If the integral $$\int_0^{10} \frac{|\sin 2\pi x|}{e^{[x]}}dx = \alpha e^{-1} + \beta e^{-\frac{1}{2}} + \gamma$$, where $$\alpha, \beta, \gamma$$ are integers and $$[x]$$ denotes the greatest integer less than or equal to $$x$$, then the value of $$\alpha + \beta + \gamma$$ is equal to:

Given,

$$I=\int_0^{10}\frac{[\sin2\pi x]}{e^{x-[x]}}dx$$

where

$$[x]$$ denotes the greatest integer function.

Since

$$x-[x]=\{x\},$$the fractional part of $$x,$$

we write

$$I=\int_0^{10}[\sin2\pi x]\cdot e^{-\{x\}}dx$$

Now both $$[\sin2\pi x]$$ and $$e^{-\{x\}}$$ have period $$1.$$

Hence,

$$I=10\int_0^1[\sin2\pi x]e^{-x}dx$$

Now analyze $$[\sin2\pi x]$$ on $$[0,1].$$

For

$$0\le x<\frac12,$$ 

 $$0\le\sin2\pi x\le1$$

Therefore,

$$[\sin2\pi x]=0$$

For $$\frac12\le x<1,$$

$$-1\le\sin2\pi x<0$$

Therefore,

$$[\sin2\pi x]=-1$$

Hence,

$$I=10\left(\int_0^{1/2}0\cdot e^{-x}dx+\int_{1/2}^1(-1)e^{-x}dx\right)$$ $$=10\left(-\int_{1/2}^1e^{-x}dx\right)$$

Using

$$\int e^{-x}dx=-e^{-x},$$

we get $$I=10\left[e^{-x}\right]_{1/2}^1$$ $$=10(e^{-1}-e^{-1/2})$$

Comparing with

$$\alpha e^{-1}+\beta e^{-1/2}+\gamma,$$

we get

$$\alpha=10,\qquad \beta=-10,\qquad \gamma=0$$

Therefore,

$$\alpha+\beta+\gamma =10-10+0$$ $$=0$$

Hence, the required answer is $$\boxed{0}$$