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new to Mathematica. I am trying to approximate to ~30 decimal places $0<t<1$ satisfying $$\int_0^1 \frac{\sqrt{1-\sqrt x}}{\arctan(t + \arctan(x))} ~dx = \frac{\pi^2}{6}$$ I used the following code:

NSolve[Integrate[Sqrt[1 - Sqrt[x]]/ArcTan[t + ArcTan[x]], {x, 0, 1}] == Pi^2/6 && t > 0 && t < 1,t]

But the computation is taking very, very long. What am I doing wrong? How can I optimize this? A similar question has likely been asked before, but I do not know what to search for. Should I use NIntegrate, somehow?

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Use FindRoot

Clear["Global`*"]

f[t_?NumericQ] := Module[{tt = SetPrecision[t, 30]},
  NIntegrate[Sqrt[1 - Sqrt[x]]/ArcTan[tt + ArcTan[x]], {x, 0, 1}, 
   WorkingPrecision -> 30]]

Plot the sides of the equation

Plot[{f[t], Pi^2/6}, {t, 0, 1},
 WorkingPrecision -> 30,
 PlotLegends -> Placed["Expressions", {.5, .7}]]

enter image description here

sol1 = FindRoot[f[t] == Pi^2/6, {t, 3/25}, WorkingPrecision -> 30]

(* {t -> 0.112689874579840073770094027065} *)

Verifying the solution

f[t] - Pi^2/6 /. sol1

(* 0.*10^-30 *)
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You can search it.

Select[Table[{t, 
NIntegrate[
 Sqrt[1. - Sqrt[x]]/ArcTan[t + ArcTan[x]], {x, 0, 1}]}, {t, 0, 1, 
0.001}], RealAbs[#[[2]] - Pi^2/6] < 0.01 &]

{{0.112, 1.64927}, {0.113, 1.643}, {0.114, 1.63679}}

so 0.112 ~ 0.114 is where the answer at.

With more search, you can get more precious answer.

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  • $\begingroup$ Thank you - how can I get the right-side elements of the list items to be more precise? That is, display more decimal places? $\endgroup$ – Descartes Before the Horse Jan 21 at 5:01
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    $\begingroup$ @TheWheelisBeforeDescartes downstair gives out better solution. Back to my answer, to get preciser answer, you can replace {t, 0, 1, 0.001} with {t, 0.112, 0.114, 0.0001} and then set the <0.01& to < 0.001& . Or you can write other code to do this ,which is kind of trouble(not recommend). $\endgroup$ – wuyudi Jan 21 at 5:34
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another option is to find a function to fit the plot, then use Solve to find the root.

ClearAll[f, x, t];
f[t_?NumericQ] := NIntegrate[Sqrt[1 - Sqrt[x]]/ArcTan[t + ArcTan[x]], {x, 0, 1}]
Plot[f[t] - Pi^2/6, {t, 0, 1}]

Mathematica graphics

So your t is somewhere around 0.1. To find it, you could fit the curve to a function and then use Solve or root finding. For example

data=Catenate@Cases[Plot[f[t]-Pi^2/6,{t,0.1,0.2}],Line[data_]:>data,Infinity];
fit=Fit[data,{1,t,t^2,t^3,t^4},t]

Mathematica graphics

 Plot[fit, {t, 0.1, 0.2}, PlotRange -> All]

Mathematica graphics

Solve[fit == 0, t, Reals]

Mathematica graphics

or

 FindRoot[fit == 0, {t, 0.1}]

Mathematica graphics

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  • $\begingroup$ Would be hard to get 30 digits of accuraccy this way though. $\endgroup$ – mmeent Jan 22 at 7:17

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