If ${\int }_0^{10}\frac{\left[\sin 2\pi x\right]}{e^{x-\left[x\right]}}dx=$ $\alpha e^{-1}+\beta e^{-\frac{1}{2}}+\gamma \text{'}$ where $\alpha , \beta , \gamma$ are integers and $\left[x\right]$ denotes the greatest integer less than or equal to x, then the value of $\alpha +\beta +\gamma$ is equal to 

$\begin{array}{rcl} & & {\int }_0^{10}\frac{\left[\sin 2\mathrm{\pi x}\right]}{{\mathrm{e}}^{\mathrm{x}-\left[\mathrm{x}\right]}}\\ & =& {\int }_0^{10}\frac{\left[\sin 2\mathrm{\pi x}\right]}{{\mathrm{e}}^{\left\{\mathrm{x}\right\}}}\end{array}$

${\mathrm{e}}^{\left\{\mathrm{x}\right\}}, \left[\sin 2 \mathrm{\pi } \mathrm{x}\right]$  are Periodic functions with Period=1

$\begin{array}{rcl} & =& 10 {\int }_0^1 \frac{\left[\sin 2 \mathrm{\pi } \mathrm{x}\right]}{{\mathrm{e}}^{\left\{\mathrm{x}\right\}}} \mathrm{dx}\\ & =& 10 \left[{\int }_0^{1/2} \frac{\left[\sin 2 \mathrm{\pi } \mathrm{x}\right]}{{\mathrm{e}}^{\mathrm{x}-\left\{\mathrm{x}\right\}}}+{\int }_{1/2}^1 \frac{\left[\sin 2 \mathrm{\pi } \mathrm{x}\right]}{{\mathrm{e}}^{\mathrm{x}-\left\{\mathrm{x}\right\}}}\right]\\ & =& 10 \left[{\int }_0^{1/2} \frac{0}{{\mathrm{e}}^{\mathrm{x}}}+{\int }_{1/2}^1 \frac{-1}{{\mathrm{e}}^{\mathrm{x}}} \mathrm{dx}\right]\\ & =& 10 {\left[{\mathrm{e}}^{-\mathrm{x}}\right]}_{1/2}^1=10\left[{\mathrm{e}}^{-1}-{\mathrm{e}}^{-11/2}\right]\end{array}$

$$\begin{gathered} \mathrm{I}=10{\mathrm{e}}^{-1}-10 {\mathrm{e}}^{-1/2} \\ \mathrm{\alpha }=10, \mathrm{\beta }=-10, \mathrm{\gamma }=0 \\ \mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=0 \end{gathered}$$