1
$\begingroup$

I have a double integral which I want to evaluate numerically by approximating this integral with its Riemann sum. The integral is of the following form $$ \int_{0}^{a}\int_{0}^{b} f(x,y)dxdy\approx\sum_{n=1}^{K}\sum_{m=1}^{J}f(n\delta_x,m\delta_y)\delta_x\delta_y$$

There are upper-bounds on error of the approximation for one integration such as $\int_{a}^{b} f(x)dx$. However I didn't find any bounds on error of double integrals.

How can I choose $\delta_x$ and $\delta_y$ such that:

$$\left|\int_{0}^{a}\int_{0}^{b} f(x,y)dxdy-\sum_{n=1}^{K}\sum_{m=1}^{J}f(n\delta_x,m\delta_y)\delta_x\delta_y\right|\leq \epsilon$$ for a predefined $\epsilon$ knowing $f(x,y)$?


PS: In my actual problem the function $f()$ is :

$$f(I,I')=Q\left(\frac{\eta T \frac{I+I'}{v\nu}}{\sqrt{m_1'I+m_0'I'+2\sigma_n^2}}\right)\frac{1}{2I} \frac{1}{\sqrt{2\pi\sigma_X^2}} \exp\left\lbrace \frac{[\ln(I)-\ln(I_0)]^2}{8\sigma_X^2} \right\rbrace\nonumber\\ \times\frac{1}{2I'} \frac{1}{\sqrt{2\pi\sigma_X^2}} \exp\left\lbrace \frac{[\ln(I')-\ln(I_0)]^2}{8\sigma_X^2} \right\rbrace$$ where $Q(.)$ is the integration of a normalized Gaussian distribution

$\endgroup$
2
$\begingroup$

The easy way out would be to do a two-step approximation: Write $$ \int_{0}^{b} f(x,y)\,dx\approx\sum_{n=1}^{K}f(n\delta x,y)\delta_x $$ and estimate the error in this approximation, then integrate over $y$ to get $$ \int_{0}^{a}\int_{0}^{b} f(x,y)\,dx\,dy\approx \int_{0}^{a}\sum_{n=1}^{K}f(n\delta x,y)\delta_x\,dy $$ noting that your error estimate from above now gets integrated from $0$ to $a$ as well; then estimate the integral on the right with yet a Riemann sum, once more applying the one-dimensional error estimates.

$\endgroup$
  • 1
    $\begingroup$ Thanks for answering. Is there any one-step approximation too? $\endgroup$ – SMA.D Jul 19 '17 at 20:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.