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With a given $x > 0$ (I think we could restrict it to $x \in [0, 3]$), I'm trying to find the following integral:

$$ \int_{z = 0}^{\min(x,1)} \int_{y = 0}^{\min(x - z, 1)} |x -z -y -1| dydz $$

However, here I'm not sure how to deal with absolute values. By the construction, the value under the modulus seems to be almost always negative when $x < 1$, and it feels like negative when $x \leq 3$, but I feel like I'm missing something.

Any hints on how should we proceed further?

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1 Answer 1

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Like you said, the hardest part here is the absolute value. We can make our lives easier by defining $u := x-z-1$, which simplifies the integrand into $|u - y|$. Since this function has a constant analytic form in the regions $y \leq u$ and $y \geq u$, we can rewrite our integral as the piecewise function:

$$ \int_{0}^{\min(x, 1)}\left[ \int_{0}^{\min(u, 1)}(u-y)\ dy + \int_{\max(u, 0)}^{\min(1+u, 1)}(y-u)\ dy \right]\ dz $$

You still have to deal with the differing behavior of these integrals for different values of $u$, but this should be more mechanical than directly reasoning about the original absolute value. Just use the predictable behavior of $\min(a,b)$ and $\max(a,b)$, along with the fact that $\int_{a}^{b}f(y)\ dy = 0$ when $a\geq b$, as seen by the fact that the region of integration $[a, b] := \left\{\ x\ |\ a\leq x\leq b \right\}$ is empty.

After solving the inner integral for each separate region of $u$ (namely, $u\leq0$, $u\geq1$, and $0\leq u\leq1$), you just replace all your $u$'s by $x-z-1$ (including in your inequalities defining the different regions), and solve the outer integral. This requires more of the same case-by-case reasoning, but by this point you'll have the hang of it :)

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