triple fractional part integral

[galactus]

check this one out:

http://integralsandseries.prophpbb.com/ ... html#p3814

I am pretty sure the curly brackets represent fractional part, thought the OP should have stated that.

I have seen the double integral version, so I am pretty sure a general form can be derived for this.

What I mean is something like:

$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\left\{\frac{x}{y}\right\}^{m}\left\{\frac{y}{z}\right\}^{n}\left\{\frac{z}{x}\right\}^{r}dxdydx=\text{a bunch of zetas}$$

[galactus]

$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\left(\left\{\frac{x}{y}\right\}\left\{\frac{y}{z}\right\}\left\{\frac{z}{x}\right\}\right)^{2}dxdydz$$

$$=\int_{0}^{1}\int_{0}^{1}\left(\left\{\frac{x}{y}\right\}\left\{\frac{1}{x}\right\}\right)^{2}dxdy$$

Break up integral at y:

$$\int_{0}^{x}\int_{0}^{1}\left(\left\{\frac{x}{y}\right\}\left\{\frac{1}{x}\right\}\right)^{2}dxdy+\int_{x}^{1}\int_{0}^{1}\left(\left\{\frac{x}{y}\right\}\left\{\frac{1}{x}\right\}\right)^{2}dxdy$$

Let $t=y/x, \;\ y=xt, \;\ dy=xdt$

$$=\int_{0}^{1}x^{2}\left\{\frac{1}{x}\right\}^{2}dx\int_{0}^{1}t^{3}\left\{\frac{1}{t}\right\}^{2}dt+\int_{1}^{1/x}t^{3}\left\{\frac{1}{t}\right\}^{2}dt\int_{0}^{1}x^{2}\left\{\frac{1}{x}\right\}^{2}dx\tag{5}$$

We can use a dummy variable and write the left integral all in terms of x:

$$\int_{0}^{1}x^{2}\left\{\frac{1}{x}\right\}^{2}dx\int_{0}^{1}x^{3}\left\{\frac{1}{x}\right\}^{2}dx\tag{1}$$

Noting the limits of integration, $[1,1/x]$, the right integral can be written as:

$$\int_{0}^{1}\left(\frac{1}{x}-1\right)x^{3}\left\{\frac{x}{y}\right\}^{2}dx$$

$$\int_{0}^{1}\left(x^{2}\left\{\frac{1}{x}\right\}^{2}-x^{3}\left\{\frac{1}{x}\right\}^{2}\right)dx\tag{2}$$

Now, use $\left\{\frac{1}{x}\right\}=\frac{1}{x}-n$ and write the integral in (1) as

$$\int_{\frac{1}{n+1}}^{1/n}x^{2}(1/x-n)^{2}dx\int_{\frac{1}{n+1}}^{1/n}x^{3}(1/x-n)^{2}dx$$

$$=\sum_{n=1}^{\infty}\frac{1}{3n(n+1)^{3}}\sum_{n=1}^{\infty}\frac{4n+1}{12n^{2}(n+1)^{4}}$$

$$=\boxed{\left(\frac{-1}{3}\zeta(3)+1-\frac{1}{3}\zeta(2)\right)\left(\frac{-1}{6}\zeta(3)+\frac{1}{2}-\frac{1}{4}\zeta(4)\right)}\tag{3}$$

Now, write (2) as $$\int_{\frac{1}{n+1}}^{1/n}(x^{2}(1/x-n)^{2}-x^{3}(1/x-n)^{2})dx$$

$$=\sum_{n=1}^{\infty}\frac{(2n+1)(2n-1)}{12n^{2}(n+1)^{4}}$$

$$=\boxed{-1/6\zeta(3)+1/4\zeta(4)-1/3\zeta(2)+1/2}\tag{4}$$

add (3)+(4) and get final result:

=$$\boxed{1/18\zeta(2)\zeta(3)+7/48\zeta(6)-1/2\zeta(2)+1/18\zeta^{2}(3)+1/12\zeta(3)\zeta(4)-1/2\zeta(3)+1}$$