Exponential fractional part

Let $\{ \cdot \}$ denote the fractional part of $x$ . Evaluate the integral

$\mathcal{J} = \int_0^1 \int_0^1 \left\{ \frac{e^x}{e^y} \right\} \, \mathrm{d}(x, y)$

Solution

By definition it holds that

$\left \{ x \right \} = x - \left \lfloor x \right \rfloor$

hence

$\iint \limits_{\left [ 0, 1 \right ]^2} \left \{ e^x e^{-y} \right \} \, \mathrm{d}(x, y)= \iint \limits_{\left [ 0, 1 \right ]^2} e^x e^{-y} \, \mathrm{d}(x, y)-\iint \limits_{\left [ 0, 1 \right ]^2} \left \lfloor e^x e^{-y} \right \rfloor \, \mathrm{d}(x, y)$

We note however that

$\left \lfloor e^x e^{-y} \right \rfloor = \left\{\begin{matrix} 0 & , & x< y \\ 2 & , & y< x - \ln 2 \\ 1 & , & \text{otherwise} \end{matrix}\right.$

Hence,

$\begin{align*} \iint \limits_{[0, 1]^2} \left \lfloor e^x e^{-y} \right \rfloor \, \mathrm{d} (x, y) &= \int_{0}^{1} \int_{x}^{1} 0 \, \mathrm{d}(y, x) + \int_{0}^{1} \int_{0}^{x} \, \mathrm{d} (y, x) + \\ & \quad \quad + \int_{\ln 2}^{1} \int_{0}^{x-\ln 2} 1 \, \mathrm{d} \left ( y, x \right ) \\ &= 1 - \ln 2 + \frac{\ln^2 2}{2} \end{align*}$