$, $psi(vec{x},,t)$ becomes just a function of $t$; you know how to integrate that. This recipe gives each $varphi(vec{x})$, thus defining the function $varphi$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I had thought of that about defining a different function g for each $vec x$, as a function of only t, then integrating g. But then how would you explain the next equation that is taking the partial derivative of $phi(vec x)$ ?
            $endgroup$
            – Bijayan Ray
            Dec 28 '18 at 5:09












          • $begingroup$
            @BijayanRag You have s function of $vec{x}$; differentiate it.
            $endgroup$
            – J.G.
            Dec 28 '18 at 7:18










          • $begingroup$
            your statement about function g (as per my last comment) is :- for all $vec x$ there exist a function g such that $psi (vec x,t)$=g(t)
            $endgroup$
            – Bijayan Ray
            Dec 28 '18 at 10:38












          • $begingroup$
            can you please tell me in formal words where are you getting a function of $vec x$ to differentiate?
            $endgroup$
            – Bijayan Ray
            Dec 28 '18 at 10:42










          • $begingroup$
            I think what I want to tell from the comments above is similar to the logical difference between uniform convergence and pointwise convergence & and also similar to logic behind existence of all directional derivative does not imply continuity
            $endgroup$
            – Bijayan Ray
            Dec 28 '18 at 10:52














          2












          2








          2





          $begingroup$

          You're confused because $psi(vec{x},,t)$ depends on multiple variables. But for each value of $vec{x}$, $psi(vec{x},,t)$ becomes just a function of $t$; you know how to integrate that. This recipe gives each $varphi(vec{x})$, thus defining the function $varphi$.






          share|cite|improve this answer









          $endgroup$



          You're confused because $psi(vec{x},,t)$ depends on multiple variables. But for each value of $vec{x}$, $psi(vec{x},,t)$ becomes just a function of $t$; you know how to integrate that. This recipe gives each $varphi(vec{x})$, thus defining the function $varphi$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 27 '18 at 17:59









          J.G.J.G.

          30.6k23149




          30.6k23149












          • $begingroup$
            I had thought of that about defining a different function g for each $vec x$, as a function of only t, then integrating g. But then how would you explain the next equation that is taking the partial derivative of $phi(vec x)$ ?
            $endgroup$
            – Bijayan Ray
            Dec 28 '18 at 5:09












          • $begingroup$
            @BijayanRag You have s function of $vec{x}$; differentiate it.
            $endgroup$
            – J.G.
            Dec 28 '18 at 7:18










          • $begingroup$
            your statement about function g (as per my last comment) is :- for all $vec x$ there exist a function g such that $psi (vec x,t)$=g(t)
            $endgroup$
            – Bijayan Ray
            Dec 28 '18 at 10:38












          • $begingroup$
            can you please tell me in formal words where are you getting a function of $vec x$ to differentiate?
            $endgroup$
            – Bijayan Ray
            Dec 28 '18 at 10:42










          • $begingroup$
            I think what I want to tell from the comments above is similar to the logical difference between uniform convergence and pointwise convergence & and also similar to logic behind existence of all directional derivative does not imply continuity
            $endgroup$
            – Bijayan Ray
            Dec 28 '18 at 10:52


















          • $begingroup$
            I had thought of that about defining a different function g for each $vec x$, as a function of only t, then integrating g. But then how would you explain the next equation that is taking the partial derivative of $phi(vec x)$ ?
            $endgroup$
            – Bijayan Ray
            Dec 28 '18 at 5:09












          • $begingroup$
            @BijayanRag You have s function of $vec{x}$; differentiate it.
            $endgroup$
            – J.G.
            Dec 28 '18 at 7:18










          • $begingroup$
            your statement about function g (as per my last comment) is :- for all $vec x$ there exist a function g such that $psi (vec x,t)$=g(t)
            $endgroup$
            – Bijayan Ray
            Dec 28 '18 at 10:38












          • $begingroup$
            can you please tell me in formal words where are you getting a function of $vec x$ to differentiate?
            $endgroup$
            – Bijayan Ray
            Dec 28 '18 at 10:42










          • $begingroup$
            I think what I want to tell from the comments above is similar to the logical difference between uniform convergence and pointwise convergence & and also similar to logic behind existence of all directional derivative does not imply continuity
            $endgroup$
            – Bijayan Ray
            Dec 28 '18 at 10:52
















          $begingroup$
          I had thought of that about defining a different function g for each $vec x$, as a function of only t, then integrating g. But then how would you explain the next equation that is taking the partial derivative of $phi(vec x)$ ?
          $endgroup$
          – Bijayan Ray
          Dec 28 '18 at 5:09






          $begingroup$
          I had thought of that about defining a different function g for each $vec x$, as a function of only t, then integrating g. But then how would you explain the next equation that is taking the partial derivative of $phi(vec x)$ ?
          $endgroup$
          – Bijayan Ray
          Dec 28 '18 at 5:09














          $begingroup$
          @BijayanRag You have s function of $vec{x}$; differentiate it.
          $endgroup$
          – J.G.
          Dec 28 '18 at 7:18




          $begingroup$
          @BijayanRag You have s function of $vec{x}$; differentiate it.
          $endgroup$
          – J.G.
          Dec 28 '18 at 7:18












          $begingroup$
          your statement about function g (as per my last comment) is :- for all $vec x$ there exist a function g such that $psi (vec x,t)$=g(t)
          $endgroup$
          – Bijayan Ray
          Dec 28 '18 at 10:38






          $begingroup$
          your statement about function g (as per my last comment) is :- for all $vec x$ there exist a function g such that $psi (vec x,t)$=g(t)
          $endgroup$
          – Bijayan Ray
          Dec 28 '18 at 10:38














          $begingroup$
          can you please tell me in formal words where are you getting a function of $vec x$ to differentiate?
          $endgroup$
          – Bijayan Ray
          Dec 28 '18 at 10:42




          $begingroup$
          can you please tell me in formal words where are you getting a function of $vec x$ to differentiate?
          $endgroup$
          – Bijayan Ray
          Dec 28 '18 at 10:42












          $begingroup$
          I think what I want to tell from the comments above is similar to the logical difference between uniform convergence and pointwise convergence & and also similar to logic behind existence of all directional derivative does not imply continuity
          $endgroup$
          – Bijayan Ray
          Dec 28 '18 at 10:52




          $begingroup$
          I think what I want to tell from the comments above is similar to the logical difference between uniform convergence and pointwise convergence & and also similar to logic behind existence of all directional derivative does not imply continuity
          $endgroup$
          – Bijayan Ray
          Dec 28 '18 at 10:52


















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