Question Number 227255 by Spillover last updated on 11/Jan/26
Answered by Spillover last updated on 11/Jan/26
Answered by Spillover last updated on 11/Jan/26
Answered by Spillover last updated on 11/Jan/26
Answered by Spillover last updated on 11/Jan/26
Answered by Spillover last updated on 11/Jan/26
Answered by Spillover last updated on 11/Jan/26
Answered by Kassista last updated on 11/Jan/26
![Observe that: ((cos x )/(x^2 +1)) is an even function ∴ ∫_(−∞) ^( 0) ((cosx)/(x^2 +1)) dx = ∫_0 ^( ∞) ((cos x)/(x^2 +1)) dx=(I/2) let f(z):= (e^(iz) /(z^2 +1)) where z∈C I=ℜ∫_(−∞) ^( ∞) (e^(iz) /(z^2 +1)) dz and define the contour C as a semi−circle in upper half plane with radius R ∮_C f(z) dz = ∮_([0, R]) f(x) dx + ∮_([−R, 0]) f(x) dx + ∮_(Arc) f(z) dz ⇒lim_(R→∞) {∮_([0,R]) f(x) dx+∮_([−∞, 0]) f(x) dx+∮_(Arc ) f(z) dz }= I +∮_(Arc) (e^(iz) /(1+z^2 )) dz Since (1/(1+z^2 )) decreases faster than (1/(∣z∣)), by Jordan′s Lemma: = I+ 0=I ∴ ∮_C f(z) dz = I = 2πiΣ_(poles) Res(f(z), pole) Note that f(z) has verticals asymptotes at z=±i however, only z=i ∈ C I=2πi lim_(z→i) (z−i).(e^(iz) /(1+z^2 )) = 2πi lim_(z→i) (1/(z+i)).e^(iz) 2πi.(e^(−1) /(2i))=(π/e) ∴ I=(π/e)](https://www.tinkutara.com/question/Q227277.png)
$$ \\ $$$${Observe}\:{that}:\:\frac{{cos}\:{x}\:}{{x}^{\mathrm{2}} +\mathrm{1}}\:{is}\:{an}\:{even}\:{function} \\ $$$$\therefore\:\int_{−\infty} ^{\:\mathrm{0}} \frac{{cosx}}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx}\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\:{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx}=\frac{{I}}{\mathrm{2}} \\ $$$${let}\:{f}\left({z}\right):=\:\frac{{e}^{{iz}} }{{z}^{\mathrm{2}} +\mathrm{1}}\:{where}\:{z}\in\mathbb{C} \\ $$$${I}=\Re\int_{−\infty} ^{\:\infty} \frac{{e}^{{iz}} }{{z}^{\mathrm{2}} +\mathrm{1}}\:{dz} \\ $$$${and}\:{define}\:{the}\:{contour}\:{C}\:{as}\:{a}\:{semi}−{circle}\:{in}\:{upper}\:{half}\:{plane} \\ $$$${with}\:{radius}\:{R} \\ $$$$ \\ $$$$\oint_{{C}} {f}\left({z}\right)\:{dz}\:=\:\oint_{\left[\mathrm{0},\:{R}\right]} {f}\left({x}\right)\:{dx}\:+\:\oint_{\left[−{R},\:\mathrm{0}\right]} {f}\left({x}\right)\:{dx}\:+\:\oint_{{Arc}} {f}\left({z}\right)\:{dz} \\ $$$$\Rightarrow\underset{{R}\rightarrow\infty} {\mathrm{lim}}\:\left\{\oint_{\left[\mathrm{0},{R}\right]} {f}\left({x}\right)\:{dx}+\oint_{\left[−\infty,\:\mathrm{0}\right]} {f}\left({x}\right)\:{dx}+\oint_{{Arc}\:} {f}\left({z}\right)\:{dz}\:\right\}=\:{I}\:+\oint_{{Arc}} \frac{{e}^{{iz}} }{\mathrm{1}+{z}^{\mathrm{2}} }\:{dz} \\ $$$${Since}\:\frac{\mathrm{1}}{\mathrm{1}+{z}^{\mathrm{2}} }\:{decreases}\:{faster}\:{than}\:\frac{\mathrm{1}}{\mid{z}\mid},\:{by}\:{Jordan}'{s}\:{Lemma}: \\ $$$$=\:{I}+\:\mathrm{0}={I} \\ $$$$\therefore\:\oint_{{C}} {f}\left({z}\right)\:{dz}\:=\:{I}\:=\:\mathrm{2}\pi{i}\underset{{poles}} {\sum}{Res}\left({f}\left({z}\right),\:{pole}\right) \\ $$$${Note}\:{that}\:{f}\left({z}\right)\:{has}\:{verticals}\:{asymptotes}\:{at}\:{z}=\pm{i} \\ $$$${however},\:{only}\:{z}={i}\:\in\:{C} \\ $$$${I}=\mathrm{2}\pi{i}\:\underset{{z}\rightarrow{i}} {\mathrm{lim}}\left({z}−{i}\right).\frac{{e}^{{iz}} }{\mathrm{1}+{z}^{\mathrm{2}} }\:=\:\mathrm{2}\pi{i}\:\underset{{z}\rightarrow{i}} {\mathrm{lim}}\:\frac{\mathrm{1}}{{z}+{i}}.{e}^{{iz}} \\ $$$$\mathrm{2}\pi{i}.\frac{{e}^{−\mathrm{1}} }{\mathrm{2}{i}}=\frac{\pi}{{e}}\:\therefore\:{I}=\frac{\pi}{{e}} \\ $$$$ \\ $$