# Evaluate $\int_0^1···\int_0^1\frac{dx_n...x_1}{(1+x_1+x_2+ · · ·+x_n)^n}$ using Fubini's Theorem

Evaluate using Fubini's Theorem

$$\displaystyle\int_{[0,1]^n}\frac{1}{(1+x_1+x_2+ · · ·+x_n)^n}d(x_1,...,x_n)$$

My idea:

Fubini’s Theorem $$n$$ times, to write it as $$n$$ interated integrals over $$[0, 1]$$:

$$\displaystyle\int_0^1···\int_0^1\frac{1}{(1+x_1+x_2+ · · ·+x_n)^n}dx_n...x_1$$

$$\displaystyle\int_0^1\frac{1}{1+x_1}\int_0^1\frac{1}{(1+x_1+x_2)^2}\int_0^1···\int_0^1\frac{1}{(1+x_1+x_2+ · · ·+x_n)^n}dx_n...x_1$$

And then I'm not sure where to go from there or if I'm on the right lines in the first place. Any help?

• Apologies, I have fixed a typo
– user415105
Apr 25, 2017 at 14:06
• I don't see how you get the iterated integral you write, but have you tried to do the calculation for $n=2$ and $n=3$? Do you see a pattern? Apr 25, 2017 at 14:40
• You can use a Laplace Transform. Apr 27, 2017 at 6:22

If $X_1,X_2,\ldots,X_n$ are independent random variables, uniformly distributed over $[0,1]$, and $S=X_1+X_2+\ldots+X_n$, your integral equals the average value of $\frac{1}{(1+S)^n}$. By the law of large numbers it is expected to be pretty small, close to $\frac{1}{\left(1+\frac{n}{2}\right)^n}$.

Its explicit value can be simply computed by integrating with respect to $x_1$ first, then $x_2$ and so on. For instance, if $n=3$

$$\begin{eqnarray*} \int_{[0,1]^3}\frac{d\mu}{(1+x_1+x_2+x_3)^3} &=& \frac{1}{2}\int_{[0,1]^2}\left[\frac{1}{(1+x_2+x_3)^2}-\frac{1}{(2+x_2+x_3)^2}\right]\,d\mu\\&=&\int_{0}^{1}\frac{dx_3}{(1+x_3)(2+x_3)(3+x_3)} \end{eqnarray*}$$ and the general case boils down to integrating $$\int_{0}^{1}\frac{dx_n}{(1+x_n)(2+x_n)\cdots(n+x_n)}$$ By partial fraction decomposition we have $$\frac{1}{(1+x_n)(2+x_n)\cdots(n+x_n)} = \frac{1}{(n-1)!}\sum_{k=0}^{n-1}\frac{(-1)^k \binom{n-1}{k}}{x_n+k+1}$$ hence the explicit value of the integral is

$$\int_{[0,1]^n}\frac{d\mu}{(1+x_1+\ldots+x_n)^n}=\frac{1}{(n-1)!}\sum_{k=0}^{n-1}(-1)^k\binom{n-1}{k}\log\left(\frac{k+2}{k+1}\right).$$

By exploiting Frullani's integral (leading to an integral representation for $\log\frac{k+2}{k+1}$) and the binomial theorem we also get the identity

$$\begin{eqnarray*} \int_{[0,1]^n}\frac{d\mu}{(1+x_1+\ldots+x_n)^n}&=&\frac{1}{(n-1)!}\int_{0}^{+\infty}\frac{(1-e^{-x})^n}{xe^x}\,dx\\ &=& \frac{1}{(n-1)!}\int_{0}^{1}\frac{t^n}{-\log(1-t)}\,dt.\end{eqnarray*}$$

from which it follows that $$\int_{[0,1]^n}\frac{d\mu}{(1+x_1+\ldots+x_n)^n} \sim \frac{1}{n!\log(n)}.$$

• Is there an easy way to see the law-of-large-numbers estimate from the final result? Using Stirling's approximation on $(n-1)!)$ doesn't quite work because it contains a factor of $e^{n}$ whereas the law-of-large-numbers estimate contains $2^n$. (Though I guess the latter may just not be precise enough to expect complete agreement.) Apr 25, 2017 at 14:52
• @Semiclassical: I think it is not completely trivial to deduce the magnitude of the integral from the last representation, but if we replace $\log\left(\frac{k+2}{k+1}\right)$ with $\frac{1}{k+1}$ the last expression simply turns into $\frac{1}{n!}$. Apr 25, 2017 at 15:01
• @Semiclassical: indeed the correct asymptotics is $\frac{1}{n!\log n}$. There is an interesting connection with en.wikipedia.org/wiki/Gregory_coefficients Apr 25, 2017 at 15:29