ROOT 5.34

gcc version 4.9.2 (Debian 4.9.2-10)

Hello, everyone! I got a problem. I wrote a function with four variables. And I have their integration ranges in the form of two arrays xmin[4] and xmax [4]. I know we can get the finite integration of this function by ROOT::Math::IntegratorMultiDim just like the below codes. However, I want to know is there any way I can eliminate the first three variable by integration and leave the fourth variable unchanged. Thus changing this four-dimention function into one-dimention function and drawing this as a TF1. My simple codes are as belows:

```
#include "TMath.h"
#include "Math/WrappedFunction.h"
#include "Math/IntegratorMultiDim.h"
#include <iostream>
Double_t func(const Double_t *x){
return x[0]*x[1]*TMath::Exp(x[2]*x[3]);
}
void test5()
{
// numeric integration
//ROOT::Math::WrappedMultiFunction<> f1(func,3);
ROOT::Math::Functor f1(&func,4);
ROOT::Math::IntegratorMultiDim ig(f1);
Double_t xmin[4] = {0.0,0.0,0.0,0.0};
Double_t xmax[4] = {2.0,3.0,4.0,4.0};
Double_t result = ig.Integral(xmin, xmax);
cout<<result<<endl;
}
```

This code is for definite integration. And I thought about one way to get it’s dimentions reduced like the following code:

```
#include "TMath.h"
#include "Math/WrappedFunction.h"
#include "Math/IntegratorMultiDim.h"
#include <iostream>
Double_t func(Double_t *x,Double_t *b){
return x[0]*x[1]*TMath::Exp(x[2]*b[0]);
}
void test5()
{
int nbins=500;
TF3* f3=new TF3("f3",func,0.,2.,0.,3.,0.,4.,1);
TH1F* h=new TH1F("h","",nbins,0.,5.);
for(int i=0;i<nbins;i++){
double E_tem=h->GetBinCenter(i+1);
f3->SetParameter(0,E_tem);
double integral;
integral=f3->Integral(0.,2.,0.,3.,0.,4.);
h->SetBinContent(i+1,integral);
}
h->Draw();
}
```

However, it will need lots of calculation for my real work and cause new problem. So I am wondering is there a simple way in root to get the dimentions reduced. Thanks very much for your attention!