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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.
letI=∫0∞e−xcos2(π[x])dxandJ=∫0∞e−xsin2(π[x])dx1)calculateI+JandI−J2)findthevaluesofIandJ.
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 .
1)I+J=∫0∞e−x{cos2(π[x])+sin2(π[x])}dx=∫0∞e−xdx=[−e−x]0+∞=1I−J=∫0∞e−x{cos2(π[x])−sin2(π[x])}dx=∫0∞e−xcos(2π[x])dx=∑n=0∞∫nn+1e−xcos(2πn)dx=∑n=0∞[−e−x]nn+1=∑n=0∞(e−n−e−(n+1))=(1−e−1)∑n=0∞(e−1)n=(1−e−1)11−e−1=1⇒I+J=1andI−J=1⇒I=1andJ=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...
I+J=∫0∞e−x{cos2(Π[x])+sin2(Π[x])}dx=∫0∞e−xdx=∣e−x−1∣0∞=−1(1e∞−1e0)=1I−J=∫0∞e−xcos(2Π[x])dxnowthevalueof[x]=01>x⩾0[x]=12>x⩾1[x]=23>x⩾2soforallvalues∞>x⩾0valueofcos(2Π[x])=1so∫01e−xdx+∫12e−xdx+∫23e−xdx+…infinity=−1{(e−1−e−0)+(e−2−e−1)+(e−3−e−2)+..=−1(−e0otherstermscancelledout…=1plscheckandcomment…

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