Exact formula or non-trivial upper bound on p-norm of $f(x)=\|x\|_2$ in $[0,1)^d$ I wonder whether one can exactly calculate the following integral in terms of $d$ and $p\geq 1$ or not, or a better bound(than the trivial one I am going to give) in terms of $d,p$:
$$\left(\int_{[0,1)^d}\|x\|_2^p\,dx\right)^{1/p},$$
where $\|x\|_2$ is the Euclidean distance from $x$ to 0. Trivially, one has $\|x\|_2\leq\sqrt{d}$, but this ignores the effect of $p$. Can we have a better explicit bound which also relies on $p$, or even an explicit formula?
 A: $\DeclareMathOperator\E{E}\DeclareMathOperator\Var{Var}\DeclareMathOperator\P{P}$Note that
$$\int_{[0,1)^d}\|x\|_2^p\,dx
=\E S_d^{p/2},\tag{1}\label{1}$$
where $S_d:=\sum_1^d U_j^2$ and the $U_j$'s are iid random variables uniformly distributed on the interval $[0,1]$.
Note next that $\E S_d=d/3$ and $\Var S_d=4d/45<d/10$. So, by Cantelli's inequality,
$$\P(S_d\ge d/6)\ge1-\frac{\Var S_d}{\Var S_d+(d/3-d/6)^2} \\ 
\ge 1-\frac{d/10}{d/10+(1/3-1/6)^2 d} \\
=1-\frac{1/10}{1/10+(1/3-1/6)^2}=:c\in(0,1).$$
So,
$$\E S_d^{p/2}\ge(d/6)^{p/2} \P(S_d\ge d/6)
\ge c(d/6)^{p/2}$$
and hence
$$\Bigl(\int_{[0,1)^d}\|x\|_2^p\,dx\Bigr)^{1/p}
=(\E S_d^{p/2})^{1/p}
\ge c^{1/p}\sqrt{d/6}
\ge c\sqrt{d/6}$$
for $p\ge1$.
So, the trivial upper bound $\sqrt d$ on $\bigl(\int_{[0,1)^d}\|x\|_2^p\,dx\bigr)^{1/p}$ is optimal up to a universal constant factor.

For $p\ge2$, one can can do with a much simpler reasoning: by Jensen's inequality,
$$\Bigl(\int_{[0,1)^d}\|x\|_2^p\,dx\Bigr)^{1/p}
\ge\Bigl(\int_{[0,1)^d}\|x\|_2^2\,dx\Bigr)^{1/2}
=\sqrt{d/3}.$$

One may also note that, for any real $p>0$, by the Fatou lemma,
$$\liminf_{d\to\infty}\Bigl(\int_{[0,1)^d}\|x\|_2^p\,dx\Bigr)^{1/p}
\Big/\sqrt{d/3}\ge1.$$
In view of \eqref{1}, this follows because, by the law of large numbers, $S_d/d\to \E U_1^2=1/3$ in probability (as $d\to\infty$).
