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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.

Update

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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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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), $$I=\sqrt3\,_3F_2\left(\frac12,\frac12,1;\frac74,\frac54;-\frac{27}4\right)$$

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