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let-I-0-e-x-cos-2-2pix-dx-and-J-0-e-x-sin-2-2pix-dx-1-calculate-I-J-and-I-J-2-find-the-value-of-I-and-J-

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Question Number 37884 by prof Abdo imad last updated on 18/Jun/18
let I = ∫_0 ^∞ e^(−[x]) cos^2 (2πx)dx and J =∫_0 ^∞ e^(−[x]) sin^2 (2πx) dx 1) calculate I +J and I −J 2) find the value of I and J .
letI=∫0∞e−[x]cos2(2πx)dxandJ=∫0∞e−[x]sin2(2πx)dx1)calculateI+JandI−J2)findthevalueofIandJ.
Commented by math khazana by abdo last updated on 20/Jun/18
1) I+J = ∫_0 ^∞ e^(−[x]) dx=Σ_(n=0) ^∞ ∫_n ^(n+1) e^(−n) dx =Σ_(n=0) ^∞ (e^(−1) )^n = (1/(1−e^(−1) )) = (e/(e−1)) . I −J = ∫_0 ^∞ e^(−[x]) cos(4πx)dx =Σ_(n=0) ^∞ ∫_n ^(n+1) e^(−n) cos(4πx)dx =Σ_(n=0) ^∞ e^(−n) ∫_n ^(n+1) cos(4πx)dx = Σ_(n=0) ^∞ e^(−n) [ (1/(4π))sin(4πx)]_n ^(n+1) = (1/(4π)) Σ_(n=0) ^∞ e^(−n) { sin(4π(n+1))−sin(4πn)} =0 ⇒ I =J = (e/(2(e−1))) .
1)I+J=∫0∞e−[x]dx=∑n=0∞∫nn+1e−ndx=∑n=0∞(e−1)n=11−e−1=ee−1.I−J=∫0∞e−[x]cos(4πx)dx=∑n=0∞∫nn+1e−ncos(4πx)dx=∑n=0∞e−n∫nn+1cos(4πx)dx=∑n=0∞e−n[14πsin(4πx)]nn+1=14π∑n=0∞e−n{sin(4π(n+1))−sin(4πn)}=0⇒I=J=e2(e−1).

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