Question Number 37812 by prof Abdo imad last updated on 17/Jun/18
![calculate ∫_0 ^∞ e^(−2x) sin{π[x]}dx .](https://www.tinkutara.com/question/Q37812.png)
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\mathrm{2}{x}} {sin}\left\{\pi\left[{x}\right]\right\}{dx}\:. \\ $$
Commented by abdo mathsup 649 cc last updated on 19/Jun/18
![∫_0 ^∞ e^(−2x) sin{π[x]}dx=Σ_(n=0) ^∞ ∫_n ^(n+1) e^(−2x) sin(nπ)dx =Σ_(n=0) ^∞ sin(nπ)∫_n ^(n+1) e^(−2x) dx =0](https://www.tinkutara.com/question/Q37897.png)
$$\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−\mathrm{2}{x}} {sin}\left\{\pi\left[{x}\right]\right\}{dx}=\sum_{{n}=\mathrm{0}} ^{\infty} \:\int_{{n}} ^{{n}+\mathrm{1}} \:{e}^{−\mathrm{2}{x}} \:{sin}\left({n}\pi\right){dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:{sin}\left({n}\pi\right)\int_{{n}} ^{{n}+\mathrm{1}} \:{e}^{−\mathrm{2}{x}} \:{dx}\:=\mathrm{0} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jun/18
![[x]=0 1>x≥0 [x]=1 2>x≥1 [x]=2 3>x≥2 thus putting the value of [x] we get intregal multiple of Π...thus value of sin{Π[x]}=0 so the intrdgation value is zero Refer floor function/greatest integer function](https://www.tinkutara.com/question/Q37857.png)
$$\left[{x}\right]=\mathrm{0}\:\:\:\mathrm{1}>{x}\geqslant\mathrm{0} \\ $$$$\left[{x}\right]=\mathrm{1}\:\:\:\mathrm{2}>{x}\geqslant\mathrm{1} \\ $$$$\left[{x}\right]=\mathrm{2}\:\:\:\mathrm{3}>{x}\geqslant\mathrm{2} \\ $$$$ \\ $$$${thus}\:{putting}\:{the}\:{value}\:{of}\:\left[{x}\right]\:\:{we}\:{get}\:{intregal} \\ $$$${multiple}\:{of}\:\Pi…{thus}\:{value}\:{of}\:{sin}\left\{\Pi\left[{x}\right]\right\}=\mathrm{0} \\ $$$${so}\:{the}\:{intrdgation}\:{value}\:{is}\:{zero} \\ $$$${Refer}\:{floor}\:{function}/{greatest}\:{integer}\:{function} \\ $$