0
$\begingroup$

I am facing some integrals of the form: $$I=\int_{0\le x\le a}\left(\int_{0\le y\le a}f(x,y)\theta(a-x-y)dy\right)dx$$ actually, I am not familiar with this kind of integral. How can I calculate it?

$\endgroup$

migrated from physics.stackexchange.com Jul 21 '17 at 13:59

This question came from our site for active researchers, academics and students of physics.

2
$\begingroup$

Since $\theta(a-x-y)$ is equal to $1$ only when $a-x-y \geq 0$ (i.e., $y \leq a-x$) and is equal to $0$ everywhere else, your integral is equivalent to integrating $f(x,y)$ over a triangular domain whose three vertices are $(0,0)$, $(0,a)$ and $(a,0)$. Therefore $$I = \int_{0}^{a}dx\int_{0}^{a-x}dy ~f(x,y)\ .$$ Hope this helps.

$\endgroup$
0
$\begingroup$

Assuming $a\geq 0$, it is by definition $ \int_0^a\mathrm{d}x \int_0^{a-x}\mathrm{d}y ~f(x,y).$

$\endgroup$

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.