Integral $\int_0^1 \int_0^1 \cdots \int_0^1\frac{x_{1}^2+x_{2}^2+\cdots+x_{n}^2}{x_{1}+x_{2}+\cdots+x_{n}}dx_{1}\, dx_{2}\cdots \, dx_{n}=?$

Question

How to evaluate this integral: $$\int_0^1 \int_0^1 \cdots \int_0^1\frac{x_{1}^2+x_{2}^2+\cdots+x_{n}^2}{x_{1}+x_{2}+\cdots+x_{n}}dx_{1}\, dx_{2}\cdots \, dx_{n}=?$$ I'm making use of the integral identity: $$\int_{0}^{+\infty }e^{-t(x_{1}+x_{2}\cdots +x_{n})}dt=\frac{1}{x_{1}+x_{2}\cdots +x_{n}}$$ and then reversing the order of integration with respect to time and space variables. But for $n=1$, then such that, $$\int_{0}^{\infty }dt\int_{0}^{1}x^{2}e^{-tx}dx=\int_{0}^{\infty }\frac{2 - e^{-t}(2 + 2t+t^2)}{t^3}dt=\int_{0}^{1}x\,dx=\frac{1}{2},$$ and $$\int_0^1 \int_0^1 \cdots \int_0^1\frac{x_{1}^2+x_{2}^2+\cdots+x_{n}^2}{x_{1}+x_{2}+\cdots+x_{n}}dx_{1}\, dx_{2}\cdots \, dx_{n}\\=n\int_{0}^{+\infty }\frac{2 - e^{-t}\left ( 2 + 2t+t^2 \right )}{t^3}\left ( \frac{1-e^{-t}}{t} \right )^{n-1}dt.$$

Answer 1

Here is another approach, which also gives the rational term.

(I) To see how it works let $n\geq 2$ and consider first the simpler case \begin{align*} \mathbb{E}\bigg(\frac{1}{X_1+\ldots+X_n}\bigg)=\int_0^\infty \bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt \end{align*} Using $\frac{1}{t^n}=\int_0^\infty \frac{z^{n-1}}{(n-1)!} e^{-zt}\,dz$ we find \begin{align*} \int_0^\infty \bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt&=\int_0^\infty\int_0^\infty \frac{z^{n-1}}{(n-1)!} e^{-zt}(1-e^{-t})^n\, dz\,dt\\ &=\int_0^\infty\int_0^\infty \frac{z^{n-1}}{(n-1)!} e^{-zt}(1-e^{-t})^n\, dt\,dz\\ &=\int_0^\infty \frac{z^{n-1}}{(n-1)!}\, \mathrm{Beta}(z,n+1)\,dz\\ &=n\,\int_0^\infty \frac{z^{n-1}}{z(z+1)\cdots(z+n)}\,dz \end{align*} The following observation will be the key:

Lemma: let $x_0,\ldots,x_n$ be distinct positive numbers and $k\leq n-1$. Then $$I_{k,n}(x_0,\ldots,x_n):=\int_0^\infty \frac{z^{k}}{(z+x_0)(z+x_1)\cdots(z+x_n)}\,dz=(-1)^{n+k+1}\Delta^n(x^k\log(x);x_0,\ldots,x_n)$$ where (for a real function $f$) $\Delta^n(f;x_0,\ldots,x_n)$ denotes the divided difference of $f$ corresponding to $x_0,\ldots,x_n$.

Recall that (Newton-interpolation)

(1) the divided differences are for $f$ and mutually distinct $x_0,\ldots,x_n$ are defined recursively by $\Delta^0(f;x_0)=f(x_0)$, $\Delta^n(f;x_0,\ldots,x_n)=\frac{\Delta^{n-1}(f; x_1,\ldots,x_n)-\Delta^{n-1}(f; x_0,\ldots,x_{n-1})}{x_n-x_0}$

(2) they are explicitly given by $$\Delta^n(f;x_0,\ldots,x_n)=\sum_{i=0}^n \frac{f(x_i)}{\prod_{j\neq i} (x_i-x_j)}\;\;(**)$$ (3) $$\Delta^n(f;x,x+1,\ldots,x+n)=\frac{1}{n!} \sum_{i=0}^n {n \choose i} (-1)^{n-i} f(x+i)$$

Proof of the lemma: For $k=0, n=1$ we have $$\int_0^\infty \frac{1}{(z+x_0)(z+x_1)}=\frac{\log(x_1)-\log(x_0)}{x_1-x_0}=\Delta^1(\log(x);,x_0,x_1)$$ For $k=0,n>1$ the repeated use of $\frac{1}{(z+a)(z+b)}=\frac{-1}{b-a}\left(\frac{1}{z+b}-\frac{1}{z+a}\right)$ shows that $(-1)^{n+1}I_{0,n}(x)=\Delta^n(\log(x),x)$. The validity for $k>0,n=k+1$ follows from $\frac{z}{z+b}=1-\frac{b}{z+b}$ and $(**)$. End of proof.

The lemma and (3) now give that \begin{align*} \int_0^\infty \bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt &=n \Delta^{n-1}(x^{n-2}\log(x);x+1,x+2,\ldots,x+n)\\ &=\frac{n}{(n-1)!}\sum_{i=0}^{n-1}{ n-1 \choose i} (-1)^{n-1-i} (i+1)^{n-2}\log(i+1) \end{align*}

(II) Now let $n\geq 1$ and consider $$ Q_{n+1}:=\mathbb{E}\bigg(\frac{X_1^2+\ldots + X_{n+1}^2}{X_1+\ldots+X_{n+1}}\bigg)$$ Write $$Q_{n+1}=(n+1)\int_0^1 u^2 \bigg(\int_0^\infty e^{-ut}\bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt\bigg)\,du$$ Proceeding as above shows that for $u>0$ \begin{align*} \int_0^\infty e^{-ut}\bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt &=\frac{n}{n!} \sum_{i=0}^n {n \choose i} (-1)^{n-i} (u+i)^{n-1}\log(u+i) \end{align*} Thus $$q_{n+1}:=\frac{Q_{n+1}}{n(n+1)}=\frac{1}{n!} \sum_{i=0}^n {n \choose i} (-1)^{n-i} \int_0^1 u^2(u+i)^{n-1}\log(u+i)\,du$$ Now expand $u^2=(u+i-i)^2$ and integrate partial to find that $q_{n+1}=L-R$ where $L=\frac{1}{n!}\sum_{i=0}^n{n\choose i}(-1)^{n-i} \ell(i),\;R=\frac{1}{n!}\sum_{i=0}^n {n \choose i}(-1)^{n-i} r(i)$ with \begin{align*} \ell(i)=&\frac{1}{n+2}\left((i+1)^{n+2}\log(i+1)-i^{n+2}\log(i)\right)\\ &-\frac{2i}{n+1}\left((i+1)^{n+1}\log(i+1)-i^{n+1}\log(i)\right)\\ &+\frac{i^2}{n}\left((i+1)^{n}\log(i+1)-i^n\log(i)\right)\\[0.2cm] r(i)=&\frac{1}{(n+2)^2}\left((i+1)^{n+2}-i^{n+2}\right)\\ &-\frac{2i}{(n+1)^2}\left((i+1)^{n+1}-i^{n+1}\right)\\ &+\frac{i^2}{n^2}\left((i+1)^{n}-i^n\right) \end{align*} Collecting terms in the logarithmic part $L$ shows that the coefficient of $\log(i)$ in $q_{n+1}$ is $c_{i,n+1} =(-1)^{n-i} \frac{i^n}{n}{ n \choose i-1}\left(\frac{n^2+3n+2-2i}{(n+2)!}\right)$, matching Fedor Petrov's answer.

The rational part $R$ can readily be summed (note that only the contributions of the powers $n$ and $n+1$ (with coefficients $-\frac{1}{n(n+1)(n+2)}$ and $\frac{2}{n(n+1)(n+2)}$) need to be considered) to give $$R=\frac{1}{n+2}-\frac{1}{n(n+1)(n+2)}\;\;,$$ so that the rational term of $Q_{n+1}$ is $-\frac{n(n+1)-1}{n+2}=-\frac{(n+2)(n-1)+1}{n+2}$, confirming Sylvain JULIEN's guess.

Answer 2

Ah, we may simply integrate by parts!

Denote $h(t)=(2 - e^{-t}( 2 + 2t+t^2))(1-e^{-t})^{n-1}$. Integrating by parts $(n+1)$ times we get $$\int_0^\infty h(t)t^{-n-2}dt=\frac1{(n+1)!}\int_0^\infty h^{(n+1)}(t)t^{-1}dt,$$ all off-integral terms vanish since $h(0)=h'(0)=\dots=h^{(n+1)}(0)=0$ and at infinity we have $h(t)=O(1)$, $h^{(i)}(t)=o(1)$ for $i>0$. We have $h^{(n+1)}(t)=\sum_{k=1}^n a_ke^{-kt}+u(t)$, where $u(t)=t\times \text{polynomial}(e^{-t},t)$. Note that $\sum a_k=h^{(n+1)}(0)=0$, thus $$\int_0^\infty \sum a_ke^{-kt}t^{-1}dt=\int_0^\infty \sum a_k(e^{-kt}-e^{-t})t^{-1}dt=-\sum a_k \log k$$ by Frullani integrals. It is easy to see that $$a_k=(-1)^{k+n+1}\binom{n-1}{k-1}k^{n-1}(n^2+n-2k).$$ It remains to evaluate $\int_0^\infty u(t) t^{-1}dt$. We have $$u(t)=-(2t+t^2)(g(t))^{(n+1)}-2(n+1)t(g(t))^{(n)},\,g(t)=e^{-t}(1-e^{-t})^{n-1}.$$ Therefore $$\int_0^\infty u(t) t^{-1}dt=-2\int g^{(n+1)}(t)dt-2(n+1)\int g^{(n)}(t)dt-\int_0^{\infty}tg^{(n+1)}(t)dt.$$ We get $\int (g^{(n)}+tg^{(n+1)})=tg^{(n)}$, and the definite integral against $(0,\infty)$ equals 0. It remains $\int_0^\infty -2g^{(n+1)}-(2n+1)g^{(n)}=2g^{(n)}(0)+(2n+1)g^{(n-1)}(0)$. We have $$g(t)=(1-t+\dots)t^{n-1}(1-t/2+\dots)^{n-1}=t^{n-1}-\frac{n+1}2t^n+\dots,$$ $g^{(n-1)}(0)=(n-1)!$, $g^{n}(0)=-\frac{(n+1)!}2$, $2g^{(n)}(0)+(2n+1)g^{(n-1)}(0)=-(n^2-n-1)(n-1)!$, confirming the guess of Sylvain JULIEN.

PREVIOUS NON COMPLETE VERSION

Still not a complete answer, but a method to be completed or improved.

Denote $f_n(t)=\frac{2 - e^{-t}\left ( 2 + 2t+t^2 \right )}{t^3}\left ( \frac{1-e^{-t}}{t} \right )^{n-1}$. Assume that we have found the coefficients $c_1,c_2,\dots,c_n$ and polynomials $g_1,\dots,g_n$ such that $$f_n(t)=2t^{-n-2}+\sum_{k=1}^n(g_k(t^{-1})e^{-kt})'+\sum_{k=1}^n c_k \frac{e^{-kt}}t.\,\,(*)$$ Then $\sum c_k=0$ (else $f_n$ would have a non-zero residue at 0, which is absurd). We have $\int_0^\infty \sum_{k=1}^n c_k \frac{e^{-kt}}t dt=-\sum c_k\log k$ by Frullani integrals. The integral of the other part of our sum $(*)$ is minus the limit at zero of the function $-\frac2{n+1}t^{-n-1}+\sum g_k(t^{-1})e^{-kt}$. The limit must exist, since the initial integral converges.

Now I cheat a bit. Note that if we write $f_n(t)=\sum_{k=0}^n q_k(1/t) e^{-kt}$ for polynomials $q_k$, then $c_k$ equals to the residue of the $q_k(1/t) e^{-kt}$, which is pretty computable. If I am not mistaken, $$q_k(t^{-1})=(-1)^kt^{-n-2}\left(2\binom{n-1}k+\binom{n-1}{k-1}(2+2t+t^2)\right)=\\=(-1)^kt^{-n-2}\left(2\binom{n}k+\binom{n-1}{k-1}(2t+t^2)\right).$$ Thus $$c_k=2(-1)^{k+n+1}\binom{n}k\frac{k^{n+1}}{(n+1)!}+2(-1)^{k+n}\binom{n-1}{k-1}\frac{k^{n}}{n!}+(-1)^{k+n+1}\binom{n-1}{k-1}\frac{k^{n-1}}{(n-1)!}=\\ =(-1)^{k+n+1}\frac{k^{n-1}(n^2+n-2k)}{n(n+1)(n-k)!(k-1)!}.$$ This matches a coefficient of $\log 3$ for $n=4$ from Shahrooz Janbaz's answer, you may check others for be sure. It remains to prove Sylvain JULIEN's guess for the rational part.

Answer 3

A little long as a comment:

Thanks to Maple, some first few terms are as follows:

$n=1:$ $I=\frac{1}{2}$

$n=2:$ $I=\frac{2^2\ln(2)}{3}-\frac{1}{3}$

$n=3:$ $I=-2^3\ln(2)+\frac{3^3\ln(3)}{4}-\frac{5}{4}$

$n=4:$ $I={2^6\ln(2)}-\frac{189\ln(3)}{5}-\frac{11}{5}$

$n=5:$ $I=-\frac{5\times 584\ln(2)}{9}+3^4\ln(3)+\frac{5^4\ln(5)}{36}-\frac{19}{6}$

$n=6:$ $I=\frac{6\times 11696\ln(2)}{63}+\frac{6\times 405\ln(3)}{14}-\frac{6\times 6250\ln(5)}{63}-\frac{29}{7}$

and so on. It seems there is some patterns which may help to obtain closed formula for the integral.