Integrate $int_{-infty}^infty [4(log r_1 - log r_2) - 2(x_1^2/r_1^2 - x_2^2/r_2^2)]^2 dx$











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As the title suggests, I am having trouble evaluating the following definite integral:



$$int_{-infty}^infty left[4left(log r_1 - log r_2right) - 2left(frac{x_1^2}{r_1^2} - frac{x_2^2}{r_2^2}right)right]^2 dx$$



where



$$begin{align}
x_1 &= x-a\
x_2 &= x+a\
r_1^2 &= x_1^2 + z^2\
r_2^2 &= x_2^2 + z^2
end{align}$$



and $a > 0$, $z > 0$.



I've started by expanding the square, which gives



$$int_{-infty}^infty left[16left(log r_1 - log r_2right)^2 - 16left(log r_1 - log r_2right)left(frac{x_1^2}{r_1^2} - frac{x_2^2}{r_2^2}right) + 4left(frac{x_1^2}{r_1^2} - frac{x_2^2}{r_2^2}right)^2right] dx.$$



I managed to integrate the rightmost term $4(x_1^2/r_1^2 - x_2^2/r_2^2)^2$ using a partial fraction decomposition. However, I'm struggling with the two remaining terms: $16(log r_1 - log r_2)^2$ and $-16(log r_1 - log r_2)(x_1^2/r_1^2 - x_2^2/r_2^2)$.



I only know that the integral converges, as I am able to approach it numericaly by fixing $a$ and $z$.



If anyone has an idea on how to proceed, or, even better, has a solution, I'll take it.



Thanks!










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    up vote
    1
    down vote

    favorite












    As the title suggests, I am having trouble evaluating the following definite integral:



    $$int_{-infty}^infty left[4left(log r_1 - log r_2right) - 2left(frac{x_1^2}{r_1^2} - frac{x_2^2}{r_2^2}right)right]^2 dx$$



    where



    $$begin{align}
    x_1 &= x-a\
    x_2 &= x+a\
    r_1^2 &= x_1^2 + z^2\
    r_2^2 &= x_2^2 + z^2
    end{align}$$



    and $a > 0$, $z > 0$.



    I've started by expanding the square, which gives



    $$int_{-infty}^infty left[16left(log r_1 - log r_2right)^2 - 16left(log r_1 - log r_2right)left(frac{x_1^2}{r_1^2} - frac{x_2^2}{r_2^2}right) + 4left(frac{x_1^2}{r_1^2} - frac{x_2^2}{r_2^2}right)^2right] dx.$$



    I managed to integrate the rightmost term $4(x_1^2/r_1^2 - x_2^2/r_2^2)^2$ using a partial fraction decomposition. However, I'm struggling with the two remaining terms: $16(log r_1 - log r_2)^2$ and $-16(log r_1 - log r_2)(x_1^2/r_1^2 - x_2^2/r_2^2)$.



    I only know that the integral converges, as I am able to approach it numericaly by fixing $a$ and $z$.



    If anyone has an idea on how to proceed, or, even better, has a solution, I'll take it.



    Thanks!










    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      As the title suggests, I am having trouble evaluating the following definite integral:



      $$int_{-infty}^infty left[4left(log r_1 - log r_2right) - 2left(frac{x_1^2}{r_1^2} - frac{x_2^2}{r_2^2}right)right]^2 dx$$



      where



      $$begin{align}
      x_1 &= x-a\
      x_2 &= x+a\
      r_1^2 &= x_1^2 + z^2\
      r_2^2 &= x_2^2 + z^2
      end{align}$$



      and $a > 0$, $z > 0$.



      I've started by expanding the square, which gives



      $$int_{-infty}^infty left[16left(log r_1 - log r_2right)^2 - 16left(log r_1 - log r_2right)left(frac{x_1^2}{r_1^2} - frac{x_2^2}{r_2^2}right) + 4left(frac{x_1^2}{r_1^2} - frac{x_2^2}{r_2^2}right)^2right] dx.$$



      I managed to integrate the rightmost term $4(x_1^2/r_1^2 - x_2^2/r_2^2)^2$ using a partial fraction decomposition. However, I'm struggling with the two remaining terms: $16(log r_1 - log r_2)^2$ and $-16(log r_1 - log r_2)(x_1^2/r_1^2 - x_2^2/r_2^2)$.



      I only know that the integral converges, as I am able to approach it numericaly by fixing $a$ and $z$.



      If anyone has an idea on how to proceed, or, even better, has a solution, I'll take it.



      Thanks!










      share|cite|improve this question













      As the title suggests, I am having trouble evaluating the following definite integral:



      $$int_{-infty}^infty left[4left(log r_1 - log r_2right) - 2left(frac{x_1^2}{r_1^2} - frac{x_2^2}{r_2^2}right)right]^2 dx$$



      where



      $$begin{align}
      x_1 &= x-a\
      x_2 &= x+a\
      r_1^2 &= x_1^2 + z^2\
      r_2^2 &= x_2^2 + z^2
      end{align}$$



      and $a > 0$, $z > 0$.



      I've started by expanding the square, which gives



      $$int_{-infty}^infty left[16left(log r_1 - log r_2right)^2 - 16left(log r_1 - log r_2right)left(frac{x_1^2}{r_1^2} - frac{x_2^2}{r_2^2}right) + 4left(frac{x_1^2}{r_1^2} - frac{x_2^2}{r_2^2}right)^2right] dx.$$



      I managed to integrate the rightmost term $4(x_1^2/r_1^2 - x_2^2/r_2^2)^2$ using a partial fraction decomposition. However, I'm struggling with the two remaining terms: $16(log r_1 - log r_2)^2$ and $-16(log r_1 - log r_2)(x_1^2/r_1^2 - x_2^2/r_2^2)$.



      I only know that the integral converges, as I am able to approach it numericaly by fixing $a$ and $z$.



      If anyone has an idea on how to proceed, or, even better, has a solution, I'll take it.



      Thanks!







      calculus integration logarithms polylogarithm






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      asked Nov 21 at 15:15









      Son Pham-Ba

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