# Incorrect result of AsymptoticIntegrate

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?

• It should be noticed that as2 = AsymptoticIntegrate[ Log[t]^2/Sqrt[t^2 + 1], {t, 2, x}, {x, Infinity, 2}] returns the input. Oct 5, 2021 at 14:50
• Log[x]^3/3/.x->1000. results in 109.873. Oct 5, 2021 at 16:14
• Mathematica 12.2 states Integrate[Log[t]^2/Sqrt[t^2 + 1], {t, 2, Infinity}] doesn't converge. If 12.3 states the same, what kind of asymptotic behavior do you expect? Oct 6, 2021 at 14:33

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])   *)

• Many thanks from me to you for your analysis. I am glad that Mathematica can do it. In 12.3.1 on Windows 10 ser4 is 1/24 (Log[4]^3 + \[Pi]^2 Log[16] + 8 Log[x]^3 + 12 (-8 HypergeometricPFQ[{1/2, 1/2, 1/2, 1/2}, {3/2, 3/2, 3/ 2}, -4] - 2 ArcSinh[2] Log[2]^2 + HypergeometricPFQ[{1/2, 1/2, 1/2}, {3/2, 3/2}, -4] Log[256] + Zeta[3])). Oct 6, 2021 at 5:46

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*)

• Integrate[Log[t]^2/Sqrt[t^2 + 1], {t, 2, x}, Assumptions -> x > 2] produces the same as your F (see the question, line 12). Oct 6, 2021 at 16:53
• Also Series[F[x], {x, Infinity, 1}] produces a part of the expression in lines 19-22 of the question. Oct 6, 2021 at 16:59
• @user64494 My answer shows another way to solve the problem, though it's not surprising to get the same answers! Oct 6, 2021 at 20:38
• It's trivial that DSolve[{f'[x] == Log[x]^2/Sqrt[x^2 + 1], f[2] == 0}, f, x] is equivalent to Integrate[Log[t]^2/Sqrt[t^2 + 1], {t, 2, x}, Assumptions -> x > 2], isn't it? Oct 8, 2021 at 8:40
• Sure it is trivial, if you know th Fundamental theorem of calculus. But my answer shows a way to solve your problem with Mathamatica version <12.3! Oct 9, 2021 at 10:18