If the integral _0^10[ 2 x]e^x-[x],dx= e^-1+ e^-1/2+, where ,, are integers and [x] denote…

Question

If the integral $\displaystyle\int_0^{10}\frac{[\sin 2\pi x]}{e^{x-[x]}}\,dx=\alpha e^{-1}+\beta e^{-1/2}+\gamma$, 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

Accepted Answer

Let $I=\displaystyle\int_0^{10}\frac{[\sin2\pi x]}{e^{x-[x]}}dx=\int_0^{10}[\sin2\pi x]\,e^{-\{x\}}dx$ since $x-[x]=\{x\}$.

The integrand $[\sin2\pi x]e^{-\{x\}}$ is periodic with period $1$. So
$$I=10\int_0^1[\sin2\pi x]e^{-\{x\}}dx=10\int_0^1[\sin2\pi x]e^{-x}dx\quad(\text{since }\{x\}=x,\ 0\le x<1)$$

On $[0,1/2)$, $\sin2\pi x\ge0\Rightarrow[\sin2\pi x]=0$. On $[1/2,1)$, $\sin2\pi x<0\Rightarrow[\sin2\pi x]=-1$.

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

Carrying out the integral: $I=10\left(e^{-1}-e^{-1/2}\right)=10e^{-1}+(-10)e^{-1/2}+0$

Comparing with $\alpha e^{-1}+\beta e^{-1/2}+\gamma$: $\alpha=10,\ \beta=-10,\ \gamma=0$

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

Solutions · 1