Question Number 37815 by prof Abdo imad last updated on 17/Jun/18
![let I = ∫_0 ^∞ e^(−x) cos^2 (π[x])dx and J = ∫_0 ^∞ e^(−x) sin^2 (π[x])dx 1) calculate I +J and I −J 2) find the values of I and J.](Q37815.png)
$${let}\:{I}\:\:=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{x}} \:{cos}^{\mathrm{2}} \left(\pi\left[{x}\right]\right){dx}\:{and} \\ $$$${J}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{x}} \:{sin}^{\mathrm{2}} \left(\pi\left[{x}\right]\right){dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{I}\:+{J}\:\:{and}\:{I}\:−{J} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{values}\:{of}\:{I}\:{and}\:{J}. \\ $$
Commented by prof Abdo imad last updated on 18/Jun/18
![1) I +J = ∫_0 ^∞ e^(−x) {cos^2 (π[x]) +sin^2 (π[x])}dx =∫_0 ^∞ e^(−x) dx=[ −e^(−x) ]_0 ^(+∞) =1 I−J =∫_0 ^∞ e^(−x) { cos^2 (π[x]) −sin^2 (π[x])}dx = ∫_0 ^∞ e^(−x) cos(2π[x])dx =Σ_(n=0) ^∞ ∫_n ^(n+1) e^(−x) cos(2πn) dx =Σ_(n=0) ^∞ [ −e^(−x) ]_n ^(n+1) =Σ_(n=0) ^∞ ( e^(−n) −e^(−(n+1)) ) = (1−e^(−1) ) Σ_(n=0) ^∞ (e^(−1) )^n =(1−e^(−1) )(1/(1−e^(−1) )) =1 ⇒ I +J =1 and I−J =1 ⇒ I =1 and J =0 .](Q37883.png)
$$\left.\mathrm{1}\right)\:{I}\:+{J}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{x}} \left\{{cos}^{\mathrm{2}} \left(\pi\left[{x}\right]\right)\:+{sin}^{\mathrm{2}} \left(\pi\left[{x}\right]\right)\right\}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}} {dx}=\left[\:−{e}^{−{x}} \right]_{\mathrm{0}} ^{+\infty} \:=\mathrm{1} \\ $$$${I}−{J}\:=\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{x}} \left\{\:{cos}^{\mathrm{2}} \left(\pi\left[{x}\right]\right)\:−{sin}^{\mathrm{2}} \left(\pi\left[{x}\right]\right)\right\}{dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{x}} {cos}\left(\mathrm{2}\pi\left[{x}\right]\right){dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\int_{{n}} ^{{n}+\mathrm{1}} \:{e}^{−{x}} \:{cos}\left(\mathrm{2}\pi{n}\right)\:{dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left[\:−{e}^{−{x}} \right]_{{n}} ^{{n}+\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(\:\:{e}^{−{n}} \:−{e}^{−\left({n}+\mathrm{1}\right)} \right) \\ $$$$=\:\left(\mathrm{1}−{e}^{−\mathrm{1}} \right)\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left({e}^{−\mathrm{1}} \right)^{{n}} \\ $$$$=\left(\mathrm{1}−{e}^{−\mathrm{1}} \right)\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}} }\:=\mathrm{1}\:\Rightarrow\:{I}\:+{J}\:=\mathrm{1}\:{and}\:{I}−{J}\:=\mathrm{1}\:\Rightarrow \\ $$$${I}\:=\mathrm{1}\:{and}\:\:{J}\:=\mathrm{0}\:. \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jun/18
![I+J=∫_0 ^∞ e^(−x) {cos^2 (Π[x])+sin^2 (Π[x])}dx =∫_0 ^∞ e^(−x) dx=∣(e^(−x) /(−1))∣_0 ^∞ =−1((1/e^∞ )−(1/e^0 ))=1 I−J=∫_0 ^∞ e^(−x) cos(2Π[x]) dx now the value of [x] =0 1>x≥0 [x]=1 2>x≥1 [x]=2 3>x≥2 so for all values ∞>x≥0 value of cos(2Π[x])=1 so ∫_0 ^1 e^(−x) dx+∫_1 ^2 e^(−x) dx+∫_2 ^3 e^(−x) dx+...infinity =−1{(e^(−1) −e^(−0) )+(e^(−2) −e^(−1) )+(e^(−3) −e^(−2) )+.. =−1(−e^0 others terms cancelled out... =1 pls check and comment...](Q37852.png)
$${I}+{J}=\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} \left\{{cos}^{\mathrm{2}} \left(\Pi\left[{x}\right]\right)+{sin}^{\mathrm{2}} \left(\Pi\left[{x}\right]\right)\right\}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {dx}=\mid\frac{{e}^{−{x}} }{−\mathrm{1}}\mid_{\mathrm{0}} ^{\infty} =−\mathrm{1}\left(\frac{\mathrm{1}}{{e}^{\infty} }−\frac{\mathrm{1}}{{e}^{\mathrm{0}} }\right)=\mathrm{1} \\ $$$${I}−{J}=\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {cos}\left(\mathrm{2}\Pi\left[{x}\right]\right)\:{dx} \\ $$$${now}\:{the}\:{value}\:{of}\:\left[{x}\right]\:=\mathrm{0}\:\:\:\:\:\mathrm{1}>{x}\geqslant\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\left[{x}\right]=\mathrm{1}\:\:\:\:\:\:\mathrm{2}>{x}\geqslant\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\left[{x}\right]=\mathrm{2}\:\:\:\:\:\mathrm{3}>{x}\geqslant\mathrm{2} \\ $$$${so}\:{for}\:{all}\:{values}\:\:\:\:\:\infty>{x}\geqslant\mathrm{0} \\ $$$${value}\:{of}\:{cos}\left(\mathrm{2}\Pi\left[{x}\right]\right)=\mathrm{1} \\ $$$${so}\:\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{x}} {dx}+\int_{\mathrm{1}} ^{\mathrm{2}} {e}^{−{x}} {dx}+\int_{\mathrm{2}} ^{\mathrm{3}} {e}^{−{x}} {dx}+...{infinity} \\ $$$$=−\mathrm{1}\left\{\left({e}^{−\mathrm{1}} −{e}^{−\mathrm{0}} \right)+\left({e}^{−\mathrm{2}} −{e}^{−\mathrm{1}} \right)+\left({e}^{−\mathrm{3}} −{e}^{−\mathrm{2}} \right)+..\right. \\ $$$$=−\mathrm{1}\left(−{e}^{\mathrm{0}} \:\:\:{others}\:{terms}\:{cancelled}\:{out}...\right. \\ $$$$=\mathrm{1} \\ $$$${pls}\:{check}\:{and}\:{comment}... \\ $$