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Mathematics Calculus Wolfram Language Numerical Computation

Speed up Nintegrate for certain range of function's input

John Wick

Posted 1 year ago

This is a part of my code SE[x_, g_, c_] := 1/x NIntegrate[Rationalize[( 6 (10^-10 Sqrt[xP])^2 10^-11 (10^-7 Sqrt[( 10^-38 c^2 xP + (10^-10 Sqrt[xP])/x)/( 0.1` + 10^-38 c^2)])^3)/((10^-38 c^2 xP + (10^-10 Sqrt[xP])/ x)^4/(0.1` + 10^-38 c^2)^4)], {xP, 1.`/g, 10^12}, PrecisionGoal -> 10.`, MinRecursion -> 3.`, MaxRecursion -> 15.`, Method -> {Automatic, "SymbolicProcessing" -> 0.`}] For which this part Sqrt[NIntegrate[(SE[x, 10^-23, 10^14])^2, {x, 10^-5, 1}]] // AbsoluteTiming Taking too much time for my range of values for `c->(10^9-10^15)` but when I decrease the value of `c` the integral becomes faster. I have to make the integral atleast `10-100` times more faster

This is a part of my code

SE[x_, g_, c_] := 1/x NIntegrate[Rationalize[(
6 (10^-10 Sqrt[xP])^2 10^-11 (10^-7 Sqrt[(
   10^-38 c^2 xP + (10^-10 Sqrt[xP])/x)/(
   0.1` + 10^-38  c^2)])^3)/((10^-38 c^2 xP + (10^-10 Sqrt[xP])/
  x)^4/(0.1` + 10^-38  c^2)^4)], {xP, 1.`/g, 10^12}, 
PrecisionGoal -> 10.`, MinRecursion -> 3.`, MaxRecursion -> 15.`, 
Method -> {Automatic, "SymbolicProcessing" -> 0.`}]

For which this part

Sqrt[NIntegrate[(SE[x, 10^-23, 10^14])^2, {x, 10^-5, 
 1}]] // AbsoluteTiming

Taking too much time for my range of values for c->(10^9-10^15) but when I decrease the value of c the integral becomes faster. I have to make the integral atleast 10-100 times more faster

POSTED BY: John Wick

6 Replies

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Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 1 year ago

If Mathematica doesn't do it automatically, you can easily work out the point where the `HeavisideTheta` discontinuity lies, and the conditions under which that point is to the left, within, or to the right of the integration interval `[1/g, 10^12]`.

POSTED BY: Gianluca Gorni

John Wick

Posted 1 year ago

But I can't assume `{g = 10^-23, c = 10^14}` as a general solution since it will give the results for a point but in principle `g,c` have to vary for a range of values `g=>(10^-2,10^-24)` and `c=>(10^10-10^15)`

POSTED BY: John Wick

Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 1 year ago

When integrating, I would recommend `UnitStep` instead of `HeavisideTheta`. This gives me a symbolic output: With[{g = 10^-23, c = 10^14}, Integrate[ Simplify[(6 (10^-10 Sqrt[ xP])^2 10^-11 (10^-7 Sqrt[(10^-38 c^2 xP + (10^-10 Sqrt[ xP])/x)/(1/10 + 10^-38 c^2)])^3)/((10^-38 c^2 xP + (10^-10 Sqrt[xP])/ x)^4/(1/10 + 10^-38 c^2)^4)* UnitStep[10^-38 c^2 xP + (10^-10 Sqrt[xP])/x - 1/g]], {xP, 1/g, 10^12}, Assumptions -> And[c > 0, x > 0, xP > 0, g > 0]]]

When integrating, I would recommend UnitStep instead of HeavisideTheta. This gives me a symbolic output:

With[{g = 10^-23, c = 10^14}, 
 Integrate[
  Simplify[(6 (10^-10 Sqrt[
           xP])^2 10^-11 (10^-7 Sqrt[(10^-38 c^2 xP + (10^-10 Sqrt[
                  xP])/x)/(1/10 + 
              10^-38 c^2)])^3)/((10^-38 c^2 xP + (10^-10 Sqrt[xP])/
           x)^4/(1/10 + 10^-38 c^2)^4)*
    UnitStep[10^-38 c^2 xP + (10^-10 Sqrt[xP])/x - 1/g]],
  {xP, 1/g, 10^12},
  Assumptions -> And[c > 0, x > 0, xP > 0, g > 0]]]

POSTED BY: Gianluca Gorni

John Wick

Posted 1 year ago

Also, one important reason I have not gone for symbolic integration is that my integrand in its complete form SE[x_, g_, c_] = 1/x Integrate[ Simplify[(6 (10^-10 Sqrt[ xP])^2 10^-11 (10^-7 Sqrt[(10^-38 c^2 xP + (10^-10 Sqrt[ xP])/x)/(1/10 + 10^-38 c^2)])^3)/((10^-38 c^2 xP + (10^-10 Sqrt[xP])/ x)^4/(1/10 + 10^-38 c^2)^4)* HeavisideTheta[10^-38 c^2 xP + (10^-10 Sqrt[xP])/x - 1/g]], {xP, 1/g, 10^12}, Assumptions -> And[c > 0, x > 0, xP > 0, g > 0]] contains a `HeavisideTheta` function

Also, one important reason I have not gone for symbolic integration is that my integrand in its complete form

SE[x_, g_, c_] = 
 1/x Integrate[
   Simplify[(6 (10^-10 Sqrt[
            xP])^2 10^-11 (10^-7 Sqrt[(10^-38 c^2 xP + (10^-10 Sqrt[
                   xP])/x)/(1/10 + 
               10^-38 c^2)])^3)/((10^-38 c^2 xP + (10^-10 Sqrt[xP])/
            x)^4/(1/10 + 10^-38 c^2)^4)*
     HeavisideTheta[10^-38 c^2 xP + (10^-10 Sqrt[xP])/x - 1/g]], {xP, 
    1/g, 10^12}, Assumptions -> And[c > 0, x > 0, xP > 0, g > 0]]

contains a HeavisideTheta function

POSTED BY: John Wick

John Wick

Posted 1 year ago

Symbolical evolution seems more faster but it gives a solution for `g>10^-12`, but my `g` can be low as `10^-24`

POSTED BY: John Wick

Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 1 year ago

Have you tried pre-calculating the first integral symbolically? SE[x_, g_, c_] = 1/x Integrate[ Simplify[(6 (10^-10 Sqrt[ xP])^2 10^-11 (10^-7 Sqrt[(10^-38 c^2 xP + (10^-10 Sqrt[ xP])/x)/(1/10 + 10^-38 c^2)])^3)/((10^-38 c^2 xP + (10^-10 Sqrt[xP])/ x)^4/(1/10 + 10^-38 c^2)^4)], {xP, 1/g, 10^12}, Assumptions -> And[c > 0, x > 0, xP > 0, g > 0]]

Have you tried pre-calculating the first integral symbolically?

SE[x_, g_, c_] = 
 1/x Integrate[
   Simplify[(6 (10^-10 Sqrt[
           xP])^2 10^-11 (10^-7 Sqrt[(10^-38 c^2 xP + (10^-10 Sqrt[
                  xP])/x)/(1/10 + 
              10^-38 c^2)])^3)/((10^-38 c^2 xP + (10^-10 Sqrt[xP])/
           x)^4/(1/10 + 10^-38 c^2)^4)], {xP, 1/g, 10^12}, 
   Assumptions -> And[c > 0, x > 0, xP > 0, g > 0]]

POSTED BY: Gianluca Gorni

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