Tricky integral relationship












1












$begingroup$


I am trying to prove that
$$begin{equation}int_x^{x+1}left(int_0^{v} (u-0)f(u)textrm{d}u+int_v^{1} (u-1)f(u)textrm{d}uright)textrm{d}v=\int_0^xint_v^{v+1}f(u)textrm{d}utextrm{d}vend{equation}.$$



With the following sympy code i verified it symbolically for a few cases.



from sympy import *



x,u,v=symbols('x,u,v')

print

f=lambda x:exp(x);

print expand(integrate(integrate((u-0)*f(u), 

(u,0,v))+integrate((u-1)*f(u),(u,v,1)),(v,x,x+1)))

print expand(integrate(integrate(f(u),(u,v,v+1)),(v,0,x)))





definite-integrals integral-transforms







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$endgroup$

















    1












    $begingroup$


    I am trying to prove that
    $$begin{equation}int_x^{x+1}left(int_0^{v} (u-0)f(u)textrm{d}u+int_v^{1} (u-1)f(u)textrm{d}uright)textrm{d}v=\int_0^xint_v^{v+1}f(u)textrm{d}utextrm{d}vend{equation}.$$



    With the following sympy code i verified it symbolically for a few cases.



    from sympy import *
    

    
x,u,v=symbols('x,u,v')
    
print
    
f=lambda x:exp(x);
    
print expand(integrate(integrate((u-0)*f(u), 
    
(u,0,v))+integrate((u-1)*f(u),(u,v,1)),(v,x,x+1)))
    
print expand(integrate(integrate(f(u),(u,v,v+1)),(v,0,x)))





    definite-integrals integral-transforms







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am trying to prove that
      $$begin{equation}int_x^{x+1}left(int_0^{v} (u-0)f(u)textrm{d}u+int_v^{1} (u-1)f(u)textrm{d}uright)textrm{d}v=\int_0^xint_v^{v+1}f(u)textrm{d}utextrm{d}vend{equation}.$$



      With the following sympy code i verified it symbolically for a few cases.



      from sympy import *
      

      
x,u,v=symbols('x,u,v')
      
print
      
f=lambda x:exp(x);
      
print expand(integrate(integrate((u-0)*f(u), 
      
(u,0,v))+integrate((u-1)*f(u),(u,v,1)),(v,x,x+1)))
      
print expand(integrate(integrate(f(u),(u,v,v+1)),(v,0,x)))





      definite-integrals integral-transforms







      share|cite|improve this question









      $endgroup$




      I am trying to prove that
      $$begin{equation}int_x^{x+1}left(int_0^{v} (u-0)f(u)textrm{d}u+int_v^{1} (u-1)f(u)textrm{d}uright)textrm{d}v=\int_0^xint_v^{v+1}f(u)textrm{d}utextrm{d}vend{equation}.$$



      With the following sympy code i verified it symbolically for a few cases.



      from sympy import *
      

      
x,u,v=symbols('x,u,v')
      
print
      
f=lambda x:exp(x);
      
print expand(integrate(integrate((u-0)*f(u), 
      
(u,0,v))+integrate((u-1)*f(u),(u,v,1)),(v,x,x+1)))
      
print expand(integrate(integrate(f(u),(u,v,v+1)),(v,0,x)))





      definite-integrals integral-transforms




      definite-integrals integral-transforms






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      share|cite|improve this question











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      share|cite|improve this question










      asked Dec 10 '18 at 8:00









      Peter SheldrickPeter Sheldrick

      6911339




      6911339






















          1 Answer
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          1












          $begingroup$

          Call $F$ the antiderivative of $f$ such that
          $F(1)=0$.



          Call $frak{F}$ the antiderivative of $F$ such that $frak{F}$$(1)=0$.



          Now show that both expressions give the same result, i.e., :



          $${frak F}(x+1)-{frak F}(x)+{frak F}(0).$$



          Method : Write your LHS integral under the form :



          $$begin{equation}int_x^{x+1}left(int_0^1 uf(u)textrm{d}u-int_v^{1}f(u)textrm{d}uright)textrm{d}vend{equation}.$$



          Then integrate inside the large parentheses by parts in the first expression, giving



          $$int_{x}^{x+1}([uF(u)]_0^1-int_0^1 F(u)du + F(v)-F(1)$$



          Taking into account $F(1)=0$ and ${frak F}(1)=0$ :



          $$=int_{x}^{x+1}({frak F}(0) + F(v))dv$$



          I let you find the rest...






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $int_0^{v} (u-0)f(u)textrm{d}u+int_v^{1} (u-1)f(u)textrm{d}u=int_0^{v} uf(u)textrm{d}u+int_v^{1} uf(u)textrm{d}u-int_v^{1}f(u)textrm{d}u$
            $endgroup$
            – Peter Sheldrick
            Dec 10 '18 at 10:10










          • $begingroup$
            You are right. Corrected.
            $endgroup$
            – Jean Marie
            Dec 10 '18 at 10:12











          Your Answer





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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Call $F$ the antiderivative of $f$ such that
          $F(1)=0$.



          Call $frak{F}$ the antiderivative of $F$ such that $frak{F}$$(1)=0$.



          Now show that both expressions give the same result, i.e., :



          $${frak F}(x+1)-{frak F}(x)+{frak F}(0).$$



          Method : Write your LHS integral under the form :



          $$begin{equation}int_x^{x+1}left(int_0^1 uf(u)textrm{d}u-int_v^{1}f(u)textrm{d}uright)textrm{d}vend{equation}.$$



          Then integrate inside the large parentheses by parts in the first expression, giving



          $$int_{x}^{x+1}([uF(u)]_0^1-int_0^1 F(u)du + F(v)-F(1)$$



          Taking into account $F(1)=0$ and ${frak F}(1)=0$ :



          $$=int_{x}^{x+1}({frak F}(0) + F(v))dv$$



          I let you find the rest...






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $int_0^{v} (u-0)f(u)textrm{d}u+int_v^{1} (u-1)f(u)textrm{d}u=int_0^{v} uf(u)textrm{d}u+int_v^{1} uf(u)textrm{d}u-int_v^{1}f(u)textrm{d}u$
            $endgroup$
            – Peter Sheldrick
            Dec 10 '18 at 10:10










          • $begingroup$
            You are right. Corrected.
            $endgroup$
            – Jean Marie
            Dec 10 '18 at 10:12
















          1












          $begingroup$

          Call $F$ the antiderivative of $f$ such that
          $F(1)=0$.



          Call $frak{F}$ the antiderivative of $F$ such that $frak{F}$$(1)=0$.



          Now show that both expressions give the same result, i.e., :



          $${frak F}(x+1)-{frak F}(x)+{frak F}(0).$$



          Method : Write your LHS integral under the form :



          $$begin{equation}int_x^{x+1}left(int_0^1 uf(u)textrm{d}u-int_v^{1}f(u)textrm{d}uright)textrm{d}vend{equation}.$$



          Then integrate inside the large parentheses by parts in the first expression, giving



          $$int_{x}^{x+1}([uF(u)]_0^1-int_0^1 F(u)du + F(v)-F(1)$$



          Taking into account $F(1)=0$ and ${frak F}(1)=0$ :



          $$=int_{x}^{x+1}({frak F}(0) + F(v))dv$$



          I let you find the rest...






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $int_0^{v} (u-0)f(u)textrm{d}u+int_v^{1} (u-1)f(u)textrm{d}u=int_0^{v} uf(u)textrm{d}u+int_v^{1} uf(u)textrm{d}u-int_v^{1}f(u)textrm{d}u$
            $endgroup$
            – Peter Sheldrick
            Dec 10 '18 at 10:10










          • $begingroup$
            You are right. Corrected.
            $endgroup$
            – Jean Marie
            Dec 10 '18 at 10:12














          1












          1








          1





          $begingroup$

          Call $F$ the antiderivative of $f$ such that
          $F(1)=0$.



          Call $frak{F}$ the antiderivative of $F$ such that $frak{F}$$(1)=0$.



          Now show that both expressions give the same result, i.e., :



          $${frak F}(x+1)-{frak F}(x)+{frak F}(0).$$



          Method : Write your LHS integral under the form :



          $$begin{equation}int_x^{x+1}left(int_0^1 uf(u)textrm{d}u-int_v^{1}f(u)textrm{d}uright)textrm{d}vend{equation}.$$



          Then integrate inside the large parentheses by parts in the first expression, giving



          $$int_{x}^{x+1}([uF(u)]_0^1-int_0^1 F(u)du + F(v)-F(1)$$



          Taking into account $F(1)=0$ and ${frak F}(1)=0$ :



          $$=int_{x}^{x+1}({frak F}(0) + F(v))dv$$



          I let you find the rest...






          share|cite|improve this answer











          $endgroup$



          Call $F$ the antiderivative of $f$ such that
          $F(1)=0$.



          Call $frak{F}$ the antiderivative of $F$ such that $frak{F}$$(1)=0$.



          Now show that both expressions give the same result, i.e., :



          $${frak F}(x+1)-{frak F}(x)+{frak F}(0).$$



          Method : Write your LHS integral under the form :



          $$begin{equation}int_x^{x+1}left(int_0^1 uf(u)textrm{d}u-int_v^{1}f(u)textrm{d}uright)textrm{d}vend{equation}.$$



          Then integrate inside the large parentheses by parts in the first expression, giving



          $$int_{x}^{x+1}([uF(u)]_0^1-int_0^1 F(u)du + F(v)-F(1)$$



          Taking into account $F(1)=0$ and ${frak F}(1)=0$ :



          $$=int_{x}^{x+1}({frak F}(0) + F(v))dv$$



          I let you find the rest...







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 10:12

























          answered Dec 10 '18 at 8:25









          Jean MarieJean Marie

          29.9k42051




          29.9k42051












          • $begingroup$
            $int_0^{v} (u-0)f(u)textrm{d}u+int_v^{1} (u-1)f(u)textrm{d}u=int_0^{v} uf(u)textrm{d}u+int_v^{1} uf(u)textrm{d}u-int_v^{1}f(u)textrm{d}u$
            $endgroup$
            – Peter Sheldrick
            Dec 10 '18 at 10:10










          • $begingroup$
            You are right. Corrected.
            $endgroup$
            – Jean Marie
            Dec 10 '18 at 10:12


















          • $begingroup$
            $int_0^{v} (u-0)f(u)textrm{d}u+int_v^{1} (u-1)f(u)textrm{d}u=int_0^{v} uf(u)textrm{d}u+int_v^{1} uf(u)textrm{d}u-int_v^{1}f(u)textrm{d}u$
            $endgroup$
            – Peter Sheldrick
            Dec 10 '18 at 10:10










          • $begingroup$
            You are right. Corrected.
            $endgroup$
            – Jean Marie
            Dec 10 '18 at 10:12
















          $begingroup$
          $int_0^{v} (u-0)f(u)textrm{d}u+int_v^{1} (u-1)f(u)textrm{d}u=int_0^{v} uf(u)textrm{d}u+int_v^{1} uf(u)textrm{d}u-int_v^{1}f(u)textrm{d}u$
          $endgroup$
          – Peter Sheldrick
          Dec 10 '18 at 10:10




          $begingroup$
          $int_0^{v} (u-0)f(u)textrm{d}u+int_v^{1} (u-1)f(u)textrm{d}u=int_0^{v} uf(u)textrm{d}u+int_v^{1} uf(u)textrm{d}u-int_v^{1}f(u)textrm{d}u$
          $endgroup$
          – Peter Sheldrick
          Dec 10 '18 at 10:10












          $begingroup$
          You are right. Corrected.
          $endgroup$
          – Jean Marie
          Dec 10 '18 at 10:12




          $begingroup$
          You are right. Corrected.
          $endgroup$
          – Jean Marie
          Dec 10 '18 at 10:12


















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