Is $\int_{x=0}^1\int_{y=0}^1\int_{z=0}^1 \frac{1}{(x-y)^2 (y-z)} dx dy dz$ finite?

Question

My question is in the title :

How could I prove that

$$ \int_{x=0}^1\int_{y=0}^1\int_{z=0}^1 \frac{1}{(x-y)^2 (y-z)} \ \text{d}z \ \text{d}y \ \text{d}x $$

is finite (if it is) ?

Thank you by advance.

Answer 1

You don't need to go through this work to be able to see it will blow up pretty badly, but it makes things a little more obvious: Let $u = x-y, v = y-z$, and $w = z$. Then, $u+v+w = x$, $v+w = y$, and so the jacobian of this transformation is $$ |J| = \left| \begin{vmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{vmatrix} \right| = 1 $$ so that the integral becomes (after determining the new limits) $$ \int_0 ^{1} \int_{-1} ^{1} \int_{-1} ^{1} \frac{du dv dw}{u^2 v} $$ which is not finite.

Answer 2

I think asking if the integral is finite is the wrong question, because it suggests that the integral makes sense as a number in $[-\infty,\infty],$ which it doesn't.

Certainly if we put absolute values on, the integral equals $\infty.$ You can see this just by noting

$$\int_0^1 \frac{1}{|y-z|}\,dz = \infty$$

for each fixed $y.$ Without the absolute values however, we can't even get off the ground. What is

$$\int_0^1 \frac{1}{y-z}\,dz $$

even supposed to mean if $y\in [0,1]?$