TITLE: Evaluate $\int_0^1\int_0^1 \left\{ \frac{e^x}{e^y} \right\}dxdy$
QUESTION [18 upvotes]: I want compute this integral
$$\int_0^1\int_0^1 \left\{  \frac{e^x}{e^y}  \right\}dxdy, $$
where $ \left\{ x \right\} $ is the fractional part function.
Following PROBLEMA 171, Prueba de a), last paragraph of page 109 and firts two paragraphs of page 110,here in spanish, I say the case $k=1$.
When I take $x=\log u$ and $y=\log v$ then I can show that 
$$\int_0^1\int_0^1 \left\{  \frac{e^x}{e^y}  \right\}dxdy=\int_1^e\int_1^e \left\{  \frac{x}{y}  \right\}\frac{1}{xy}dxdy=I_1+I_2$$
since following the strategy in cited problem and take $t=\frac{1}{u}$
$$I_1:=\int_1^e\int_1^x \left\{  \frac{x}{y}  \right\}\frac{1}{xy}dydx=\int_1^e\frac{1}{x}\int_{\frac{1}{x}}^1 \left\{  \frac{1}{t}  \right\}\frac{dt}{t}dx=\int_1^e\int_1^x\frac{ \left\{ u \right\} }{u}dudx,$$
and since if there are no mistakes 
$$ \int_1^x\frac{ \left\{ u \right\} }{u}du =
\begin{cases}
x-1-\log x,  & \text{if $1\leq x<2$} \\
1+\log 2+(x-2)-2\log x, & \text{if $2\leq x\leq e$}
\end{cases}$$
then $$I_1=\int_1^2\frac{1}{x}(x-1-\log x)dx+\int_1^2\frac{1}{x}(1+\log 2+(x-2)-2\log x)dx,$$
It is $I_1=-3+\log 2-\frac{\log^22}{2}+e$. On the other hand following the cited problem, since $y>x$ then $ \left\{ \frac{x}{y} \right\}=  \frac{x}{y}$ and the second integral is computed as 
$$I_2:=\int_1^e\int_x^e \left\{  \frac{x}{y}  \right\}\frac{1}{xy}dydx=\int_1^e\int_x^e  \frac{x}{y} \frac{1}{y^2}dydx.$$
Thus I've computed $I_2=\frac{1}{e}$.

Question. I would to know if my computations with the fractional part function $ \left\{ x \right\} $ were rights (the evaluation of $ \int_1^x\frac{ \left\{ u \right\} }{u}du$ and $I_1$). Can you compute $$\int_0^1\int_0^1 \left\{  \frac{e^x}{e^y}  \right\}^kdxdy$$ for the case $k=1$? (At least this case to see it as a proof verification of my computations; your are welcome if you provide us similar identities for integers $k\geq 1$, as in the cited problem). Thanks in advance.

REPLY [9 votes]: Let $u=x-y$ and $v=x+y$. Then
$$
\left\{(x,y):0\le x,y\le1\right\}
=\left\{(u,v):0\le\left|u\right|\le1,\left|v-1\right|\le1-\left|u\right|\right\}
$$
The change of coordinates makes things a bit easier
$$
\begin{align}
&\int_0^1\int_0^1\left\{\frac{e^x}{e^y}\right\}\,\mathrm{d}x\,\mathrm{d}y\\
&=\int_0^1\int_0^1\left\{e^{x-y}\right\}\,\mathrm{d}x\,\mathrm{d}y\\
&=\frac12\int_{-1}^0\int_{-u}^{2+u}\left\{e^u\right\}\,\mathrm{d}v\,\mathrm{d}u
+\frac12\int_0^1\int_u^{2-u}\left\{e^u\right\}\,\mathrm{d}v\,\mathrm{d}u\\
&=\int_{-1}^0(1+u)\left\{e^u\right\}\,\mathrm{d}u
+\int_0^1(1-u)\left\{e^u\right\}\,\mathrm{d}u\\
&=\int_0^1(1-u)\left\{e^{-u}\right\}\,\mathrm{d}u
+\int_0^1(1-u)\left\{e^u\right\}\,\mathrm{d}u\\
&=\int_0^1(1-u)\,e^{-u}\,\mathrm{d}u
+\int_0^{\log(2)}(1-u)\left(e^u-1\right)\,\mathrm{d}u
+\int_{\log(2)}^1(1-u)\left(e^u-2\right)\,\mathrm{d}u\\
&=\int_0^1(1-u)\left(e^{-u}+e^u\right)\,\mathrm{d}u
-\int_0^{\log(2)}(1-u)\,\mathrm{d}u
-2\int_{\log(2)}^1(1-u)\,\mathrm{d}u\\[3pt]
&=\left[(2-u)e^u+ue^{-u}\right]_0^1
-\left[u-\tfrac12u^2\right]_0^{\log(2)}
-2\left[u-\tfrac12u^2\right]_{\log(2)}^1\\[6pt]
&=\left[(2-u)e^u+ue^{-u}\right]_0^1
-\left[u-\tfrac12u^2\right]_0^1
-\left[u-\tfrac12u^2\right]_{\log(2)}^1\\[9pt]
&=2\cosh(1)-2-\tfrac12-\tfrac12+\log(2)-\tfrac12\log(2)^2\\[12pt]
&=2\cosh(1)-3+\log(2)-\tfrac12\log(2)^2
\end{align}
$$