Numerically integrate bivariate function

[sunrah]

What methods are available for integrating, e.g.
$\int^{\infty}_{0} f(x) dx \int^{x}_{0} g(x,y) dy$

numerically without resorting to symbolic integration. Thanks

 

[RUber]

Assuming that the integral exists and there is a sufficiently large x for which the function is approximately zero, then you could look at this as a bounded integral.
$\int_0^L\int_0^x f(x) g(x,y) dy dx$
Numerically, you would partition the interval from 0 to L with N+1 points $\{x_i\}_{i=0}^N$, such that $x_0 = 0$ and $x_N = L.$
Then you would have a region of integration that grows like half a square.

If you apply a midpoint rule for a simple approximation, this might look like ( with $dx_i = x_i - x_{i-1}$) :
$\int_0^L\int_0^x f(x) g(x,y) dy dx \approx \sum_{ i = 1 }^N dx_i f\left( \frac{x_i + x_{i-1}}{2}\right) \sum_{j=1}^i dx_j g\left(\frac{x_i + x_{i-1}}{2}, \frac{x_j + x_{j-1}}{2} \right)$

Does that help?

 

[pasmith]

I think I would first move to polar coordinates, so that $$I = \int_0^\infty \int_0^x f(x)g(x,y)\,dy\,dx = <br /> \int_0^\infty \int_0^{\pi/4} f(r\cos\theta) g(r\cos\theta, r\sin\theta) r\,d\theta\,dr.$$ Then I'd set $r = \mathrm{arctanh}(s)$ so that $$I = \int_0^1 \int_0^{\pi/4} f(\mathrm{arctanh}(s)\cos\theta) g(\mathrm{arctanh}(s)\cos\theta, \mathrm{arctanh}(s)\sin\theta) \frac{\mathrm{arctanh}(s)}{1 - s^2}\,d\theta\,ds.$$
Now I'm integrating over a finite rectangle, and I probably want to use an algorithm which doesn't require me to evaluate the integrand at $s = 1$.

Alternatively, I could split the integral as $$\int_0^1 \int_0^{\pi/4} f(r\cos\theta) g(r\cos\theta, r\sin\theta) r\,d\theta\,dr + \int_1^\infty \int_0^{\pi/4} f(r\cos\theta) g(r\cos\theta, r\sin\theta) r\,d\theta\,dr$$
and in the second integral make the substitution $z = 1/r$, to obtain $$I = \int_0^1 \int_0^{\pi/4} f(r\cos\theta) g(r\cos\theta, r\sin\theta) r<br /> + f(r^{-1}\cos\theta)g(r^{-1}\cos\theta,r^{-1}\sin\theta)r^{-3}\,d\theta\,dr$$ and this time I probably want to use an algorithm which doesn't require me to evaluate the integrand at $r = 0$.

 

[sunrah]

use an algorithm which doesn't require me to evaluate the integrand at $s = 1$

Thanks, the change of variables is intuitive, but I don't know any algorithms?
Also, in the last step in your alternative procedure shouldn't the second integral be over z not r?

 

[Ssnow]

For algorithms I suggest the Burden Faires, 8th edition, there is a section regarding the approximations of single and double integrals ...