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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*}

Summing up,

    \[\iint \limits_{[0, 1]^2} \left \{ e^x e^{-y} \right \} \, \mathrm{d}(x, y) = \frac{1}{e} +e + \ln 2 - \frac{\ln^2 2}{2} - 3\]

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