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Question Number 97707 by Power last updated on 09/Jun/20
Answered by smridha last updated on 09/Jun/20
=(1/2)∫_(−∞) ^(+∞) (e^(ix) /(x^2 +1^2 ))dx +(1/2)∫_(+∞) ^(−∞) (e^(−(−ix)) /((−x)^2 +1^2 ))d(−x) =∫_(−∞) ^(+∞) (e^(ix) /(x^2 +1^2 ))dx=∮_C (e^(iz) /(z^2 +1^2 ))dz =∮_C (e^(iz) /((z+i)(z−i)))dz now time for calculate the residues residue at z_1 =i lim_(z→i) (((z−i)e^(iz) )/((z−i)(z+i)))=(e^(−1) /(2i)) but z_2 =−i this pole does not enclosed by the cantour(C) now: I=∮_C f(z)dz=2𝛑iΣenclosed resudue by the cantour =2𝛑i.(e^(−1) /(2i))=𝛑e^(−1)
$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{\boldsymbol{{e}}^{\boldsymbol{{ix}}} }{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }\boldsymbol{{dx}}\:+\frac{\mathrm{1}}{\mathrm{2}}\int_{+\infty} ^{−\infty} \frac{\boldsymbol{{e}}^{−\left(−\boldsymbol{{ix}}\right)} }{\left(−\boldsymbol{{x}}\right)^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }\boldsymbol{{d}}\left(−\boldsymbol{{x}}\right) \\ $$$$=\int_{−\infty} ^{+\infty} \frac{\boldsymbol{{e}}^{\boldsymbol{{ix}}} }{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }\boldsymbol{{dx}}=\oint_{\boldsymbol{{C}}} \frac{\boldsymbol{{e}}^{\boldsymbol{{iz}}} }{\boldsymbol{{z}}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }\boldsymbol{{dz}} \\ $$$$=\oint_{\boldsymbol{{C}}} \frac{\boldsymbol{{e}}^{\boldsymbol{{iz}}} }{\left(\boldsymbol{{z}}+\boldsymbol{{i}}\right)\left(\boldsymbol{{z}}−\boldsymbol{{i}}\right)}\boldsymbol{{dz}} \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{time}}\:\boldsymbol{{for}}\:\boldsymbol{{calculate}}\:\boldsymbol{{the}}\:\boldsymbol{{residues}} \\ $$$$\boldsymbol{{residue}}\:\boldsymbol{{at}}\:\boldsymbol{{z}}_{\mathrm{1}} =\boldsymbol{{i}} \\ $$$$\:\:\:\:\:\:\underset{\boldsymbol{{z}}\rightarrow\boldsymbol{{i}}} {\mathrm{lim}}\frac{\left(\boldsymbol{{z}}−\boldsymbol{{i}}\right)\boldsymbol{{e}}^{\boldsymbol{{iz}}} }{\left(\boldsymbol{{z}}−\boldsymbol{{i}}\right)\left(\boldsymbol{{z}}+\boldsymbol{{i}}\right)}=\frac{\boldsymbol{{e}}^{−\mathrm{1}} }{\mathrm{2}\boldsymbol{{i}}} \\ $$$$\boldsymbol{{but}}\:\boldsymbol{{z}}_{\mathrm{2}} =−\boldsymbol{{i}}\:\boldsymbol{{this}}\:\boldsymbol{{pole}}\:\boldsymbol{{does}}\:\boldsymbol{{not}}\:\boldsymbol{{enclosed}} \\ $$$$\boldsymbol{{by}}\:\boldsymbol{{the}}\:\boldsymbol{{cantour}}\left(\boldsymbol{{C}}\right) \\ $$$$\boldsymbol{{now}}: \\ $$$$\boldsymbol{{I}}=\oint_{\boldsymbol{{C}}} \boldsymbol{{f}}\left(\boldsymbol{{z}}\right)\boldsymbol{{dz}}=\mathrm{2}\boldsymbol{\pi}{i}\Sigma\boldsymbol{{enclosed}}\:\boldsymbol{{resudue}}\:\boldsymbol{{by}}\:\boldsymbol{{the}}\:\boldsymbol{{cantour}} \\ $$$$=\mathrm{2}\boldsymbol{\pi}{i}.\frac{\boldsymbol{{e}}^{−\mathrm{1}} }{\mathrm{2}\boldsymbol{{i}}}=\boldsymbol{\pi}{e}^{−\mathrm{1}} \\ $$
Answered by smridha last updated on 09/Jun/20
Method (2) let F(t)=2∫_0 ^(+∞) ((cos(tx))/(x^2 +1))dx[given integrand is even] now by Laplace Transformation f(s)=L[f(t)].....(i) so f(s)=∫_0 ^∞ e^(−st) [2∫_0 ^∞ ((cos(tx))/(x^2 +1))dx]dt =2∫_0 ^∞ (1/(x^2 +1))[∫_0 ^∞ e^(−st) cos(tx)dt]dx =2∫_0 ^∞ (s/((x^2 +1)(s^2 +x^2 )))dx =((2s)/((s^2 −1)))[∫_0 ^∞ ((1/((x^2 +1)))−(1/((s^2 +x^2 ))))dx] =((2s)/((s^2 −1)))[tan^(−1) (x)−(1/s)tan^(−1) ((x/s))]_0 ^∞ =(𝛑/((s−(−1)))) now from (i) F(t)=L^(−1) [f(s)] =𝛑.L^(−1) [(1/((s−(−1))))] =𝛑e^(−1) now I=F(1)=(𝛑/e)
$$\boldsymbol{{M}}{e}\boldsymbol{{thod}}\:\left(\mathrm{2}\right) \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{F}}\left({t}\right)=\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \frac{\boldsymbol{{cos}}\left(\boldsymbol{{tx}}\right)}{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}}\boldsymbol{{dx}}\left[\boldsymbol{{given}}\:\boldsymbol{{integrand}}\:\boldsymbol{{is}}\:\boldsymbol{{even}}\right] \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{by}}\:\boldsymbol{{L}}{a}\boldsymbol{{place}}\:\boldsymbol{{Transformation}} \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{s}}\right)=\boldsymbol{{L}}\left[\boldsymbol{{f}}\left(\boldsymbol{{t}}\right)\right]…..\left(\boldsymbol{{i}}\right) \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{f}}\left(\boldsymbol{{s}}\right)=\int_{\mathrm{0}} ^{\infty} \boldsymbol{{e}}^{−\boldsymbol{{st}}} \left[\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{cos}}\left(\boldsymbol{{tx}}\right)}{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}}\boldsymbol{{dx}}\right]\boldsymbol{{dt}}\: \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}}\left[\int_{\mathrm{0}} ^{\infty} \boldsymbol{{e}}^{−\boldsymbol{{st}}} \boldsymbol{{cos}}\left(\boldsymbol{{tx}}\right)\boldsymbol{{dt}}\right]\boldsymbol{{dx}} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{s}}}{\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}\right)\left(\boldsymbol{{s}}^{\mathrm{2}} +\boldsymbol{{x}}^{\mathrm{2}} \right)}\boldsymbol{{dx}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{2}\boldsymbol{{s}}}{\left(\boldsymbol{{s}}^{\mathrm{2}} −\mathrm{1}\right)}\left[\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)}−\frac{\mathrm{1}}{\left(\boldsymbol{{s}}^{\mathrm{2}} +\boldsymbol{{x}}^{\mathrm{2}} \right)}\right)\boldsymbol{{dx}}\right] \\ $$$$=\frac{\mathrm{2}\boldsymbol{{s}}}{\left(\boldsymbol{{s}}^{\mathrm{2}} −\mathrm{1}\right)}\left[\mathrm{tan}^{−\mathrm{1}} \left(\boldsymbol{{x}}\right)−\frac{\mathrm{1}}{\boldsymbol{{s}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\boldsymbol{{x}}}{\boldsymbol{{s}}}\right)\right]_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\boldsymbol{\pi}}{\left({s}−\left(−\mathrm{1}\right)\right)} \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{from}}\:\left(\boldsymbol{{i}}\right)\:\boldsymbol{{F}}\left(\boldsymbol{{t}}\right)=\boldsymbol{{L}}^{−\mathrm{1}} \left[\boldsymbol{{f}}\left(\boldsymbol{{s}}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\boldsymbol{\pi}.\boldsymbol{{L}}^{−\mathrm{1}} \left[\frac{\mathrm{1}}{\left(\boldsymbol{{s}}−\left(−\mathrm{1}\right)\right)}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\boldsymbol{\pi{e}}^{−\mathrm{1}} \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{I}}=\boldsymbol{{F}}\left(\mathrm{1}\right)=\frac{\boldsymbol{\pi}}{\boldsymbol{{e}}} \\ $$
Commented by Power last updated on 09/Jun/20
thanks sir
$$\mathrm{thanks}\:\mathrm{sir} \\ $$
Commented by smridha last updated on 09/Jun/20
also thankful because this is very significant integral.
$$\boldsymbol{{also}}\:\boldsymbol{{thankful}}\:\boldsymbol{{because}}\:\boldsymbol{{this}}\:\boldsymbol{{is}}\: \\ $$$$\boldsymbol{{very}}\:\boldsymbol{{significant}}\:\boldsymbol{{integral}}. \\ $$
Answered by abdomathmax last updated on 09/Jun/20
A =∫_(−∞) ^(+∞) ((cosx)/(x^2 +1))dx ⇒ A = Re(∫_(−∞) ^(+∞) (e^(ix) /(x^(2 ) +1))dx) let ϕ(z) = (e^(iz) /(z^2 +1)) ⇒ϕ(z) =(e^(iz) /((z−i)(z+i))) residus tbeorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) =2iπ×(e^(−1) /(2i)) =πe^(−1) ⇒ A =(π/e)
$$\mathrm{A}\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cosx}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx}\:\Rightarrow\:\mathrm{A}\:=\:\mathrm{Re}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{e}^{\mathrm{ix}} }{\mathrm{x}^{\mathrm{2}\:} +\mathrm{1}}\mathrm{dx}\right) \\ $$$$\mathrm{let}\:\varphi\left(\mathrm{z}\right)\:=\:\frac{\mathrm{e}^{\mathrm{iz}} }{\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{e}^{\mathrm{iz}} }{\left(\mathrm{z}−\mathrm{i}\right)\left(\mathrm{z}+\mathrm{i}\right)} \\ $$$$\mathrm{residus}\:\mathrm{tbeorem}\:\mathrm{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\varphi,\mathrm{i}\right)\:=\mathrm{2i}\pi×\frac{\mathrm{e}^{−\mathrm{1}} }{\mathrm{2i}} \\ $$$$=\pi\mathrm{e}^{−\mathrm{1}} \:\Rightarrow\:\mathrm{A}\:=\frac{\pi}{\mathrm{e}} \\ $$

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