Integral separation

Question

I have an integral of the form: $$ \int_Y{\int_X{\mathbb{1}_{(\alpha x +y \geq \gamma)} f(x) f(y) dx dy}}$$ where $\alpha, \gamma \in \mathbb{R}$.

Is it possible to separate the integral in this case?

Answer 1

Suppose we want to evaluate the following integral

$$I = \int\!\!\!\int\limits_\Omega {{\bf{1}}(A)f(x)f(y)dxdy}\tag{1}$$

Where $\Omega$ is the domain of integration in the $xy$ plane and $A$ is another region in that plane. According to the definition of the indicator function we have

$${\bf{1}}(A) = \left\{ \matrix{ 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x,y} \right) \in A \hfill \cr 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x,y} \right) \notin A \hfill \cr} \right.\tag{2}$$

Now, combining $(1)$ and $(2)$ you can write

$$I = \int\!\!\!\int\limits_{\Omega \cap A} {f(x)f(y)dxdy} \tag{3}$$

That's all you can do based on the assumptions mentioned in the question!

Now, if the region ${\Omega \cap A}$ can be written as

$$\Omega \cap A = \left\{ {\left( {x,y} \right)|{x_1} < x < {x_2},{y_1} \le y < {y_2}} \right\}\tag{4}$$

then $(3)$ can be simplified as follows

$$\eqalign{ & I = \int_{{x_1}}^{{x_2}} {\int_{{y_1}}^{{y_2}} {f(x)f(y)dydx} } \cr & \,\,\, = \int_{{x_1}}^{{x_{}}} {f(x)\left( {\int_{{y_1}}^{{y_2}} {f(y)dy} } \right)dx} \cr & \,\,\, = \left( {\int_{{y_1}}^{{y_2}} {f(y)dy} } \right)\left( {\int_{{x_1}}^{{x_{}}} {f(x)dx} } \right) \cr}\tag{5}$$

Also if $A \subset \Omega $ then you have

$$\eqalign{ & \Omega \cap A = A = \left\{ {\left( {x,y} \right)|\alpha x + y \ge \gamma } \right\} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ = \left\{ {\left( {x,y} \right)| - \infty < x < + \infty ,\gamma - \alpha x \le y < + \infty } \right\} \cr}\tag{6}$$

and hence you can write $(3)$ as

$$\eqalign{ & I = \int_{ - \infty }^{ + \infty } {\int_{\gamma - \alpha x}^{ + \infty } {f(x)f(y)dydx} } \cr & \,\,\, = \int_{ - \infty }^{ + \infty } {f(x)\left( {\int_{\gamma - \alpha x}^{ + \infty } {f(y)dy} } \right)dx} \cr} \tag{7}$$

since the limits of integration are not constant in $(7)$ you cannot just simplify it to products of one-dimensional integrals like $(5)$.

Answer 2

This is what you can do :

$$\int_Y\int_X 1_{(\alpha x +y \le \gamma)}f(x)f(y)dxdy=\int_Yf(y)\{ \int_{x \ge \frac{\gamma -y}{\alpha}}f(x)dx \}dy$$

First integrate over $x$ and then over $y$, I don't think you can rewrite this as a product of integrals without other informations on $f$.

Answer 3

You can go one step further in case $X=Y=\mathbb{R}$ and $f(x)$ and $xf(x)$ are absolutely integrable and vanish as $|x|\rightarrow \infty $ then

\begin{eqnarray*} \int_{\mathbb{R}}dyf(y)\int_{\frac{\gamma -y}{\alpha }}dxf(x) &=&\left[ yf(y)\int_{\frac{\gamma -y}{\alpha }}dxf(x)\right] _{y=-\infty }^{+\infty }-\int_{\mathbb{R}}dyyf(y)\partial _{y}\int_{\frac{\gamma -y}{\alpha }}dxf(x) \\ &=&-\int_{\mathbb{R}}dyyf(y)\frac{1}{\alpha }f(\frac{\gamma -y}{\alpha })\end{eqnarray*}