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 .](https://www.tinkutara.com/question/Q37884.png)
$${let}\:{I}\:\:=\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left[{x}\right]} \:{cos}^{\mathrm{2}} \left(\mathrm{2}\pi{x}\right){dx}\:{and}\: \\ $$$${J}\:=\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left[{x}\right]} \:{sin}^{\mathrm{2}} \left(\mathrm{2}\pi{x}\right)\:{dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{I}\:+{J}\:{and}\:{I}\:−{J} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\:{I}\:{and}\:{J}\:. \\ $$
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))) .](https://www.tinkutara.com/question/Q37973.png)
$$\left.\mathrm{1}\right)\:{I}+{J}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−\left[{x}\right]} {dx}=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\int_{{n}} ^{{n}+\mathrm{1}} {e}^{−{n}} {dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left({e}^{−\mathrm{1}} \right)^{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}} }\:=\:\frac{{e}}{{e}−\mathrm{1}}\:. \\ $$$${I}\:−{J}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left[{x}\right]} \:{cos}\left(\mathrm{4}\pi{x}\right){dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\int_{{n}} ^{{n}+\mathrm{1}} \:{e}^{−{n}} \:{cos}\left(\mathrm{4}\pi{x}\right){dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{n}} \:\:\int_{{n}} ^{{n}+\mathrm{1}} \:{cos}\left(\mathrm{4}\pi{x}\right){dx} \\ $$$$=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{n}} \left[\:\frac{\mathrm{1}}{\mathrm{4}\pi}{sin}\left(\mathrm{4}\pi{x}\right)\right]_{{n}} ^{{n}+\mathrm{1}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}\pi}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{n}} \left\{\:{sin}\left(\mathrm{4}\pi\left({n}+\mathrm{1}\right)\right)−{sin}\left(\mathrm{4}\pi{n}\right)\right\} \\ $$$$=\mathrm{0}\:\Rightarrow\:{I}\:={J}\:\:\:\:=\:\frac{{e}}{\mathrm{2}\left({e}−\mathrm{1}\right)}\:. \\ $$$$ \\ $$