# Integration by parts for a double integral.

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.

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

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 $g(x,y)$.