# Split up a double integral

Is it true that

$$\int_a^b \int_c^d f(x)g(y)dydx = \int_a^b f(x)dx \cdot \int_c^dg(y)dy$$

My intuition says it is true, but I also have the feeling that I am missing something, but I cannot prove it. In my application $$a,b,c,d$$ are length variables. $$f(x)$$ and $$g(y)$$ are sinusoidal functions.

I would like to prove the general case, can somebody help me out?

• both functions are independent so it can be split up like this Jan 9 '19 at 11:34
• in the same way $\left(\int_a^bf(x)dx\right)^2=\int_a^b\int_a^bf(x)f(y)dxdy$ Jan 9 '19 at 11:35

Note that for any integrable $$f \colon \def\R{\mathbb R}\R \to \R$$ and any $$\zeta \in \R$$ we have $$\int_\R \zeta f(x)\, dx = \zeta \int_\R f(x) \, dx$$ Now note that with $$\zeta := \int_\R f(y)\, dy$$ which depends on $$f$$, but since $$f$$ is a constant this gives $$\int_\R \left(\int_\R f(y)\, dy\right)\, f(x)\, dx = \int_\R f(y)\, dy \cdot \int_\R f(x)\, dx$$ Now, for every fixed $$x \in \R$$, just apply the above result, now for $$\zeta = f(x)$$ (not dependent on $$y$$): $$\left(\int_\R f(y)\, dy\right)\, f(x) = \int_\R f(x)f(y)\, dy$$ Thus $$\int_\R \int_\R f(y)f(x)\, dy\, dx = \int_\R f(y)\, dy \cdot \int_\R f(x)\, dx$$
Here is a try: $$I=\int_a^b\int_c^df(x)g(y)dxdy$$ where $$\int f(x)dx=F(x)$$ and $$\int g(y)dy=G(y)$$ we can start by saying: $$I=\int_a^b\int_c^df(x)g(y)dxdy=\int_c^d\left[F(x)\right]_a^b g(y)dy=\left[F(x)\right]_a^b \left[G(x)\right]_c^d$$ as $$\left[F(x)\right]_a^b$$ is a constant. They are separable
The integral is linear so for any constant $$K$$ you have: $$\int_a^b K \, f(x)\,\mbox{d}x=K\int_a^b f(x)\,\mbox{d}x$$
and with respect to $$y$$, $$f(x)=K_1$$ is a constant and likewise the expression $$\int_c^d g(y)\,\mbox{d}y = K_2$$ is a constant: $$\int_a^b \int_c^d \overbrace{f(x)}^{K_1} g(y)\,\mbox{d}y\,\mbox{d}x =\int_a^b \left( \overbrace{f(x)}^{K_1}\underbrace{\int_c^d g(y)\,\mbox{d}y}_{K_2} \right)\,\mbox{d}x=\underbrace{\int_c^d g(y)\,\mbox{d}y}_{K_2}\int_a^b f(x)\,\mbox{d}x$$