1
$\begingroup$

How to compute the

$$\int_{0}^{1} \int_{0}^{1} \frac{(\log(1+x^2)-\log(1+y^2))^2 }{|x-y|^{2}}dx dy.$$ Is it possible to compute the integral analytically upto some terms. I believe it should involve hypergeometric series. Any ideas are welcome.

$\endgroup$
4
  • $\begingroup$ what does the title have in common with the question? $\endgroup$ Commented Apr 13, 2019 at 13:50
  • 1
    $\begingroup$ They both contain the letter "L" twice. $\endgroup$
    – Nik Weaver
    Commented Apr 13, 2019 at 14:57
  • $\begingroup$ @NikWeaver You mean "both numbers of occurrences of 'L' are perfect squares?". Anyway, the power series decomposition is, indeed, a reasonable way to compute it. I suspect that Mathematica formula was derived that way too. $\endgroup$
    – fedja
    Commented May 13, 2019 at 14:57
  • $\begingroup$ @fedja: I think my comment was accurate at the time ... just being a little snarky. $\endgroup$
    – Nik Weaver
    Commented May 13, 2019 at 16:11

1 Answer 1

3
$\begingroup$

With some effort, Mathematica evaluates this as

$$\int_{0}^{1} \int_{0}^{1} \frac{[\ln(1+x^2)-\ln(1+y^2)]^2 }{(x-y)^{2}}dx dy=2 \sqrt{2} \; _4F_3\left(\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2};\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2};\tfrac{1}{2}\right)+(4 \pi-\ln 2) C-2 i \,\text{Li}_3\left(\tfrac{1}{2}+\tfrac{1}{2}i\right)+2 i \left[\,\text{Li}_2\left(1-e^{i\pi/4}\right) - \text{Li}_2\left(-e^{i\pi/4}\right)\right] \ln 2-\left(\tfrac{69}{8}-\tfrac{35 }{32}i\right) \zeta (3)-\tfrac{23}{192} \pi ^3+\left(\tfrac{7}{2}-\tfrac{7 }{8}i\right) \pi +\tfrac{1}{24} i \ln ^3 2-\tfrac{7}{16} \pi \ln ^2 2-\left(\tfrac{1}{12}+\tfrac{9 }{32}i\right) \pi ^2 \ln 2+\tfrac{1}{2} \pi \ln \left(1+e^{i\pi/4}\right) \ln 2+\left(\tfrac{7}{8}+\tfrac{11 }{8}i\right) \ln 5-\left(\tfrac{3}{2}+\tfrac{7 }{4}i\right) \arctan\tfrac{1}{2}-\left(\tfrac{11}{2}-\tfrac{7 }{2}i\right) \arctan(1+i)-\tfrac{17}{4} \arctan 2=0.572532$$ with $C$ Catalan's constant and $\text{Li}_n$ the polylog.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .