Best way to integrate a function

ROOT
1

Hey,

I want to integrate the function below:

  (4*lambda*(1-z*z)) /( ((lambda+1)*(lambda+1) - (lambda-1)*(lambda-1)*z*z)*((lambda+1)*(lambda+1) - (lambda-1)*(lambda-1)*z*z));

I want to integrate over various z values, but plot as a function of lambda. I’m not sure how best to go about this. I have a range of z values i.e 0.2, 0.4, 0.6, 0.8 and 1.0. and Lambda is between 1.0001 and 1000.1. I’m not sure how best to approach this. Any advice would be appreciated.

Thanks

2

Hi,

you can create a TF1 and then use the TF1::Integrate method. An example integrating f = a + b*x for different values of a and b:

double func(double* x, double* p) {
    double xx = x[0];
    double a = p[0];
    double b = p[1];

    return a + b * xx;
}

void root_int() {
    TF1 f("func", func, 0, 1, 2);
    f.SetParameters(0, 1); // Set a = 0, b = 1
    auto i = f.Integral(0, 1); // integrate between 0 and 1
    std::cout << i << std::endl; // i = 0.5

    f.SetParameters(0, 2); // Set a = 0, b = 2
    i = f.Integral(0, 1); // integrate between 0 and 1
    std::cout << i << std::endl; // i = 1


    // etc...
}
3

Hey,

I don’t quite follow this. I had initially thought of using a TF1 and figured integrating over z would be easy, but I wasn’t sure then if it was possible to plot as a function of lambda. I don’t see how this is achieved via your example? Unless I’m missing something…

Edit: I realised I wasn’t that clear. I want to take my function, and integrate over a particular z value and then plot that integrated function over the whole range of lambda (between 1.001 and 1000.1)

4

The easiest would probably be to save the values of lambda and the corresponding integral value in two vectors, and then make a plot using TGraph. I suppose you can also create a second function that computes the value of the integral, and plot that directly, e.g.:

double func(double* x, double* p) {
    double xx = x[0];
    double a = p[0];
    double b = p[1];

    return a + b * xx;
}

double integral(double* x, double* p) {
    double b = x[0];
    TF1 f("func", func, 0, 1, 2);
    f.SetParameters(0, b); // Set a = 0, b = b
    return f.Integral(0, 1);
}

void root_int() {

    TF1* g = new TF1("integral", integral, 0, 5, 0);
    g->Draw();
}
5

Hey,

Thanks, I will try and adapt this to my needs as I hope to plot all the integrated z values as a function of lambda on the same plot.

6

Hmmm, I don’t think it’s doing the integration right :confused:

7

What concerns “intergration”, see for example:

8

I changed the example code, to something of my own to understand exactly what is going on:

 double fsin(double *x, double *par){
 double y = x[0]; 
 double a = par[0];
 return a*sin(y);
}


double Integral(double *x, double *par){
 double a = x[0];
 TF1 f("fsin", fsin, 0, 10, 1); //min, max, no.parameters
  f.SetParameter(0,a);  //setting the variable to be integrated
  return f.Integral(0,5);  //Integrate 
  }

void Integrations2(){
 
  TF1* g = new TF1("Integral", Integral, 0, 10, 0);
  g->Draw();

   }

The graph that I got for the integration doesn’t seem right. It shouldn’t be a straight line but a curve. So this is why I don’t think it’s working.

9

Isn’t a straight line correct in this case? If you integrate f = asin(x) you get F = -acos(x), so if you integrate between x1 and x2 you will get I = -a*(cos(x2) - cos(x1)). So if x1 and x2 are the same in all cases you will get a line with slope cos(x1) - cos(x2) as a function of a.

10

Ahhh, massive maths fail on my part! Yes, you are correct. Thank you for this!

11

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