Integration numerical with variable limits

I am trying to find a double integration numerically where the inner integral has variable limits while the outer integral has scalar limits and not the other way round. As I understand integral2 allows you do the reverse of what I want i.e you can have the outer variable limits non scalar while the inner should be scalar.

    fxy=@(x,y)1/x+1/y
    xmin=0;
    xmax=@(y)2*y;
    ymin=0;
    ymax=+inf;
    integral2(fxy,xmin,xmax,ymin,ymax)
I get the following error 
Error using integral2 (line 76)
XMAX must be a floating point scalar.

Answer 1

[Mike Hosea]

The INTEGRAL2 interface is set up to calculate an iterated integral where the integral over the second argument of the integrand function f is the inner integral. It does not matter what you call your variables, whether it be x, y, t, p, r, or whatever. The first argument is the outer integral, and the second argument is the inner integral. If your problem happens to be formulated so that the inner integral variable is called x and the outer integral variable is called y, but your integrand is already defined so that x is the first argument and y is the second, then you just do this:

integral2(@(y,x)f(x,y),ymin,ymax,xmin,xmax)

Your example isn't integrable, or I'd demonstrate.

Answer 2

[Alberto]

There is a general solution using symbolic tools.

syms x y
fxy=1./x+1./y;
xmin=0;
xmax=2*y;
ymin=0;
ymax=+inf

int(fxy,x,xmin, xmax)                  % integration dx
int(int(fxy,x,xmin, xmax),y,ymin,ymax) % double integration

Answer 3

[Andrei Bobrov]

in your case:

out = integral2(fxy,ymin,ymax,xmin,xmax);

[DM]

But I don't think you can reverse the limits of integration without reversing the integration operation, i.e in my case the double integral is over dxdy and you changed the limits in reverse way

[DM]

What you suggested is performing double integral over dxdy but with the wrong limits since you reversed them @andrei bobrov