       Re: Calculate a numerical integral with enough precision

• To: mathgroup at smc.vnet.net
• Subject: [mg114726] Re: Calculate a numerical integral with enough precision
• From: Andrew Moylan <amoylan at wolfram.com>
• Date: Tue, 14 Dec 2010 06:56:05 -0500 (EST)

```This is a tricky one. I don't know any automatic methods for these kind of integrals.

I found one manual method which works however:

The new "LevinRule" method for NIntegrate in Mathematica 8 can efficiently handle integer powers of oscillatory functions. Therefore you can try replacing Exp with its power series:

Exp[Sin[1/x]] = Sin[1/x] + Sin[1/x]^2 / 2! + ...

In:= terms =
Table[NIntegrate[Sin[1/x]^n / n!, {x, 0, 2 \[Pi]}], {n, 0, 20}]

Out= {6.28319, 2.26277, 0.706044, 0.135229, 0.0326695, \
0.00439035, 0.000818095, 0.0000821659, 0.0000121744, 9.68476*10^-7,
1.18363*10^-7, 7.7766*10^-9, 8.07019*10^-10, 4.51197*10^-11,
4.06465*10^-12, 1.97691*10^-13, 1.57263*10^-14, 6.76684*10^-16,
4.81812*10^-17, 1.85861*10^-18, 1.19749*10^-19}

The terms shrink fast:

ListLogPlot[terms]

Therefore I suspect all the digits shown here are correct:

In:= Total[terms] // InputForm

Out//InputForm=
9.425202795524033

A different way may be to use a contour integral if you can find the right contour.

----- Original Message -----
From: "Alexei Boulbitch" <alexei.boulbitch at iee.lu>
To: mathgroup at smc.vnet.net
Sent: Monday, December 13, 2010 10:34:34 PM
Subject: [mg114726] [mg114709] Re: Calculate a numerical integral with enough precision

NIntegrate[Exp[Sin[1/x]], {x, 0, \[Pi]/2}, WorkingPrecision -> 12,
Exclusions -> (x == 0), PrecisionGoal -> 15] // Quiet

3.09892833696

* /Subject/: [mg114668] Calculate a numerical integral with enough
precision
* /From/: alphatest <iliurarfwpuap at mailinator.com>
* /Date/: Sun, 12 Dec 2010 05:45:17 -0500 (EST)

------------------------------------------------------------------------

How can we calculate the following integral up to 10-12 decimal places?

integrate exp(sin(1/x)) , x=0..pi/2

It's as if no integration method or precision option lets you calculate more than 5-6 decimal places. Is it possible?

--
Alexei Boulbitch, Dr. habil.
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