Show that the function $u \in L^n(B^0(0,1))$ for $n > 1$. I am working on Exercise 5.10.14 in Evans's Partial Differential Equations. I am trying to show that the function
$$
u(x) := \log\Bigg(\log\Bigg(1 + \frac{1}{|x|}\Bigg)\Bigg)
$$
belongs to $L^n(B^0(0,1))$ with $n >1$,  where $B^0(0,1)$ is the open unit ball.

By applying polar coordinates, we have.
\begin{align*}
\int_{B^0(0,1))}\log\Bigg(\log\Bigg(1 + \frac{1}{|x|}\Bigg)\Bigg)^n\text{d}x &= \int_0^1\int_{\partial B^0(0,r))}\log\Bigg(\log\Bigg(1 + \frac{1}{|x|}\Bigg)\Bigg)^n\text{d}S(x)dr\\
&= \int_0^1 \log\Bigg(\log\Bigg(1 + \frac{1}{r}\Bigg)\Bigg)^n\Big(n\alpha(n)r^{n-1}\Big)\text{d}r\\
&=n\alpha(n) \int_0^1 \log\Bigg(\log\Bigg(1 + \frac{1}{r}\Bigg)\Bigg)^nr^{n-1}\text{d}r\\
\end{align*}
From there, I thought about applying the change of variables $w = \log\Bigg(1 + \frac{1}{r}\Bigg)$ which after a bit of work, I think yields
$$
-n\alpha(n)\int_{w(0)}^{w(1)} (r\log(w))^n\text{d}w
$$
I think this is infinite though. Any thoughts, hints, ideas?

Since $\log(1 + t) \le t$ for all $t \ge 0$, then $$\int_0^1\log^n\left[\log\left(1 + \frac{1}{r}\right)\right] r^{n-1}\, dr \le \int_0^1 \log^n\left(\frac{1}{r}\right)\, r^{n-1}\, dr$$ Using the substitution $r = e^{-v}$ we have $$\int_0^1 \log^n\left(\frac{1}{r}\right) r^{n-1}\, dr = \int_0^\infty v^n e^{-nv}\, dv = \frac{n!}{n^{n+1}}$$ so it follows that $u\in L^n(B^0(0,1))$.
