Who's up to the challenge? 29

The substitution $u = 1 - xy$ gives $\int_0^1 x\left\{ \frac{1}{1-xy}\right\}\,dy \; = \; \int_{1-x}^1 \left\{ u^{-1}\right\}\,du$ for any $0 < x < 1$ , and hence $\begin{array}{rcl} \displaystyle I & = & \displaystyle \int_0^1 \int_0^1 x \left\{ \frac{1}{1-xy}\right\}\,dx\,dy \; = \; \int_0^1\,dx \int_{1-x}^1 \left\{ u^{-1}\right\}\,du \\ & = & \displaystyle \int_0^1 \,du \int_{1-u}^1 \left\{ u^{-1}\right\}\,dx \; = \; \int_0^1 u \left\{ u^{-1}\right\}\,du \; = \; \int_1^\infty \left\{v\right\} v^{-3}\,dv \end{array}$ using the substitution $u = v^{-1}$ . Thus $\begin{array}{rcl} I & = & \displaystyle \sum_{n=1}^\infty \int_0^1 \frac{v}{(n+v)^3}\,dv\; = \; \sum_{n=1}^\infty \left[ - \frac{1}{n+v} + \frac{n}{2(n+v)^2}\right]_0^1 \\ & = & \displaystyle \sum_{n=1}^\infty \left( \frac{1}{2n} - \frac{1}{2(n+1)} - \frac{1}{2(n+1)^2}\right) \; = \; \displaystyle \frac12 - \frac12\sum_{n=2}^\infty\frac{1}{n^2} \; = \; 1 - \tfrac12\zeta(2) \end{array}$ making the answer $1 \times 2 \times 2 = \boxed{4}$ .

Let us substitute $xy=t \implies x\,dy = dt$ Hence we have the integral:- $\displaystyle \int_{0}^{1}\int_{0}^{x} \left\{\frac{1}{1-t}\right\} \text{dt dx}$

Now let us change the order of integration(i.e we want to perform the integral wrt $x$ first and then $t$ )

Hence we have :-

$\displaystyle \int_{0}^{1}\int_{1}^{t} \left\{\frac{1}{1-t}\right\}\text{dx dt} = \int_{0}^{1} (1-t)\left\{\frac{1}{1-t}\right\} dt$

Now substituting $z=1-t$ we have :-

$\displaystyle \int_{0}^{1} z\left\{\frac{1}{z}\right\} dz$ .

We know $\{x\} = x - \lfloor x\rfloor$ . Hence we have :-

$\displaystyle \int_{0}^{1} dz - \int_{0}^{1} z\left\lfloor \frac{1}{z}\right\rfloor dz$

Now the integral $\displaystyle \int_{0}^{1} z\left\lfloor \frac{1}{z}\right\rfloor dz$ can be represented as an infinite sum.

For $r\in \N$

When $\displaystyle \frac{1}{r+1}< z<\frac{1}{r}$ then $\displaystyle r<\frac{1}{z}<r+1 \implies \left\lfloor\frac{1}{z}\right\rfloor = r$ . So we have :-

$\displaystyle \sum_{r=1}^{\infty}\int_{\frac{1}{r+1}}^{\frac{1}{r}} r\cdot z dz = \frac{1}{2}\sum_{r=1}^{\infty} r\left(\frac{1}{r^{2}} -\frac{1}{(r+1)^{2}}\right)$

$\displaystyle = \frac{1}{2}\sum_{r=1}^{\infty}\left(\frac{1}{r} - \frac{1}{r+1} +\frac{1}{(r+1)^{2}}\right) = \frac{\zeta(2)}{2}$

Hence our answer is $\displaystyle 1-\frac{\zeta(2)}{2}$ .