Fractional Part integral $I=\int_{0}^{1}\int_{0}^{1} \left\{\frac{x}{y}\right\}\mathrm{d}x\mathrm{d}y$ Let $$I=\int_{0}^{1}\int_{0}^{1} \left\{\frac{x}{y}\right\}\mathrm{d}x\mathrm{d}y.$$ When I tried computing the integral I seem to be getting a different answer to Wolramaplha, and can't find a similar integral anywhere on MSE or the internet.
Here's how I did it $$\int_{0}^{1}\left(\int_{0}^{1} \left\{\frac{x}{y}\right\}\mathrm{d}y\right)\mathrm{d}x$$, Lets evaluate, (Note: $0<x<1$) $$I_1=\int_{0}^{1} \left\{\frac{x}{y}\right\}\mathrm{d}y =\int_{1}^{\infty}\frac{xy-\lfloor xy \rfloor  }{y^2}\mathrm{d}y=x(1-\gamma)$$
Therefore $I=\frac{(1-\gamma)}{2}$.
However Wolfram gives $I= 0.458868$.
Can someone help, or if my answer is wrong then provide a solution.
 A: Here is a different way of doing this. I assume the unstated fact, apparently obvious, that $\left\{\frac xy\right\}$ is the fractional part of $\frac xy$.
Then
\begin{aligned}
I&=\int_0^1\int_0^1\left\{\frac xy\right\}dxdy=\int_0^1\left[
\int_x^1\frac xy dy+\int_0^x\left\{\frac xy\right\}dy\right]dx\\
&=-\int_0^1 x\ln x dx+\int_0^1\int_0^x\left\{\frac xy\right\}dydx
\end{aligned}
We now divide the interval from 0 to $x$ in the intervals $\frac x{n+1}<y<\frac xn$. Hence, in this interval, we have $n<\frac xy<n+1$. Therefore,
in each such interval, $\frac xy=n+\left\{\frac xy\right\}$.
Thus, we obtain
\begin{aligned}
I &=\frac14+\int_0^1\sum_{n=1}^\infty\int_{\frac x{n+1}}^{\frac xn}\left\{\frac xy\right\}dydx=\frac14+\int_0^1\sum_{n=1}^\infty\int_{\frac x{n+1}}^{\frac xn}\left[\frac xy-n\right]dydx\\
&=\frac 14+\int_0^1xdx\sum_{n=1}^\infty\left[\ln\left(\frac{n+1}n\right)-n\left(\frac1n-\frac1{n+1}\right)\right]\\
&=\frac 14+\frac12\sum_{n=1}^\infty\left[\ln\left(1+\frac1n\right)-\frac1{n+1}\right]
\end{aligned}
This is the same result obtained by @dan_fulea
