Recently I came across with respect to this post of mine hyperbolic solution to the cubic equation for one real root given by $$ t=-2\sqrt \frac {p}{3} \sinh \left( \frac {1}{3} \sinh^{-1} \left( \frac {3q}{2p} \sqrt \frac {3}{p} \right) \right) $$ Intuitively I sought to find the related definitely integral, $$ I=\int^{1}_{-1} \frac {1}{3} \sinh^{-1} \left( \frac {3\sqrt 3}{2} (1-t^2) \right) dt $$ Unfortunately, there was no closed form solution. However, the Integral is amazingly near $\sqrt 2$. $$ I=0.8285267994716327, \frac {I}{2} +1=1.4142633998 $$ To investigated more, I tried a heuristic expansion of the integral into Egyptian fractions. Although it gets problematic after the 4th term, The first four terms are, $$ \frac {I}{2} +1 = 1+ \frac {1}{2} - \frac {1}{12}-\frac {1}{416} $$ Here the denominators can be given by, $$ a_n = \sum_{k=0}^{n} { }^nC_k (2^n - 2^kq)^{n-k}q^k , q=\sqrt 2 $$ (Likewise, the denominators in the expansion for $\sqrt 2$ are related to Pell numbers, which makes me believe that my integral too is somewhat related to the numbers $a_n$.) Therefore, I am finding either a closed form or possibly a fast converging infinite series solution to the integral, just any of these. Thanks for any help.

The indefinite integral

For $t=\sin z$ and applying integration by parts, I get another, somewhat simpler, indefinite integral, $$ \frac {\sin z}{3} \sinh^{-1} \left( \frac {3\sqrt 3}{2} \cos^2 z \right) + 2\sqrt 3 \int \frac {\sin^2 z \cos z dz}{\sqrt {27\cos^4 z + 4}} $$ Then again I am stuck. Moreover, this expression ensures that my definite integral is an improper one.


A closed solution in terms of incomplete elliptic integrals with complex arguments is, as given by a user in the comments section, $$ \frac {4}{9} (9+2\sqrt 3 i) \left[ F \left( \sin^{-1} \sqrt {\frac {3}{31}(9+2\sqrt 3 i)} ; \frac {1}{31} (23-12\sqrt 3 i) \right)- E \left( \sin^{-1} \sqrt {\frac {3}{31}(9+2\sqrt 3 i)} ; \frac {1}{31} (23-12\sqrt 3 i) \right) \right] $$ However, I am still wondering how to transform this into a real number, especially the $a_n$ connection of the integral is fascinating my mind.

  • $\begingroup$ Have you tried integration by parts? with $u = $ your integrand and $v' = 1 $ ? $\endgroup$ – user619699 Mar 18 at 12:32
  • $\begingroup$ With CAS I have solution by elliptic integrals. $\endgroup$ – Mariusz Iwaniuk Mar 18 at 16:47
  • $\begingroup$ @Mariusz Iwaniuk, would you please bother to tell me the result, for it would be then somewhat easier to find the steps to solution than what I am doing now (I am firing bullets in the dark!!). $\endgroup$ – Awe Kumar Jha Mar 18 at 17:02
  • 1
    $\begingroup$ Mathematica code,answer is: 4/9 Sqrt[9 + 2 I Sqrt[3]] (-EllipticE[ArcSin[Sqrt[3/31 (9 + 2 I Sqrt[3])]], 1/31 (23 - 12 I Sqrt[3])] + EllipticF[ArcSin[Sqrt[3/31 (9 + 2 I Sqrt[3])]], 1/31 (23 - 12 I Sqrt[3])]) $\endgroup$ – Mariusz Iwaniuk Mar 18 at 17:03
  • 3
    $\begingroup$ Using the substitution $y^2=1-x^2$ the equivalent integral is $$\frac{2}{3} \int_0^1 \frac{y \sinh ^{-1}\left(a y^2\right)}{\sqrt{1-y^2}} \, dy$$, where $a=\frac{3\sqrt{3}}{2}$. Mathematical 11.3 then gives the answer $$\frac{4}{9}\, a \,\, _3F_2\left(\frac{1}{2},\frac{1}{2},1;\frac{5}{4},\frac{7}{4};-a^2\right),$$ for $a>0$. $\endgroup$ – James Arathoon Mar 19 at 10:50

We have from symmetry that $$I=\frac23\int_0^1\sinh^{-1}\left[\frac{3\sqrt3}2(1-x^2)\right]dx$$ So we define $$f(a)=\int_0^1\sinh^{-1}[a(1-x^2)]dx$$ Then we recall that $$\sinh^{-1}(x)=x\,_2F_1\left(\frac12,\frac12;\frac32;-x^2\right)=\sum_{n\geq0}(-1)^n\frac{(1/2)_n^2}{(3/2)_n}\frac{x^{2n+1}}{n!}$$ so $$\sinh^{-1}[a(1-x^2)]=a(1-x^2)\,_2F_1\left(\frac12,\frac12;\frac32;-a^2(1-x^2)^2\right)\\ =\sum_{n\geq0}(-1)^n\frac{a^{2n+1}}{n!}\frac{(1/2)_n^2}{(3/2)_n}(1-x^2)^{2n+1}$$ so $$f(a)=\sum_{n\geq0}(-1)^n\frac{a^{2n+1}}{n!}\frac{(1/2)_n^2}{(3/2)_n}\int_0^1(1-x^2)^{2n+1}dx$$ For this integral, we use $x=\sin(t)$: $$j_n=\int_0^1(1-x^2)^{2n+1}dx=\int_0^{\pi/2}\cos(t)^{4n+3}dt$$ I leave it as a challenge to you to show that $$\int_0^{\pi/2}\sin(t)^a\cos(t)^bdt=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}$$ choosing $b=4n+3$, $a=0$ we have $$j_n=\frac{\Gamma(1/2)\Gamma(2n+2)}{2\Gamma(2n+5/2)}$$ Then defining $$t_n=\frac{(1/2)_n^2}{(3/2)_n}j_n$$ we have $$\frac{t_{n+1}}{t_n}=\frac{(n+\frac12)^2(n+1)}{(n+\frac74)(n+\frac54)}$$ Which gives $$f(a)=a\,_3F_2\left(\frac12,\frac12,1;\frac74,\frac54;-a^2\right)$$ And since $I=\frac23f(3\sqrt3/2)$ we have (assuming I've made no mistakes), $$I=\sqrt3\,_3F_2\left(\frac12,\frac12,1;\frac74,\frac54;-\frac{27}4\right)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.