Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to compute the integral

$\int_{[0,1]^m} \sqrt{\sum_{j=1}^m u_j^2} \ du_1 \dots du_m$.

Wolfram Alpha gives an exact result for $m = 2$, but fails to do so for higher values. I am particularly interested in the asymptotics.

I'm sure the value is a standard result, but don't know where to find it (no, I don't have a copy of G&R handy).

Edit/followup: I'm also interested in the integral $\int_{p \ge 0, \sum_{j=1}^m p_j = 1} \sqrt{\sum_{j=1}^m p_j^2} \ dp_1\dots dp_{m-1}$, i.e. the expected 2-norm of a probability distribution on $[m]$ chosen uniformly at random.

share|cite|improve this question
up vote 2 down vote accepted

Well, a simple asymptotic is (by probability) $\sqrt{m/3}$

This can be obtained by considering that we want the expectation of

$w = \sqrt{x_1^2 +... x_m^2} = \sqrt{y_1 +... y_2}$

with $y_i = x_i^2$, so that $\displaystyle f_y(y) = \frac{y^{1/2}}{2}$, and $E(y)=1/3$, $\sigma^2(y)=4/45$.

Then, the variable $z = (y_1 +... y_2)/m$ will approach a gaussian with $\mu_z = 1/3$, $\sigma^2_z=\frac{4}{m 45}$.

And $E(w) = \sqrt{m} E(\sqrt{z})$ which in first approximation is $E(w) \approx \sqrt{m} \sqrt{\mu_z} = \sqrt{m/3}$

One can get better aproximations with the higher moments, for example:

$E(w) \approx \mu_z^{1/2} - \frac{1}{8}\mu_z^{-3/2}\sigma_z^2 = \sqrt{\frac{m}{3}} - \frac{1}{30}\sqrt{\frac{3}{m}} = \sqrt{\frac{m}{3}}\Bigl(1 - \frac{1}{10 m}\Bigr)$

(errors aside).

Are you looking for something more precise?

UPDATE: By a similar reasoning, the additional problem (if I understand it right) gives $\displaystyle \sqrt{\frac{4}{3 m}}$

UPDATE2: About the second order approximation:

In general, when we want to find moments of $Y=g(X)$ with $g(.)$ nonlinear but smooth, we can do a Taylor expansion around its media:

$Y = g(\mu_x) + g'(\mu_x)(X-\mu_x) + \frac{1}{2!} g''(\mu_x)(X-\mu_x)^2 + ...$

So that

$\displaystyle \mu_y = E(Y) = g(\mu_x) + g''(\mu_x)\frac{\sigma_x^2}{2} +...$

This, BTW, justifies the intuitive notion that $E(g(.)) \approx g(E(.))$ if the variance is small (equality applies only if $g(.)$ is linear, of course) and allows to estimate (and to some extent correct) the error of the approximation.

share|cite|improve this answer
$f_Y (y) = y^{-1/2}/2$, of course. However, to find ${\rm E}(Y)$ and ${\rm Var}(Y)$ you don't need $f_Y (y)$, as ${\rm E}(Y^n)={\rm E}(X^{2n})=\int_0^1 {x^{2n} dx} = \frac{1}{{2n + 1}}$, $n=1,2$. – Shai Covo Jan 25 '11 at 20:40
Can you please elaborate on the better approximation? – Shai Covo Jan 25 '11 at 21:13
@Shai: you're right. about the better approximation, I updated the answer – leonbloy Jan 25 '11 at 21:43
Thanks. – Shai Covo Jan 25 '11 at 21:49

As for the asymptotics, suppose that $U_i$ are i.i.d. uniform$(0,1)$ rv's. Then the integral can be written as $$ I_m = {\rm E}\bigg(\sqrt {U_1^2 + \cdots + U_m^2 } \bigg) = \sqrt m {\rm E}\bigg(\sqrt {\frac{{U_1^2 + \cdots + U_m^2 }}{m}}\bigg ). $$ Since ${\rm E}(U_1^2)=1/3$, the strong law of large numbers suggests that $$ I_m \sim \sqrt {\frac{m}{3}}. $$ This agrees with the upper bound of $\sqrt {m/3}$ obtained from $$ {\rm E}\bigg(\sqrt {U_1^2 + \cdots + U_m^2 } \bigg) < \sqrt {{\rm E}(U_1^2 + \cdots + U_m^2 )} = \sqrt {\frac{m}{3}}. $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.