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An integral related to the distance distribution between points inside the unit disk



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Average distance between two points in a circular diskMap 2D points inside a closed curve to unit diskFinite number of points inside a diskIs it true that a arbitrary 3D rotation can be composed with two rotations constrained to have their axes in the same plane?Distance between the nail and the center of the diskUpper bound on the minimum distance between $N$ points chosen inside the unit circle?Chord Length Distribution in Two-Dimensional Disk(s)Optimal placement of points inside a diskWhat's the geometry of $|r - r_1| + |r - r_2| = k$ where $k > |r_1 - r_2|$ in $2$ dimensions?Distance distribution between two points










3












$begingroup$


At the moment I am interested in some applications of geometric probability theory, like finding the average distance $langle Drangle$ of two points randomly picked from the unit disk. It is not too difficult to set up an integral which correctly describes this distance:



Pick two points $X_1,X_2$ inside the unit disk (one of them can, by symmetry, chosen to be located on the y-axis) and denote by $r_j=|X_j|$ their distance from the origin.



Then $r_j$ are indenpendently distributed random variables with probability distribution $dp_j=2r_j$. Furthermore, the angle between these two points $phi$ is uniformly distributed with pd $dp_phi=1/pi$.



From elementary geometry (law of cosines) is easy to see that $D^2=r_1^2+r_2^2-2 r_1r_2cos(phi)$ and therefore ($A$ is a cylinder with $r=1/2, z=1$ )



$$
langle Drangle=int_ADdp_1dp_2dp_phi=
\frac4piint_0^1int_0^1int_0^pir_1 r_2sqrtr_1^2+r_2^2-2 r_1r_2cos(phi)dr_1dr_2dphi
$$



After struggeling for a few hours, i was googeling around to find out that




$$langle Drangle=frac12845pi quadquad (star)$$




Fighting a bit longer, i am now ready to surrender and asking you




How on Earth/Mars/Beteigeuze $(star)$ can be proven without plodding through a swamp of nightmarish antiderivatives?




I tried a lot, from bruteforce methods over various changes of coordinate systems to series expansions but everything ended up in a big mess (interestingly for 3 dimensions the calculations go through smoothly)...



Edit:



Just in case somebody needs a bit of visual input, i created a Desmos chart



setup



Edit2:










share|cite|improve this question











$endgroup$
















    3












    $begingroup$


    At the moment I am interested in some applications of geometric probability theory, like finding the average distance $langle Drangle$ of two points randomly picked from the unit disk. It is not too difficult to set up an integral which correctly describes this distance:



    Pick two points $X_1,X_2$ inside the unit disk (one of them can, by symmetry, chosen to be located on the y-axis) and denote by $r_j=|X_j|$ their distance from the origin.



    Then $r_j$ are indenpendently distributed random variables with probability distribution $dp_j=2r_j$. Furthermore, the angle between these two points $phi$ is uniformly distributed with pd $dp_phi=1/pi$.



    From elementary geometry (law of cosines) is easy to see that $D^2=r_1^2+r_2^2-2 r_1r_2cos(phi)$ and therefore ($A$ is a cylinder with $r=1/2, z=1$ )



    $$
    langle Drangle=int_ADdp_1dp_2dp_phi=
    \frac4piint_0^1int_0^1int_0^pir_1 r_2sqrtr_1^2+r_2^2-2 r_1r_2cos(phi)dr_1dr_2dphi
    $$



    After struggeling for a few hours, i was googeling around to find out that




    $$langle Drangle=frac12845pi quadquad (star)$$




    Fighting a bit longer, i am now ready to surrender and asking you




    How on Earth/Mars/Beteigeuze $(star)$ can be proven without plodding through a swamp of nightmarish antiderivatives?




    I tried a lot, from bruteforce methods over various changes of coordinate systems to series expansions but everything ended up in a big mess (interestingly for 3 dimensions the calculations go through smoothly)...



    Edit:



    Just in case somebody needs a bit of visual input, i created a Desmos chart



    setup



    Edit2:










    share|cite|improve this question











    $endgroup$














      3












      3








      3


      2



      $begingroup$


      At the moment I am interested in some applications of geometric probability theory, like finding the average distance $langle Drangle$ of two points randomly picked from the unit disk. It is not too difficult to set up an integral which correctly describes this distance:



      Pick two points $X_1,X_2$ inside the unit disk (one of them can, by symmetry, chosen to be located on the y-axis) and denote by $r_j=|X_j|$ their distance from the origin.



      Then $r_j$ are indenpendently distributed random variables with probability distribution $dp_j=2r_j$. Furthermore, the angle between these two points $phi$ is uniformly distributed with pd $dp_phi=1/pi$.



      From elementary geometry (law of cosines) is easy to see that $D^2=r_1^2+r_2^2-2 r_1r_2cos(phi)$ and therefore ($A$ is a cylinder with $r=1/2, z=1$ )



      $$
      langle Drangle=int_ADdp_1dp_2dp_phi=
      \frac4piint_0^1int_0^1int_0^pir_1 r_2sqrtr_1^2+r_2^2-2 r_1r_2cos(phi)dr_1dr_2dphi
      $$



      After struggeling for a few hours, i was googeling around to find out that




      $$langle Drangle=frac12845pi quadquad (star)$$




      Fighting a bit longer, i am now ready to surrender and asking you




      How on Earth/Mars/Beteigeuze $(star)$ can be proven without plodding through a swamp of nightmarish antiderivatives?




      I tried a lot, from bruteforce methods over various changes of coordinate systems to series expansions but everything ended up in a big mess (interestingly for 3 dimensions the calculations go through smoothly)...



      Edit:



      Just in case somebody needs a bit of visual input, i created a Desmos chart



      setup



      Edit2:










      share|cite|improve this question











      $endgroup$




      At the moment I am interested in some applications of geometric probability theory, like finding the average distance $langle Drangle$ of two points randomly picked from the unit disk. It is not too difficult to set up an integral which correctly describes this distance:



      Pick two points $X_1,X_2$ inside the unit disk (one of them can, by symmetry, chosen to be located on the y-axis) and denote by $r_j=|X_j|$ their distance from the origin.



      Then $r_j$ are indenpendently distributed random variables with probability distribution $dp_j=2r_j$. Furthermore, the angle between these two points $phi$ is uniformly distributed with pd $dp_phi=1/pi$.



      From elementary geometry (law of cosines) is easy to see that $D^2=r_1^2+r_2^2-2 r_1r_2cos(phi)$ and therefore ($A$ is a cylinder with $r=1/2, z=1$ )



      $$
      langle Drangle=int_ADdp_1dp_2dp_phi=
      \frac4piint_0^1int_0^1int_0^pir_1 r_2sqrtr_1^2+r_2^2-2 r_1r_2cos(phi)dr_1dr_2dphi
      $$



      After struggeling for a few hours, i was googeling around to find out that




      $$langle Drangle=frac12845pi quadquad (star)$$




      Fighting a bit longer, i am now ready to surrender and asking you




      How on Earth/Mars/Beteigeuze $(star)$ can be proven without plodding through a swamp of nightmarish antiderivatives?




      I tried a lot, from bruteforce methods over various changes of coordinate systems to series expansions but everything ended up in a big mess (interestingly for 3 dimensions the calculations go through smoothly)...



      Edit:



      Just in case somebody needs a bit of visual input, i created a Desmos chart



      setup



      Edit2:







      real-analysis integration geometry probability-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 31 at 22:06







      tired

















      asked Mar 31 at 13:49









      tiredtired

      10.6k12045




      10.6k12045




















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Due to a powerful theorem due to Crofton it is possible to proof $(star)$ by indirect means. The theorem states for our particular problem (we introduce variable disk radius $R$)



          $fracdlangle DrangledR=2(langle Drangle_1-langle Drangle)frac2R quad (starstar)$



          Here $langle Drangle_1$ is the average distance if one of the points sits at the circumference of the disk
          ($2/R$ comes from the quotient of the infinitesimal change of volume and the disk volume)
          which is easy enough to calculate since this distance is just the average length of the inscribed arc length of a second circle intersecting our disk at some distance $y$ away from our point at the circumference and is given by $2yTheta(y)$ where $Theta(y)=arccos(y/2R)$ is the angle betweeen the Diameter and the line connecting the two points on the circle
          $$
          langle Drangle_1= frac2pi R^2 int_0^2Ryyarccos(y/2R)dy=frac329pi
          $$



          Putting this back into $(starstar)$ and solving the ODE demanding $langle Drangle_R=0=0$ yields




          $$
          langle Drangle=frac12845pi R
          $$




          QED



          The direct Evaluation of the integral in my Question nevertheless stays a mystery



          Some visualisation:



          enter image description here






          share|cite|improve this answer











          $endgroup$













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            1 Answer
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            1 Answer
            1






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            active

            oldest

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            active

            oldest

            votes









            1












            $begingroup$

            Due to a powerful theorem due to Crofton it is possible to proof $(star)$ by indirect means. The theorem states for our particular problem (we introduce variable disk radius $R$)



            $fracdlangle DrangledR=2(langle Drangle_1-langle Drangle)frac2R quad (starstar)$



            Here $langle Drangle_1$ is the average distance if one of the points sits at the circumference of the disk
            ($2/R$ comes from the quotient of the infinitesimal change of volume and the disk volume)
            which is easy enough to calculate since this distance is just the average length of the inscribed arc length of a second circle intersecting our disk at some distance $y$ away from our point at the circumference and is given by $2yTheta(y)$ where $Theta(y)=arccos(y/2R)$ is the angle betweeen the Diameter and the line connecting the two points on the circle
            $$
            langle Drangle_1= frac2pi R^2 int_0^2Ryyarccos(y/2R)dy=frac329pi
            $$



            Putting this back into $(starstar)$ and solving the ODE demanding $langle Drangle_R=0=0$ yields




            $$
            langle Drangle=frac12845pi R
            $$




            QED



            The direct Evaluation of the integral in my Question nevertheless stays a mystery



            Some visualisation:



            enter image description here






            share|cite|improve this answer











            $endgroup$

















              1












              $begingroup$

              Due to a powerful theorem due to Crofton it is possible to proof $(star)$ by indirect means. The theorem states for our particular problem (we introduce variable disk radius $R$)



              $fracdlangle DrangledR=2(langle Drangle_1-langle Drangle)frac2R quad (starstar)$



              Here $langle Drangle_1$ is the average distance if one of the points sits at the circumference of the disk
              ($2/R$ comes from the quotient of the infinitesimal change of volume and the disk volume)
              which is easy enough to calculate since this distance is just the average length of the inscribed arc length of a second circle intersecting our disk at some distance $y$ away from our point at the circumference and is given by $2yTheta(y)$ where $Theta(y)=arccos(y/2R)$ is the angle betweeen the Diameter and the line connecting the two points on the circle
              $$
              langle Drangle_1= frac2pi R^2 int_0^2Ryyarccos(y/2R)dy=frac329pi
              $$



              Putting this back into $(starstar)$ and solving the ODE demanding $langle Drangle_R=0=0$ yields




              $$
              langle Drangle=frac12845pi R
              $$




              QED



              The direct Evaluation of the integral in my Question nevertheless stays a mystery



              Some visualisation:



              enter image description here






              share|cite|improve this answer











              $endgroup$















                1












                1








                1





                $begingroup$

                Due to a powerful theorem due to Crofton it is possible to proof $(star)$ by indirect means. The theorem states for our particular problem (we introduce variable disk radius $R$)



                $fracdlangle DrangledR=2(langle Drangle_1-langle Drangle)frac2R quad (starstar)$



                Here $langle Drangle_1$ is the average distance if one of the points sits at the circumference of the disk
                ($2/R$ comes from the quotient of the infinitesimal change of volume and the disk volume)
                which is easy enough to calculate since this distance is just the average length of the inscribed arc length of a second circle intersecting our disk at some distance $y$ away from our point at the circumference and is given by $2yTheta(y)$ where $Theta(y)=arccos(y/2R)$ is the angle betweeen the Diameter and the line connecting the two points on the circle
                $$
                langle Drangle_1= frac2pi R^2 int_0^2Ryyarccos(y/2R)dy=frac329pi
                $$



                Putting this back into $(starstar)$ and solving the ODE demanding $langle Drangle_R=0=0$ yields




                $$
                langle Drangle=frac12845pi R
                $$




                QED



                The direct Evaluation of the integral in my Question nevertheless stays a mystery



                Some visualisation:



                enter image description here






                share|cite|improve this answer











                $endgroup$



                Due to a powerful theorem due to Crofton it is possible to proof $(star)$ by indirect means. The theorem states for our particular problem (we introduce variable disk radius $R$)



                $fracdlangle DrangledR=2(langle Drangle_1-langle Drangle)frac2R quad (starstar)$



                Here $langle Drangle_1$ is the average distance if one of the points sits at the circumference of the disk
                ($2/R$ comes from the quotient of the infinitesimal change of volume and the disk volume)
                which is easy enough to calculate since this distance is just the average length of the inscribed arc length of a second circle intersecting our disk at some distance $y$ away from our point at the circumference and is given by $2yTheta(y)$ where $Theta(y)=arccos(y/2R)$ is the angle betweeen the Diameter and the line connecting the two points on the circle
                $$
                langle Drangle_1= frac2pi R^2 int_0^2Ryyarccos(y/2R)dy=frac329pi
                $$



                Putting this back into $(starstar)$ and solving the ODE demanding $langle Drangle_R=0=0$ yields




                $$
                langle Drangle=frac12845pi R
                $$




                QED



                The direct Evaluation of the integral in my Question nevertheless stays a mystery



                Some visualisation:



                enter image description here







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Apr 1 at 0:03

























                answered Mar 31 at 23:39









                tiredtired

                10.6k12045




                10.6k12045



























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