# How might one obtain a high-precision estimate of the integral over [0,1] of a specific function?

I have the following function

(1/(35 \[Pi]^3 (-1 +
t^2)^13))65536 t (-(-1 + t) (1 + t) (363 + 10310 t^2 + 58673 t^4 +
101548 t^6 + 58673 t^8 + 10310 t^10 + 363 t^12) +
140 (1 + t^2) (1 + 48 t^2 + 393 t^4 + 832 t^6 + 393 t^8 + 48 t^10 +
t^12) Log[t]) (t - t^3 + (-1 + t^4) ArcTanh[t] +
t^2 (-4 PolyLog[2, t] + PolyLog[2, t^2]))^2


and would like to obtain a high-precision estimate of its integral over [0,1] (which I would then input, say to WolframAlpha, to check for a possible exact value). I've encountered numerical problems near t = 1.

• How high a precision do you need? Dec 15, 2016 at 3:22
• Thanks bbgodfrey! Well, I'm basically (perhaps wildly) speculating that there might be some underlying exact value of note (maybe even a rational number), based on some prior considerations. So, if WolframAlpha, say, doesn't indicate such a value for a given precision, I could just try to get more numerical precision and try again. Of course, such a process could not go on for ever. I would imagine that perhaps an estimate accurate to say 15 places might settle the issue. Dec 15, 2016 at 3:35
• Try 0.450160328041638434796903399409, but I am not aware of any capabilities that WolframAlpha has that Mathematica does not. Dec 15, 2016 at 3:37
• Can you maybe describe the provenance of this function? Dec 15, 2016 at 6:48
• For a possible exact value: Can Mathematica propose an exact value based on an approximate one? and an example application. Try also RootApproximant with ToRadicals. Dec 15, 2016 at 7:55

f[t_] = (1/(35 \[Pi]^3 (-1 + t^2)^13)) 65536 t (-(-1 + t) (1 +
t) (363 + 10310 t^2 + 58673 t^4 + 101548 t^6 + 58673 t^8 +
10310 t^10 + 363 t^12) +
140 (1 + t^2) (1 + 48 t^2 + 393 t^4 + 832 t^6 + 393 t^8 +
48 t^10 + t^12) Log[t]) (t - t^3 + (-1 + t^4) ArcTanh[t] +
t^2 (-4 PolyLog[2, t] + PolyLog[2, t^2]))^2;


High WorkingPrecision is required to Plot

Plot[f[t], {t, 0, 1}, WorkingPrecision -> 30]


Table[
{wp, NIntegrate[f[t], {t, 0, 1},
WorkingPrecision -> wp,
MaxRecursion -> Infinity] //
N[#, 15] &}, {wp, 15, 40,
5}] //
Grid // Quiet


So the integral stabilizes to 15 decimal places at a WorkingPrecision of around 35

(int = NIntegrate[f[t], {t, 0, 1}, WorkingPrecision -> 35,
MaxRecursion -> Infinity]) // N[#, 15] & // Quiet

(*  0.450160328041638  *)


Using Round to rationalize

intr = int // Round[#, 10^-15] &

(*  225080164020819/500000000000000  *)

intr // N[#, 15] &

(*  0.450160328041638  *)

intr2 = Rationalize[int, 10*^-16]

(*  18677965/41491806  *)

intr2 // N[#, 15] &

(*  0.450160328041638  *)