# Simple Integral

Hi there,

This might result being a very dumb mistake from myself, but I do not understand what is happening with a simple piece of code. I am trying to get the integral of a Gaussian distribution (this is not my final goal, but it is a test to ensure my final answer will be correct) using the following structure:

``````    double pi = TMath::Pi();
double Lon = 1.0;
double sig = 0.1;
double Lmin = -15.0+Lon;
double Lmax = +15.0+Lon;

TF1 *Prob = new TF1("Prob",Form("((1.0/(%f*sqrt(2*%f)))*exp(-0.5*pow((x-%f)/(%f),2)))",sig,pi,Lon,sig),Lmin,Lmax);

double average = 0.0;
average = Prob->Integral(Lmin, Lmax);    ``````

Then, when I print the value of average, I get crazy values (I am expecting the result to be one).
Am I making a mistake? Or something I don’t understand…?

And just to remark: when I test the defined function by printing some values (using the Eval function), it actually gives a Gaussian.

Thank you for your help and apologise for this “trivial” issue.

Mario AAO.

I guess it’s a deficiency in ROOT.
Your function is always 0 except for a very sharp peak around x=1. This fools the “integrator” (it fails to “detect” this peak).

Hi,

I think I disagree. You see, if I print out some values of the function by using

``````    for (int i = 0 ; i < 10 ; i++) {
double fac = i/5.0;
double testVal = Prob->Eval(fac*Lon);
std::cout << "L = " << fac*Lon << " The test value is = " << testVal << std::endl;
}``````

the result is

``````L = 0 The test value is = 7.6946e-22
L = 0.2 The test value is = 5.05227e-14
L = 0.4 The test value is = 6.07588e-08
L = 0.6 The test value is = 0.0013383
L = 0.8 The test value is = 0.53991
L = 1 The test value is = 3.98942
L = 1.2 The test value is = 0.53991
L = 1.4 The test value is = 0.0013383
L = 1.6 The test value is = 6.07588e-08
L = 1.8 The test value is = 5.05227e-14``````

I could also plot it, but I think these values make clear that the function is working as expected, i.e., as a Gaussian with a peak in L=1 and a width of 0.1.

What do you think?

Thanks,

Mario AAO.

Hello again,

So, I solved the problem, although I must say I don’t know the reason for this.

What happens is that

does not accept Lmin and Lmax the way I passed them to it, even though I have defined them as doubles, as you can see in my initial message.

I finally changed to

and it works (everything else is just the same as in my initial message).

Does anyone know why this happen?

Cheers,

Mario AAO.

You “solved” nothing. Try simply:
double Lon = 2; // 2, 3, 4, …

[quote=“Pepe Le Pew”]You “solved” nothing. Try simply:
double Lon = 2; // 2, 3, 4, …[/quote]

You are right!

But it only makes things more unclear, don’t you think?
In order to get it working properly, I have to use either

``````    double mx = +15.0;
double mn = -15.0;
double testInt = Prob->Integral(mn, mx);``````

or

``    double testInt = Prob->Integral(-15.0, 15.0);``

As long as the value of Lon is in the integration interval, it will give the correct answer. So, I did not solve the problem and my question remains: what is the problem with my initial procedure (and the second one, for that matters)?

Is there any kind of “limitations” on the Integral() function? May be I cannot pass any operator as a parameter to this function (and why does it work for Lon=1.0)?

Thanks,

Mario AAO.