Excluding a point in double integral

Question

I have a double integral as below $$\int_{x=-\infty}^{\infty}\int_{y=-\infty, y\neq x}^{\infty} f(x)g(y)\,dx\,dy$$ is this equal to $$\int_{x=-\infty}^{\infty}\int_{y=-\infty}^{\infty} f(x)g(y)\,dx\,dy$$ or is it equal to $$\int_{x=-\infty}^{\infty}\int_{y=-\infty}^{\infty} f(x)g(y)\,dx\,dy-\int_{x=-\infty}^{\infty}f(x)g(x)\,dx$$ Thanks.

Answer 1

The diagonal $x=y$ that you're excluding has zero measure under $dxdy$, so excluding it will not change the value of the intergal, nor whether it exists.

Answer 2

You may write

$$\int_{x=-\infty}^{\infty}\int_{y=-\infty, y\neq x}^{\infty} f(x)g(y)\,dx\,dy \\=\int_{x=-\infty}^{\infty}\int_{y=-\infty}^{\infty} f(x)g(y)\,dx\,dy-\int_{x=-\infty}^{\infty}\int_{y=x}^{x} f(x)g(y)\,dx\,dy.$$

Answer 3

Remember that an integral is the summation of MANY infinitesimal slices.

Taking out the slices that result in $y=x$ will have no effect.

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We can also say that the set has zero measure.

So it is equal to the $\displaystyle \int_{x=-\infty}^{\infty}\int_{y=-\infty}^{\infty} f(x)g(y)\,dx\,dy$