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?



Exact formula or non-trivial upper bound on p-norm of $f(x)=\|x\|_2$ in $[0,1)^d$
$\DeclareMathOperator\E{E}\DeclareMathOperator\Var{Var}\DeclareMathOperator\P{P}$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](https://en.wikipedia.org/wiki/Cantelli%27s_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.




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




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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 $d\to\infty$).


