Changing order of integration:$int_0^inftyint_{-infty}^{-y}f(x)mathrm dxmathrm...












2












$begingroup$



Why does $$int_{0}^{infty} int_{-infty}^{-y} f(x)mathrm dx mathrm dy Rightarrow int_{-infty}^{0} int_{0}^{-x} f(x) mathrm dy mathrm dx$$




The title is pretty self explanatory. I couldn't see how to properly change the order of the left integeral to the right one.



I'd love to hear your thoughts, thanks.






integration multivariable-calculus change-of-variable







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$endgroup$








  • 1




    $begingroup$
    Have you tried drawing the region?
    $endgroup$
    – Tom Himler
    Jan 14 at 21:36






  • 1




    $begingroup$
    The first thing that comes to mind is Paul's Online Math Notes. They're very good at explaining double integrals over general regions. tutorial.math.lamar.edu/Classes/CalcIII/DIGeneralRegion.aspx
    $endgroup$
    – Tom Himler
    Jan 14 at 21:46






  • 1




    $begingroup$
    Apply Fubini's theorem to $$int_{-infty}^{infty} int_{-infty}^{infty}f(x) , mathbf{1}_{{x leqslant -y }}mathbf{1}_{{x leqslant 0 }}, mathbf{1}_{{y geqslant 0 }},dx , dy$$
    $endgroup$
    – RRL
    Jan 14 at 21:57
















2












$begingroup$



Why does $$int_{0}^{infty} int_{-infty}^{-y} f(x)mathrm dx mathrm dy Rightarrow int_{-infty}^{0} int_{0}^{-x} f(x) mathrm dy mathrm dx$$




The title is pretty self explanatory. I couldn't see how to properly change the order of the left integeral to the right one.



I'd love to hear your thoughts, thanks.






integration multivariable-calculus change-of-variable







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Have you tried drawing the region?
    $endgroup$
    – Tom Himler
    Jan 14 at 21:36






  • 1




    $begingroup$
    The first thing that comes to mind is Paul's Online Math Notes. They're very good at explaining double integrals over general regions. tutorial.math.lamar.edu/Classes/CalcIII/DIGeneralRegion.aspx
    $endgroup$
    – Tom Himler
    Jan 14 at 21:46






  • 1




    $begingroup$
    Apply Fubini's theorem to $$int_{-infty}^{infty} int_{-infty}^{infty}f(x) , mathbf{1}_{{x leqslant -y }}mathbf{1}_{{x leqslant 0 }}, mathbf{1}_{{y geqslant 0 }},dx , dy$$
    $endgroup$
    – RRL
    Jan 14 at 21:57














2












2








2


0



$begingroup$



Why does $$int_{0}^{infty} int_{-infty}^{-y} f(x)mathrm dx mathrm dy Rightarrow int_{-infty}^{0} int_{0}^{-x} f(x) mathrm dy mathrm dx$$




The title is pretty self explanatory. I couldn't see how to properly change the order of the left integeral to the right one.



I'd love to hear your thoughts, thanks.






integration multivariable-calculus change-of-variable







share|cite|improve this question











$endgroup$





Why does $$int_{0}^{infty} int_{-infty}^{-y} f(x)mathrm dx mathrm dy Rightarrow int_{-infty}^{0} int_{0}^{-x} f(x) mathrm dy mathrm dx$$




The title is pretty self explanatory. I couldn't see how to properly change the order of the left integeral to the right one.



I'd love to hear your thoughts, thanks.






integration multivariable-calculus change-of-variable




integration multivariable-calculus change-of-variable






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Jan 14 at 21:44









Zacky

7,89511062




7,89511062










asked Jan 14 at 21:33









superuser123superuser123

48628




48628








  • 1




    $begingroup$
    Have you tried drawing the region?
    $endgroup$
    – Tom Himler
    Jan 14 at 21:36






  • 1




    $begingroup$
    The first thing that comes to mind is Paul's Online Math Notes. They're very good at explaining double integrals over general regions. tutorial.math.lamar.edu/Classes/CalcIII/DIGeneralRegion.aspx
    $endgroup$
    – Tom Himler
    Jan 14 at 21:46






  • 1




    $begingroup$
    Apply Fubini's theorem to $$int_{-infty}^{infty} int_{-infty}^{infty}f(x) , mathbf{1}_{{x leqslant -y }}mathbf{1}_{{x leqslant 0 }}, mathbf{1}_{{y geqslant 0 }},dx , dy$$
    $endgroup$
    – RRL
    Jan 14 at 21:57














  • 1




    $begingroup$
    Have you tried drawing the region?
    $endgroup$
    – Tom Himler
    Jan 14 at 21:36






  • 1




    $begingroup$
    The first thing that comes to mind is Paul's Online Math Notes. They're very good at explaining double integrals over general regions. tutorial.math.lamar.edu/Classes/CalcIII/DIGeneralRegion.aspx
    $endgroup$
    – Tom Himler
    Jan 14 at 21:46






  • 1




    $begingroup$
    Apply Fubini's theorem to $$int_{-infty}^{infty} int_{-infty}^{infty}f(x) , mathbf{1}_{{x leqslant -y }}mathbf{1}_{{x leqslant 0 }}, mathbf{1}_{{y geqslant 0 }},dx , dy$$
    $endgroup$
    – RRL
    Jan 14 at 21:57








1




1




$begingroup$
Have you tried drawing the region?
$endgroup$
– Tom Himler
Jan 14 at 21:36




$begingroup$
Have you tried drawing the region?
$endgroup$
– Tom Himler
Jan 14 at 21:36




1




1




$begingroup$
The first thing that comes to mind is Paul's Online Math Notes. They're very good at explaining double integrals over general regions. tutorial.math.lamar.edu/Classes/CalcIII/DIGeneralRegion.aspx
$endgroup$
– Tom Himler
Jan 14 at 21:46




$begingroup$
The first thing that comes to mind is Paul's Online Math Notes. They're very good at explaining double integrals over general regions. tutorial.math.lamar.edu/Classes/CalcIII/DIGeneralRegion.aspx
$endgroup$
– Tom Himler
Jan 14 at 21:46




1




1




$begingroup$
Apply Fubini's theorem to $$int_{-infty}^{infty} int_{-infty}^{infty}f(x) , mathbf{1}_{{x leqslant -y }}mathbf{1}_{{x leqslant 0 }}, mathbf{1}_{{y geqslant 0 }},dx , dy$$
$endgroup$
– RRL
Jan 14 at 21:57




$begingroup$
Apply Fubini's theorem to $$int_{-infty}^{infty} int_{-infty}^{infty}f(x) , mathbf{1}_{{x leqslant -y }}mathbf{1}_{{x leqslant 0 }}, mathbf{1}_{{y geqslant 0 }},dx , dy$$
$endgroup$
– RRL
Jan 14 at 21:57










2 Answers
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$begingroup$

The easiest way to perform a change of the order of integration in the multivariable setting is via Iverson's bracket. This is the indicator function such that $$[P] =begin{cases} 1 & P text{ is true}\
0 & text{else}.end{cases}$$ The change of variable arises from reinterpreting the system of inequalities (see below).



With the Iverson notation, one can remove the boundaries from the integral and implement it in the integrand, i.e.,
$$int_{0}^{infty} int_{-infty}^{-y} f(x,y)dx ,dy =iint_{mathbb{R}^2} f(x,y)Bigl[(ygeq 0) text{ and } (xleq-y) Bigr]dx, dy ,.$$



Now in order to perform the change of the order of integration, you have to reinterpret Iverson's bracket. You have to figure out what condition $$P=(ygeq0) text{ and } (xleq-y)$$ poses on $x$ first.



The maximal value that $x$ can achieve is $0$ (when $y=0$). The second condition in $P$ is equivalent to
$$ x leq -y Leftrightarrow y leq -x ,.$$
The first condition demands that $y>0$. Together, we have that
lf that $P$ is equivalent to
$$ P Leftrightarrow (x<0) text{ and } (0 < y < -x ),.$$



So we find
$$int_{0}^{infty} int_{-infty}^{-y} f(x,y)dx, dy =iint_{mathbb{R}^2} f(x,y)Bigl[(x<0) text{ and } (0 < y < -x )Bigr]dx ,dy = int_{-infty}^0 int_{0}^{-x} f(x,y) ,dy,dx ,.$$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    The inequalities given with the first double integral are $0leq y leq infty$ and $-infty leq x leq -y$.



    If you plot the funtions and the regions you get the region:



    enter image description here



    As you can see, if you we let the $x$ be constant limits instead of $y$ the same region can be defined as $-infty leq x leq 0$ and $0leq y leq -x$. Which gives you the bounds in your question.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
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      2 Answers
      2






      active

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      active

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      active

      oldest

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      3












      $begingroup$

      The easiest way to perform a change of the order of integration in the multivariable setting is via Iverson's bracket. This is the indicator function such that $$[P] =begin{cases} 1 & P text{ is true}\
      0 & text{else}.end{cases}$$       The change of variable arises from reinterpreting the system of inequalities (see below).



      With the Iverson notation, one can remove the boundaries from the integral and implement it in the integrand, i.e.,
      $$int_{0}^{infty} int_{-infty}^{-y} f(x,y)dx ,dy =iint_{mathbb{R}^2} f(x,y)Bigl[(ygeq 0) text{ and } (xleq-y) Bigr]dx, dy ,.$$



      Now in order to perform the change of the order of integration, you have to reinterpret Iverson's bracket. You have to figure out what condition $$P=(ygeq0) text{ and } (xleq-y)$$ poses on $x$ first.



      The maximal value that $x$ can achieve is $0$ (when $y=0$). The second condition in $P$ is equivalent to
      $$ x leq -y Leftrightarrow y leq -x ,.$$
      The first condition demands that $y>0$. Together, we have that
      lf that $P$ is equivalent to
      $$ P Leftrightarrow (x<0) text{ and } (0 < y < -x ),.$$



      So we find
      $$int_{0}^{infty} int_{-infty}^{-y} f(x,y)dx, dy =iint_{mathbb{R}^2} f(x,y)Bigl[(x<0) text{ and } (0 < y < -x )Bigr]dx ,dy = int_{-infty}^0 int_{0}^{-x} f(x,y) ,dy,dx ,.$$






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        The easiest way to perform a change of the order of integration in the multivariable setting is via Iverson's bracket. This is the indicator function such that $$[P] =begin{cases} 1 & P text{ is true}\
        0 & text{else}.end{cases}$$         The change of variable arises from reinterpreting the system of inequalities (see below).



        With the Iverson notation, one can remove the boundaries from the integral and implement it in the integrand, i.e.,
        $$int_{0}^{infty} int_{-infty}^{-y} f(x,y)dx ,dy =iint_{mathbb{R}^2} f(x,y)Bigl[(ygeq 0) text{ and } (xleq-y) Bigr]dx, dy ,.$$



        Now in order to perform the change of the order of integration, you have to reinterpret Iverson's bracket. You have to figure out what condition $$P=(ygeq0) text{ and } (xleq-y)$$ poses on $x$ first.



        The maximal value that $x$ can achieve is $0$ (when $y=0$). The second condition in $P$ is equivalent to
        $$ x leq -y Leftrightarrow y leq -x ,.$$
        The first condition demands that $y>0$. Together, we have that
        lf that $P$ is equivalent to
        $$ P Leftrightarrow (x<0) text{ and } (0 < y < -x ),.$$



        So we find
        $$int_{0}^{infty} int_{-infty}^{-y} f(x,y)dx, dy =iint_{mathbb{R}^2} f(x,y)Bigl[(x<0) text{ and } (0 < y < -x )Bigr]dx ,dy = int_{-infty}^0 int_{0}^{-x} f(x,y) ,dy,dx ,.$$






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          The easiest way to perform a change of the order of integration in the multivariable setting is via Iverson's bracket. This is the indicator function such that $$[P] =begin{cases} 1 & P text{ is true}\
          0 & text{else}.end{cases}$$           The change of variable arises from reinterpreting the system of inequalities (see below).



          With the Iverson notation, one can remove the boundaries from the integral and implement it in the integrand, i.e.,
          $$int_{0}^{infty} int_{-infty}^{-y} f(x,y)dx ,dy =iint_{mathbb{R}^2} f(x,y)Bigl[(ygeq 0) text{ and } (xleq-y) Bigr]dx, dy ,.$$



          Now in order to perform the change of the order of integration, you have to reinterpret Iverson's bracket. You have to figure out what condition $$P=(ygeq0) text{ and } (xleq-y)$$ poses on $x$ first.



          The maximal value that $x$ can achieve is $0$ (when $y=0$). The second condition in $P$ is equivalent to
          $$ x leq -y Leftrightarrow y leq -x ,.$$
          The first condition demands that $y>0$. Together, we have that
          lf that $P$ is equivalent to
          $$ P Leftrightarrow (x<0) text{ and } (0 < y < -x ),.$$



          So we find
          $$int_{0}^{infty} int_{-infty}^{-y} f(x,y)dx, dy =iint_{mathbb{R}^2} f(x,y)Bigl[(x<0) text{ and } (0 < y < -x )Bigr]dx ,dy = int_{-infty}^0 int_{0}^{-x} f(x,y) ,dy,dx ,.$$






          share|cite|improve this answer









          $endgroup$



          The easiest way to perform a change of the order of integration in the multivariable setting is via Iverson's bracket. This is the indicator function such that $$[P] =begin{cases} 1 & P text{ is true}\
          0 & text{else}.end{cases}$$           The change of variable arises from reinterpreting the system of inequalities (see below).



          With the Iverson notation, one can remove the boundaries from the integral and implement it in the integrand, i.e.,
          $$int_{0}^{infty} int_{-infty}^{-y} f(x,y)dx ,dy =iint_{mathbb{R}^2} f(x,y)Bigl[(ygeq 0) text{ and } (xleq-y) Bigr]dx, dy ,.$$



          Now in order to perform the change of the order of integration, you have to reinterpret Iverson's bracket. You have to figure out what condition $$P=(ygeq0) text{ and } (xleq-y)$$ poses on $x$ first.



          The maximal value that $x$ can achieve is $0$ (when $y=0$). The second condition in $P$ is equivalent to
          $$ x leq -y Leftrightarrow y leq -x ,.$$
          The first condition demands that $y>0$. Together, we have that
          lf that $P$ is equivalent to
          $$ P Leftrightarrow (x<0) text{ and } (0 < y < -x ),.$$



          So we find
          $$int_{0}^{infty} int_{-infty}^{-y} f(x,y)dx, dy =iint_{mathbb{R}^2} f(x,y)Bigl[(x<0) text{ and } (0 < y < -x )Bigr]dx ,dy = int_{-infty}^0 int_{0}^{-x} f(x,y) ,dy,dx ,.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 14 at 22:00









          FabianFabian

          20k3774




          20k3774























              2












              $begingroup$

              The inequalities given with the first double integral are $0leq y leq infty$ and $-infty leq x leq -y$.



              If you plot the funtions and the regions you get the region:



              enter image description here



              As you can see, if you we let the $x$ be constant limits instead of $y$ the same region can be defined as $-infty leq x leq 0$ and $0leq y leq -x$. Which gives you the bounds in your question.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                The inequalities given with the first double integral are $0leq y leq infty$ and $-infty leq x leq -y$.



                If you plot the funtions and the regions you get the region:



                enter image description here



                As you can see, if you we let the $x$ be constant limits instead of $y$ the same region can be defined as $-infty leq x leq 0$ and $0leq y leq -x$. Which gives you the bounds in your question.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  The inequalities given with the first double integral are $0leq y leq infty$ and $-infty leq x leq -y$.



                  If you plot the funtions and the regions you get the region:



                  enter image description here



                  As you can see, if you we let the $x$ be constant limits instead of $y$ the same region can be defined as $-infty leq x leq 0$ and $0leq y leq -x$. Which gives you the bounds in your question.






                  share|cite|improve this answer









                  $endgroup$



                  The inequalities given with the first double integral are $0leq y leq infty$ and $-infty leq x leq -y$.



                  If you plot the funtions and the regions you get the region:



                  enter image description here



                  As you can see, if you we let the $x$ be constant limits instead of $y$ the same region can be defined as $-infty leq x leq 0$ and $0leq y leq -x$. Which gives you the bounds in your question.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 14 at 22:45









                  Tom HimlerTom Himler

                  945314




                  945314






























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