Is it possible to have the asymptotics of this function?

Question

Working the problem of $$I_n=\int_0^1 \frac{\tan ^{-1}\left(x^n\right)}{\sqrt{1-x^n}} \, dx$$ which I have not be able to compute with Mathematica. A tedious work gave the result $$I_n=\frac{\sqrt{\pi }}{n}\,\,\frac{\Gamma \left(\frac{n+1}{n}\right)}{\Gamma \left(\frac{3 n+2}{2 n}\right)}\,\,\,\, _4F_3\left(\frac{1}{2},1,\frac{n+1}{2 n},\frac{2 n+1}{2 n};\frac{3}{2},\frac{3 n+2}{4 n},\frac{5 n+2}{4 n};-1\right)$$ which works for any $n$ (even complex).

The syntax is

  (Sqrt[Pi]*Gamma[(1 + n)/n]*HypergeometricPFQ[{1/2, 1, 1/2 + 1/(2*n), 
   1 + 1/(2*n)}, {3/2, 3/4 + 1/(2*n), 5/4 + 1/(2*n)}, -1])/  (n*Gamma[(2 + 3*n)/(2*n)])

Since Mathematica was not able to compute the integral AsymptoticIntegrate is of no help.

Empirically, it seems that $n^2 I_n$ is very close to linearity.

Is there any hope to obtain the asymptotics ?

Thanks in advance.

Answer 1

The following proves your expectations.

FullSimplify[Series[Sqrt[Pi]*Gamma[(1 + n)/n]*
HypergeometricPFQ[{1/2, 1, 1/2 + 1/(2*n), 1 + 1/(2*n)},
{3/2,3/4 + 1/(2*n), 5/4 + 1/(2*n)}, -1]/(n*
Gamma[(2 + 3*n)/(2*n)]), {n, Infinity, 2}] // Normal]

$$\frac{\text{HypergeometricPFQ}^{(\{0,0,0,0\},\{0,0,1\},0)}\left(\left\{\frac{1}{2},1,\frac{1}{2},1\right\},\left\{\frac{3}{2},\frac{3}{4},\frac{5}{4}\right\},-1\right)+\text{HypergeometricPFQ}^{(\{0,0,0,0\},\{0,1,0\},0)}\left(\left\{\frac{1}{2},1,\frac{1}{2},1\right\},\left\{\frac{3}{2},\frac{3}{4},\frac{5}{4}\right\},-1\right)+\text{HypergeometricPFQ}^{(\{0,0,0,1\},\{0,0,0\},0)}\left(\left\{\frac{1}{2},1,\frac{1}{2},1\right\},\left\{\frac{3}{2},\frac{3}{4},\frac{5}{4}\right\},-1\right)+\text{HypergeometricPFQ}^{(\{0,0,1,0\},\{0,0,0\},0)}\left(\left\{\frac{1}{2},1,\frac{1}{2},1\right\},\left\{\frac{3}{2},\frac{3}{4},\frac{5}{4}\right\},-1\right)+2 (n-2+\log (4)) \, _4F_3\left(\frac{1}{2},\frac{1}{2},1,1;\frac{3}{4},\frac{5}{4},\frac{3}{2};-1\right)}{n^2} $$

Expand[N[%]]

$$\frac{1.74825}{n}-\frac{1.17554}{n^2} $$

Answer 2

Another way to get @user64494 result:

Normal@Series[ArcTan[x^n]/Sqrt[1 - x^n], {n, Infinity, 2}]
(*(I Log[1 - I x^n])/(2 Sqrt[1 - x^n]) - (I Log[1 + I x^n])/( 2 Sqrt[1 - x^n])*)

Simplify[Integrate[(I Log[1 - I x^n])/(2 Sqrt[1 - x^n]), {x, 0, 1}, GenerateConditions -> False] + Integrate[-((I Log[1 + I x^n])/(2 Sqrt[1 - x^n])), {x, 0, 1}, GenerateConditions -> False]];
Asymptotic[%, {n, Infinity, 2}] // N // Chop

(*-(1.17554/n^2) + 1.74825/n*)

Answer 3

With AsymptoticIntegrate works fine:

$Version
(*"13.3.0 for Microsoft Windows (64-bit) (June 3, 2023)"*)
 
AsymptoticIntegrate[ArcTan[x^n]/Sqrt[1 - x^n] // TrigToExp, {x, 0, 1}, {n, Infinity, 2}]

 (*((-I)*(ArcSin[(-1)^(1/4)]^2 + ArcSinh[(-1)^(1/4)]^2))/n + 
 (Sqrt[Pi]*(((2*I)*EulerGamma*(ArcSin[(-1)^(1/4)]^2 + 
 ArcSinh[(-1)^(1/4)]^2))/Sqrt[Pi] + 
 Derivative[{0, 0, 0}, {0, 1}, 0][HypergeometricPFQRegularized][{1, 1, 1}, {2, 3/2}, -I] + 
 Derivative[{0, 0, 0}, {0, 1}, 0][HypergeometricPFQRegularized][{1, 1, 1}, {2, 3/2}, I] + 
 Derivative[{0, 0, 1}, {0, 0}, 0][HypergeometricPFQRegularized][{1, 1, 1}, {2, 3/2}, -I] + 
 Derivative[{0, 0, 1}, {0, 0}, 0][HypergeometricPFQRegularized][{1, 1, 1}, {2, 3/2}, I]))/(2*n^2)*)

% // N
(*-((1.17554 + 0. I)/n^2) + (1.74825 + 0. I)/n*)

To compute integral we use a trick: $$\int_0^1 \frac{\tan ^{-1}\left(x^n\right)}{\sqrt{1-x^n}} \, dx$$

Integrate[Integrate[D[ArcTan[a x^n]/Sqrt[1 - x^n], a], {x, 0, 1},  
GenerateConditions -> False], {a, 0, 1}] // FunctionExpand

(* (Sqrt[\[Pi]]Gamma[1/n] HypergeometricPFQ[{1/2, 1, 1/2 + 1/(2 n), 1 + 1/(2 n)}, {3/2, 
 3/4 + 1/(2 n), 5/4 + 1/(2 n)}, -1])/(n^2 Gamma[3/2 + 1/n])*)

Answer 4

The solution has already been posted; here are some semi-explicit forms of the series-expansion coefficients:

Sum[(-1)^q/((q + 1/2) Binomial[2 q + 1/2, 1/2]), {q, 0, ∞}]
(*    2 HypergeometricPFQ[{1/2, 1/2, 1, 1}, {3/4, 5/4, 3/2}, -1]    *)

NSum[-((-1)^q/((q + 1/2) Binomial[2 q + 1/2, 1/2])) *
     (HarmonicNumber[2 q + 1/2] - HarmonicNumber[2 q]), {q, 0, ∞}]
(*    -1.17554    *)