2
$\begingroup$

I have a complicated double integral that has the following form $$I=\int_y\int_x f(x,y) g(x,y) \, dx dy$$

Suppose I know that this integral $$\int_x g(x,y)\, dx = w(y)$$

beacuse it is easier for me to integrate then the whole product $f(x,y)g(x,y)$. How can I perform integration by parts?

Update

Can I perform integration by parts ? The answer below seems to suggest that it is possible but ends up with zero and I don't understand why.

Thanks.

$\endgroup$
  • $\begingroup$ No, the integral of a product is not the product of the integrals. $\endgroup$ – David Peterson Feb 27 '15 at 2:03
  • $\begingroup$ @DavidP thanks, do you agree with the answer below? $\endgroup$ – Tyrone Feb 27 '15 at 2:43
  • $\begingroup$ You can perform integration by parts. The integral is obviously not zero in general (it is more complicated for multiple variables) $\endgroup$ – David Peterson Feb 27 '15 at 4:09
  • $\begingroup$ @DavidP do you mean this might complicate things. $\endgroup$ – Tyrone Feb 28 '15 at 23:20
1
$\begingroup$

You need to be more clear about your double integral. Say you have $$ \int_c^d \left(\int_a^b f(x,y)g(x,y)dx\right) dy$$ And you need to know the antiderivative of $g(x,y)$ with respect to $x$. So the information $\int_X g(x,y)dx=w(y)$ is not enough. Because this is not an antiderivative of $g$ with respect to the $x$ direction. Instead, you need to have $$ \int_a^x g(s,y)dy=h(x,y)$$ That is, $$ \int_a^b g(x,y)dy=h(b,y)-h(a,y)=h(b,y)=w(y)$$ Under this assumption, you can mimic the way how we do integration by parts in 1 dimensional. $$ \int_c^d\left(\int_a^b f(x,y)g(x,y)dx\right)dy=\int_c^d\left(f(x,y)h(x,y)\Big|_a^b-\int_a^b h(x,y)\frac{\partial}{\partial x}f(x,y)dx\right)dy$$ $$ =\int_c^d f(b,y)w(y)dy-\int_c^d\left(\int_a^b h(x,y)\frac{\partial}{\partial x}f(x,y)dx\right)dy$$

If you know the antiderivative of $h(x,y)$ with respect to $y$, i.e. $$ \int_c^y h(x,t)dt=\int_c^y\int_a^x g(s,t)ds\, dt=l(x,y)$$ you can keep going: $$ \int_c^d\left(\int_a^b f(x,y)g(x,y)dx\right)dy=\int_c^d f(b,y)w(y)dy-\int_c^d\left(\int_a^b h(x,y)\frac{\partial}{\partial x}f(x,y)dx\right)dy=\int_c^d f(b,y)w(y)dy-\int_a^b\left(l(x,d)\frac{\partial}{\partial x}f(x,d)-\int_c^d l(x,y)\frac{\partial^2}{\partial x\partial y}f(x,y)dy\right)dx=...$$

It all depends on what information you have for $f(x,y)$ and $g(x,y)$.

$\endgroup$

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.