Incorrect result of AsymptoticIntegrate

Question

The result in 12.3.1 on Windows 10 of

as1 = AsymptoticIntegrate[Log[t]^2/Sqrt[t^2 + 1], {t, 2, x}, {x, Infinity, 1}]

-ArcSinh[2] Log[2]^2 + ArcSinh[x] Log[x]^2

(The same result is obtained with the Assumptions->x>0 option.) is not confirmed numerically in view of discordance between

as1 /. x -> 1000.0000000000000000000000000

361.9993064078083517853350815

and

NIntegrate[Log[t]^2/Sqrt[t^2 + 1], {t, 2, 1000}]

109.662

Usually AsymptoticIntegrate works as the composition of Series with Integrate. Let us try it:

Integrate[Log[t]^2/Sqrt[t^2 + 1], {t, 2, x}, Assumptions -> x > 2]

-4 HypergeometricPFQ[{1/2, 1/2, 1/2, 1/2}, {3/2, 3/2, 3/2}, -4] + 2 x HypergeometricPFQ[{1/2, 1/2, 1/2, 1/2}, {3/2, 3/2, 3/2}, -x^2] - ArcSinh[2] Log[2]^2 + HypergeometricPFQ[{1/2, 1/2, 1/2}, {3/2, 3/2}, -4] Log[16] - 2 x HypergeometricPFQ[{1/2, 1/2, 1/2}, {3/2, 3/2}, -x^2] Log[x] + ArcSinh[x] Log[x]^2

Series[%, {x, Infinity, 2}]

$\frac{1}{24} \left(-96 \, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2},\frac{3}{2};-4\right)+24 \log (16) \, _3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2};-4\right)+8 \log ^3(x)+12 \log (4) \log ^2(x)+12 \gamma \log ^2(x)+12 \psi ^{(0)}\left(\frac{1}{2}\right) \log ^2(x)-\gamma ^3-2 \gamma \pi ^2-\psi ^{(0)}\left(\frac{1}{2}\right)^3-3 \gamma \psi ^{(0)}\left(\frac{1}{2}\right)^2-3 \gamma ^2 \psi ^{(0)}\left(\frac{1}{2}\right)-2 \pi ^2 \psi ^{(0)}\left(\frac{1}{2}\right)-\psi ^{(2)}\left(\frac{1}{2}\right)+\psi ^{(2)}(1)-24 \log ^2(2) \sinh ^{-1}(2)\right)+\frac{2 \log ^2(x)+2 \log (x)+1}{8 x^2}+O\left(\left(\frac{1}{x}\right)^3\right)$

Normal[%] /. x -> 1000.000000000000000000000000

109.6620820304208683244267272

It looks like a happy end. However, there is a simple elementary asymptotic Log[x]^3/3 which can be verified by

Limit[D[Integrate[Log[t]^2/Sqrt[t^2 + 1], {t, 2, x}], x]/D[Log[x]^3/3, x], x -> Infinity]

1

Can a more exact elementary asymptotic be derived with Mathematica?

Answer 1

It can be shown that difference between the exact integral and Log[x]^3/3 goes to the constant dlim = 0.210576 . So substract this dlim from Log[x]^3/3 to get an ideal asymptote at infinity, but not so good at low x.

int1[x_] = 
FullSimplify[
Integrate[Log[t]^2/Sqrt[t^2 + 1], {t, 2, x}, Assumptions -> x > 2], 
Assumptions -> x > 2]

(*   -4 HypergeometricPFQ[{1/2, 1/2, 1/2, 1/2}, {3/2, 3/2, 3/2}, -4] + 
2 x HypergeometricPFQ[{1/2, 1/2, 1/2, 1/2}, {3/2, 3/2, 3/2}, -x^2] - 
ArcSinh[2] Log[2]^2 + Log[2] (ArcCsch[2] + Log[2])^2 - 
ArcSinh[2]^2 Log[4] + ArcSinh[2] Log[4] Log[3 + Sqrt[5]] - 
ArcSinh[x]^2 Log[x] + ArcSinh[x] Log[x]^2 + 1/6 \[Pi]^2 Log[2 x] - 
2 ArcSinh[x] Log[x] Log[1 - x + Sqrt[1 + x^2]] - 
Log[4] PolyLog[2, 2 - Sqrt[5]] + 
Log[4] PolyLog[2, -(1/(1 + Sqrt[5]))] + 
2 Log[x] PolyLog[2, x - Sqrt[1 + x^2]] - 
2 Log[x] PolyLog[2, 1 + x - Sqrt[1 + x^2]]   *)

LogLinearPlot[{0.210576, Log[x]^3/3 - int1[x]}, {x, 2, 10^6}, 
PlotPoints -> 20, MaxRecursion -> 1, PlotStyle -> {Red, Blue}]

dlim = Limit[Log[x]^3/3 - int1[x], x -> \[Infinity]] //    FullSimplify

(*   4 HypergeometricPFQ[{1/2, 1/2, 1/2, 1/2}, {3/2, 3/2, 3/2}, -4] - 
1/3 \[Pi]^2 Log[2] - ArcCsch[2]^2 Log[2] - 2 ArcCsch[2] Log[2]^2 + 
ArcSinh[2] Log[2]^2 - (4 Log[2]^3)/3 + ArcSinh[2]^2 Log[4] - 
ArcSinh[2] Log[4] Log[3 + Sqrt[5]] + Log[4] PolyLog[2, 2 -  Sqrt[5]] -
Log[4] PolyLog[2, -(1/(1 + Sqrt[5]))] - Zeta[3]/2   *)

dlim // N      (*   0.210576   *)

Edit

By the way, you don't have to guess Log[x]^3/3 , you get it if you take Series of int1 at infinity and FullSimplify. Then the only term dependend on x is this Log-term and the constant is dlim.

ser4 = Series[int1[x], {x, \[Infinity], 0}] // Normal // 
 FullSimplify[#, x > 2] &

(*   1/6 (-24 HypergeometricPFQ[{1/2, 1/2, 1/2, 1/2},  {3/2, 3/2, 3/
  2}, -4] + 8 Log[2]^3 + \[Pi]^2 Log[4] + 
Log[64] (ArcCsch[2] (ArcCsch[2] + Log[4]) + 
  ArcSinh[2] Log[3 - Sqrt[5]]) 
+ 2 Log[x]^3 
+ 12 Log[2] (-PolyLog[2, 2 - Sqrt[5]] + 
  PolyLog[2, -(1/(1 + Sqrt[5]))]) + 3 Zeta[3])   *)

Answer 2

Sorry I'm a little bit late:

From the definition f[x] == Integrate[Log[t]^2/Sqrt[t^2 + 1], {t, 2, x}] it follows

f[2]==0 
f'[x]==Log[x]^2/Sqrt[x^2 + 1] 

which is solved directly

F = Values@DSolve[{f'[x] == Log[x]^2/Sqrt[x^2 + 1], f[2] == 0}, f, x][[1, 1]]
(*Function[{x}, -4 HypergeometricPFQ[{1/2, 1/2, 1/2, 1/2}, {3/2, 3/2, 3/
 2}, -4] +  2 x HypergeometricPFQ[{1/2, 1/2, 1/2, 1/2}, {3/2, 3/2, 3/
 2}, -x^2] + 4 HypergeometricPFQ[{1/2, 1/2, 1/2}, {3/2, 3/2}, -4] Log[2] -ArcSinh[2] Log[2]^2 - 2 x HypergeometricPFQ[{1/2, 1/2, 1/2}, {3/2, 3/2}, -x^2] Log[x] +ArcSinh[x] Log[x]^2]*)

Asymptotically we get a Series in Log[x]

Series[F[x], {x, Infinity, 1}]
(*-0.210576 + 0.333333 Log[x]^3+O[1/x]^2*)