Let $f(u)=C(I(\lVert u\rVert\leq 2L)+\lVert u-L\rVert^{-v}I(\lVert u\rVert>2L))$, where $f: \mathbb{R}^d\to\mathbb{R}, I(\cdot)$ is the indicator function, $\lVert x\rVert=\max(\lvert x_1\rvert,\dotsc,\rvert x_d\lvert)$, and $C,L,v<\infty$ positive constants.
Show that $\int_{R^d}\lvert f(u)\rvert du\leq M<\infty$.
I founded that $\int_{\mathbb{R}^d}I(\lVert u\rVert\leq 2L)du\leq (4L)^d$. The problem is to bound the second term, which seems integrable since , intuitively, as the coordinates go to infinity, the distance to $L$ increases, and then the term $\lVert u-L\rVert^{-v}$ decreases. I'm unable to make this argument explicit. Can you help me?
*In addition,$\lVert u\rVert>2L$ implies that $u$ is outside the square centered at zero with sides $4L$. Then $\lVert u-L\rVert$ is strictly greater than $L$.
You need $v>d$, then for $|u|>2L$, $|u-L|\geq|u|-L$, so the second term can be controlled by $\displaystyle\int_{|u|>2L}1/(|u|-L)^{v}$. Now we apply spherical, the integral is then $\omega_{n-1}\displaystyle\int_{2L}^{\infty}r^{d-1}/(r-L)^{v}dr\leq\omega_{n-1}2^{d-1}\int_{L}^{\infty}r^{d-1}/r^{v}<\infty$ since $v>d$.