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calculste-I-0-ch-cos-2x-dx-x-2-4-and-J-0-cos-2chx-dx-x-2-4-compare-I-and-J-

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Question Number 90042 by abdomathmax last updated on 21/Apr/20
calculste I =∫_0 ^(+∞) ((ch(cos(2x))dx)/(x^2 +4)) and J =∫_0 ^∞ ((cos(2chx)dx)/(x^2 +4)) compare I and J
$${calculste}\:{I}\:=\int_{\mathrm{0}} ^{+\infty} \:\frac{{ch}\left({cos}\left(\mathrm{2}{x}\right)\right){dx}}{{x}^{\mathrm{2}} \:+\mathrm{4}} \\ $$$${and}\:{J}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left(\mathrm{2}{chx}\right){dx}}{{x}^{\mathrm{2}} \:+\mathrm{4}} \\ $$$${compare}\:{I}\:{and}\:{J} \\ $$
Commented by mathmax by abdo last updated on 21/Apr/20
I =∫_0 ^∞ ((ch(cos(2x))dx)/(x^2 +4)) ⇒2I =∫_(−∞) ^(+∞) ((ch(cos(2x))dx)/(x^2 +4)) let ϕ(z) =((ch(cos(2z)))/(z^2 +4)) ⇒ϕ(z) =((ch(cos(2z)))/((z−2i)(z+2i))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,2i) =2iπ×((ch(cos(4i)))/(4i)) =(π/2)ch(ch(−4)) =(π/2)ch(ch(4) =(π/4)(e^(ch(4)) +e^(−ch(4)) ) ⇒ I =(π/8)( e^(ch(4)) +e^(−ch(4)) ) also we have 2J =∫_(−∞) ^(+∞) ((cos(2chx)dx)/(x^2 +4)) =Re(∫_(−∞) ^(+∞) (e^(i2ch(x)) /(x^2 +4))dx) let g(z) =(e^(i2ch(z)) /(z^2 +4)) ⇒g(z) =(e^(i2ch(z)) /((z−2i)(z+2i))) ∫_(−∞) ^(+∞) g(z)dz =2iπ Res(g,2i) =2iπ×(e^(i2ch(2i)) /(4i)) =(π/2) e^(2icos(2)) =(π/2){ cos(2cos2) +isin(2cos2)) ⇒ J =(π/4)cos(2cos2)
$${I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{ch}\left({cos}\left(\mathrm{2}{x}\right)\right){dx}}{{x}^{\mathrm{2}} \:+\mathrm{4}}\:\Rightarrow\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{ch}\left({cos}\left(\mathrm{2}{x}\right)\right){dx}}{{x}^{\mathrm{2}} \:+\mathrm{4}} \\ $$$${let}\:\varphi\left({z}\right)\:=\frac{{ch}\left({cos}\left(\mathrm{2}{z}\right)\right)}{{z}^{\mathrm{2}} \:+\mathrm{4}}\:\Rightarrow\varphi\left({z}\right)\:=\frac{{ch}\left({cos}\left(\mathrm{2}{z}\right)\right)}{\left({z}−\mathrm{2}{i}\right)\left({z}+\mathrm{2}{i}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,\mathrm{2}{i}\right)\:=\mathrm{2}{i}\pi×\frac{{ch}\left({cos}\left(\mathrm{4}{i}\right)\right)}{\mathrm{4}{i}} \\ $$$$=\frac{\pi}{\mathrm{2}}{ch}\left({ch}\left(−\mathrm{4}\right)\right)\:=\frac{\pi}{\mathrm{2}}{ch}\left({ch}\left(\mathrm{4}\right)\:=\frac{\pi}{\mathrm{4}}\left({e}^{{ch}\left(\mathrm{4}\right)} \:+{e}^{−{ch}\left(\mathrm{4}\right)} \right)\:\Rightarrow\right. \\ $$$${I}\:=\frac{\pi}{\mathrm{8}}\left(\:{e}^{{ch}\left(\mathrm{4}\right)} \:+{e}^{−{ch}\left(\mathrm{4}\right)} \right)\:{also}\:{we}\:{have} \\ $$$$\mathrm{2}{J}\:=\int_{−\infty} ^{+\infty} \:\frac{{cos}\left(\mathrm{2}{chx}\right){dx}}{{x}^{\mathrm{2}} \:+\mathrm{4}}\:={Re}\left(\int_{−\infty} ^{+\infty} \:\frac{{e}^{{i}\mathrm{2}{ch}\left({x}\right)} }{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx}\right)\:{let} \\ $$$${g}\left({z}\right)\:=\frac{{e}^{{i}\mathrm{2}{ch}\left({z}\right)} }{{z}^{\mathrm{2}} \:+\mathrm{4}}\:\Rightarrow{g}\left({z}\right)\:=\frac{{e}^{{i}\mathrm{2}{ch}\left({z}\right)} }{\left({z}−\mathrm{2}{i}\right)\left({z}+\mathrm{2}{i}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:{g}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({g},\mathrm{2}{i}\right)\:=\mathrm{2}{i}\pi×\frac{{e}^{{i}\mathrm{2}{ch}\left(\mathrm{2}{i}\right)} }{\mathrm{4}{i}} \\ $$$$=\frac{\pi}{\mathrm{2}}\:{e}^{\mathrm{2}{icos}\left(\mathrm{2}\right)} \:=\frac{\pi}{\mathrm{2}}\left\{\:{cos}\left(\mathrm{2}{cos}\mathrm{2}\right)\:+{isin}\left(\mathrm{2}{cos}\mathrm{2}\right)\right)\:\Rightarrow \\ $$$${J}\:=\frac{\pi}{\mathrm{4}}{cos}\left(\mathrm{2}{cos}\mathrm{2}\right) \\ $$

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