2D numerical integration with infinite limit (C++) - c++

Question

In order to integrate a two dimensional function of the form
$$\int_{1}^\infty \int_{-\sqrt{x^2-1}}^{\sqrt{x^2-1}} e^{-x} \rm{d}y \rm{d}x,$$
I have been attempting to use the following code (written in C++) taken mostly from the Numerical Recipes book which calls a gaussian quadrature routine for the integration:
static float xsav;
static float (*nrfunc)(float,float);
float quad2d(float (*func)(float, float), float x1, float x2)
{
float qgaus(float (*func)(float), float a, float b);
float f1(float x);
nrfunc=func;
return qgaus(f1,x1,x2);
}
float f1(float x)
{
float qgaus(float (*func)(float), float a, float b);
float f2(float y);
float yy1(float),yy2(float);
xsav=x;
return qgaus(f2,yy1(x),yy2(x));
}
float f2(float y)
{
return (*nrfunc)(xsav,y);
}
This code works fine for two dimensional integrals with finite limits, but fails as the outer limit is taken to infinity. To account for this, I have attempted to use a change of variables:
#define FUNC(x) ((*funk)(-log(x))/(x))
float qgaus(float (*funk)(float), float aa, float bb)
{
int j;
float xr,xm,dx,s,a,b;
b=exp(-aa);
a=0.0;
static float x[]={0.0,0.1488743389,0.4333953941,
0.6794095682,0.8650633666,0.9739065285};
static float w[]={0.0,0.2955242247,0.2692667193,
0.2190863625,0.1494513491,0.0666713443};
xm=0.5*(b+a);
xr=0.5*(b-a);
s=0;
for (j=1;j<=5;j++)
{
dx=xr*x[j];
s += w[j]*(FUNC(xm+dx)+FUNC(xm-dx));
}
return s *= xr;
}
float f(float x, float y)
{
float a = exp(-x);
return a;
}
float yy1(float x)
{
float y = -sqrt(x*x-1);
return y;
}
float yy2(float x)
{
float y = sqrt(x*x-1);
return y;
}
static float xsav;
static float (*nrfunc)(float, float);
float quad2d(float (*func)(float, float), float x1, float x2)
{
float qgaus(float (*func)(float), float aa, float bb);
float f1(float x);
nrfunc=func;
float t = qgaus(f1,x1,x2);
return t;
}
float f1(float x)
{
float qgaus(float (*func)(float), float aa, float bb);
float f2(float y);
float yy1(float);
float yy2(float);
xsav=x;
float r = qgaus(f2,yy1(x),yy2(x));
return r;
}
float f2(float y)
{
float k = (*nrfunc)(xsav,y);
return k;
}
int main ()
{
float z;
z = quad2d(f, 1.0, 20.0);
cout << z << endl;
}
but this still doesn't give the right answer. It should be
$2 \times \rm{BesselK}[1,1] \approx 1.20381$
but instead gives
2.15501
Any suggestions on how I could modify this code to account for the infinite limit would be greatly appreciated!