Function $f ( x )=\frac{[\sin 2 \pi x ]}{ e ^{\{ x \}}}$ is periodic with
period $'1'$ Therefore
$I=10 \int_{0}^{1} \frac{[\sin 2 \pi x ]}{ e ^{\{ x \}}} d x$
$=10 \int_{0}^{1} \frac{[\sin 2 \pi x ]}{ e ^{ x }} d x$
$=10\left(\int_{0}^{1 / 2} \frac{[\sin 2 \pi x ]}{ e ^{ x }} d x +\int_{1 / 2}^{1} \frac{[\sin 2 \pi x ]}{ e ^{ x }} dx \right)$
$=10\left(0+\int_{1 / 2}^{1} \frac{(-1)}{ e ^{ x }} dx \right)$
$=-10 \int_{1 / 2}^{1} e ^{- x } dx$
$=10\left( e ^{-1}- e ^{-1 / 2}\right)$
Now,
$10 \cdot e ^{-1}-10 \cdot e ^{-1 / 2}=\alpha e ^{-1}+\beta e ^{-1 / 2}+\gamma($ given $)$
$\Rightarrow \alpha=10, \beta=-10, \gamma=0$
$\Rightarrow \alpha+\beta+\gamma=0$
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