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)
$$ \\ $$$${Observe}\:{that}:\:\frac{{cos}\:{x}\:}{{x}^{\mathrm{2}} +\mathrm{1}}\:{is}\:{an}\:{even}\:{function} \\ $$$$\therefore\:\int_{−\infty} ^{\:\mathrm{0}} \frac{{cosx}}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx}\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\:{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx}=\frac{{I}}{\mathrm{2}} \\ $$$${let}\:{f}\left({z}\right):=\:\frac{{e}^{{iz}} }{{z}^{\mathrm{2}} +\mathrm{1}}\:{where}\:{z}\in\mathbb{C} \\ $$$${I}=\Re\int_{−\infty} ^{\:\infty} \frac{{e}^{{iz}} }{{z}^{\mathrm{2}} +\mathrm{1}}\:{dz} \\ $$$${and}\:{define}\:{the}\:{contour}\:{C}\:{as}\:{a}\:{semi}−{circle}\:{in}\:{upper}\:{half}\:{plane} \\ $$$${with}\:{radius}\:{R} \\ $$$$ \\ $$$$\oint_{{C}} {f}\left({z}\right)\:{dz}\:=\:\oint_{\left[\mathrm{0},\:{R}\right]} {f}\left({x}\right)\:{dx}\:+\:\oint_{\left[−{R},\:\mathrm{0}\right]} {f}\left({x}\right)\:{dx}\:+\:\oint_{{Arc}} {f}\left({z}\right)\:{dz} \\ $$$$\Rightarrow\underset{{R}\rightarrow\infty} {\mathrm{lim}}\:\left\{\oint_{\left[\mathrm{0},{R}\right]} {f}\left({x}\right)\:{dx}+\oint_{\left[−\infty,\:\mathrm{0}\right]} {f}\left({x}\right)\:{dx}+\oint_{{Arc}\:} {f}\left({z}\right)\:{dz}\:\right\}=\:{I}\:+\oint_{{Arc}} \frac{{e}^{{iz}} }{\mathrm{1}+{z}^{\mathrm{2}} }\:{dz} \\ $$$${Since}\:\frac{\mathrm{1}}{\mathrm{1}+{z}^{\mathrm{2}} }\:{decreases}\:{faster}\:{than}\:\frac{\mathrm{1}}{\mid{z}\mid},\:{by}\:{Jordan}'{s}\:{Lemma}: \\ $$$$=\:{I}+\:\mathrm{0}={I} \\ $$$$\therefore\:\oint_{{C}} {f}\left({z}\right)\:{dz}\:=\:{I}\:=\:\mathrm{2}\pi{i}\underset{{poles}} {\sum}{Res}\left({f}\left({z}\right),\:{pole}\right) \\ $$$${Note}\:{that}\:{f}\left({z}\right)\:{has}\:{verticals}\:{asymptotes}\:{at}\:{z}=\pm{i} \\ $$$${however},\:{only}\:{z}={i}\:\in\:{C} \\ $$$${I}=\mathrm{2}\pi{i}\:\underset{{z}\rightarrow{i}} {\mathrm{lim}}\left({z}−{i}\right).\frac{{e}^{{iz}} }{\mathrm{1}+{z}^{\mathrm{2}} }\:=\:\mathrm{2}\pi{i}\:\underset{{z}\rightarrow{i}} {\mathrm{lim}}\:\frac{\mathrm{1}}{{z}+{i}}.{e}^{{iz}} \\ $$$$\mathrm{2}\pi{i}.\frac{{e}^{−\mathrm{1}} }{\mathrm{2}{i}}=\frac{\pi}{{e}}\:\therefore\:{I}=\frac{\pi}{{e}} \\ $$$$ \\ $$
Answered by breniam last updated on 12/Jan/26
![I(t)=∫_(−∞) ^∞ ((cos(xt))/(x^2 +1))dx=∫_(−∞) ^∞ [((cos(tx))/(x^2 +1))−((cos(0x))/(x^2 +1))]dx+∫_(−∞) ^∞ (1/(x^2 +1))dx= π+∫_(−∞) ^∞ [((cos(xy))/(x^2 +1))]_(y→0^+ ) ^(y→t^− ) dx=π−∫_(−∞) ^∞ ∫_0 ^t ((xsin(xy))/(x^2 +1))dydx= π−∫_0 ^t ∫_(−∞) ^∞ ((xsin(xy))/(x^2 +1))dxdy=π−∫_0 ^t ∫_(−∞) ^∞ ((x^2 sin(xy))/(x(x^2 +1)))dxdy= π−∫_0 ^t ∫_(−∞) ^∞ ((sin(xy))/x)dxdy+∫_0 ^t ∫_(−∞) ^∞ ((sin(xy))/(x(x^2 +1)))dxdy ∫_0 ^t ∫_(−∞) ^∞ ((sin(xy))/x)dxdy={x^− =xy}= ∫_0 ^t ∫_(−∞) ^∞ ((sin(x))/x)dxdy=∫_0 ^t ∫_(−∞) ^0 ((sin(x))/x)dxdy+∫_0 ^t ∫_0 ^∞ ((sin(x))/x)dxdy= ={x^− =−x}=2∫_0 ^t ∫_0 ^∞ ((sin(x))/x)dxdy=π∫_0 ^t dy=πt ∫_0 ^t ∫_(−∞) ^∞ ((sin(xy))/(x(x^2 +1)))dxdy=∫_0 ^t ∫_(−∞) ^∞ [((sin(xy))/(x(x^2 +1)))−((sin(x0))/(x(x^2 +1)))]dxdy= ∫_0 ^t ∫_(−∞) ^∞ [((sin(xz))/(x(x^2 +1)))]_(z→0^+ ) ^(z→y^− ) dxdy= ∫_0 ^t ∫_(−∞) ^∞ ∫_0 ^y ((cos(xz))/(x^2 +1))dzdxdy=∫_0 ^t ∫_0 ^y ∫_(−∞) ^∞ ((cos(xz))/(x^2 +1))dxdzdy ∫_(−∞) ^∞ ((cos(xt))/(x^2 +1))dx=π−πt+∫_0 ^t ∫_0 ^y ∫_(−∞) ^∞ ((cos(xz))/(x^2 +1))dxdzdy J(t)=∫_0 ^t ∫_0 ^y ∫_(−∞) ^∞ ((cos(xz))/(x^2 +1))dxdzdy J′′(t)−J(t)=−πt+π J′′(t)−J′(t)+J′(t)−J(t)=[J′(t)−J(t)]′+J′(t)−J(t)=−πt+π K(t)=J′(t)−J(t) K′(t)+K(t)=−πt+π L(t)=e^t K(t) K(t)=e^(−t) L(t) K′(t)=−e^(−t) L(t)+e^(−t) L′(t) e^(−t) L′(t)=−πt+π L′(t)=π(e^t −te^t ) ∫L′(t)dt=L(t)=π∫(e^t −te^t )dt=π(e^t −te^t +e^t )+A=π(2e^t −te^t )+A K(t)=π(2−t)+Ae^(−t) =J′(t)−J(t) M(t)=e^(−t) J(t) J(t)=e^t M(t) J′(t)=e^t M(t)+e^t M′(t) e^t M′(t)=π(2−t)+Ae^(−t) M′(t)=π(2e^(−t) −te^(−t) )+Ae^(−2t) ∫M′(t)dt=M(t)=π∫(2e^(−t) −te^(−t) )dt+Ae^(−2t) =π(−2e^(−t) +te^(−t) +e^(−t) )+Ae^(−2t) +B =π(te^(−t) −e^(−t) )+Ae^(−2t) +B=M(t) J(t)=π(t−1)+Ae^(−t) +Be^t =∫_0 ^t ∫_0 ^y ∫_(−∞) ^∞ ((cos(xz))/(x^2 +1))dxdzdy J(0)=0⇒−π+A+B=0 J′(t)=∫_0 ^t ∫_(−∞) ^∞ ((cos(xy))/(x^2 +1))dxdy=π−Ae^(−t) +Be^t J′(0)=0⇒π−A+B=0 A=π B=0 J′(t)=∫_0 ^t ∫_(−∞) ^∞ ((cos(xy))/(x^2 +1))dxdy=π−πe^(−t) J′′(t)=∫_(−∞) ^∞ ((cos(xt))/(x^2 +1))dx=πe^(−t) J′′(1)=∫_(−∞) ^∞ ((cos(x))/(x^2 +1))dx=(π/e)](https://www.tinkutara.com/question/Q227287.png)
$$ \\ $$$$ \\ $$$${I}\left({t}\right)=\underset{−\infty} {\overset{\infty} {\int}}\frac{\mathrm{cos}\left({xt}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{x}=\underset{−\infty} {\overset{\infty} {\int}}\left[\frac{\mathrm{cos}\left({tx}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{cos}\left(\mathrm{0}{x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\right]\mathrm{d}{x}+\underset{−\infty} {\overset{\infty} {\int}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{x}= \\ $$$$\pi+\underset{−\infty} {\overset{\infty} {\int}}\left[\frac{\mathrm{cos}\left({xy}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\right]_{{y}\rightarrow\mathrm{0}^{+} } ^{{y}\rightarrow{t}^{−} } \mathrm{d}{x}=\pi−\underset{−\infty} {\overset{\infty} {\int}}\underset{\mathrm{0}} {\overset{{t}} {\int}}\frac{{x}\mathrm{sin}\left({xy}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{y}\mathrm{d}{x}= \\ $$$$\pi−\underset{\mathrm{0}} {\overset{{t}} {\int}}\underset{−\infty} {\overset{\infty} {\int}}\frac{{x}\mathrm{sin}\left({xy}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{x}\mathrm{d}{y}=\pi−\underset{\mathrm{0}} {\overset{{t}} {\int}}\underset{−\infty} {\overset{\infty} {\int}}\frac{{x}^{\mathrm{2}} \mathrm{sin}\left({xy}\right)}{{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{d}{x}\mathrm{d}{y}= \\ $$$$\pi−\underset{\mathrm{0}} {\overset{{t}} {\int}}\underset{−\infty} {\overset{\infty} {\int}}\frac{\mathrm{sin}\left({xy}\right)}{{x}}\mathrm{d}{x}\mathrm{d}{y}+\underset{\mathrm{0}} {\overset{{t}} {\int}}\underset{−\infty} {\overset{\infty} {\int}}\frac{\mathrm{sin}\left({xy}\right)}{{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{d}{x}\mathrm{d}{y} \\ $$$$\underset{\mathrm{0}} {\overset{{t}} {\int}}\underset{−\infty} {\overset{\infty} {\int}}\frac{\mathrm{sin}\left({xy}\right)}{{x}}\mathrm{d}{x}\mathrm{d}{y}=\left\{\overset{−} {{x}}={xy}\right\}= \\ $$$$\underset{\mathrm{0}} {\overset{{t}} {\int}}\underset{−\infty} {\overset{\infty} {\int}}\frac{\mathrm{sin}\left({x}\right)}{{x}}\mathrm{d}{x}\mathrm{d}{y}=\underset{\mathrm{0}} {\overset{{t}} {\int}}\underset{−\infty} {\overset{\mathrm{0}} {\int}}\frac{\mathrm{sin}\left({x}\right)}{{x}}\mathrm{d}{x}\mathrm{d}{y}+\underset{\mathrm{0}} {\overset{{t}} {\int}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{sin}\left({x}\right)}{{x}}\mathrm{d}{x}\mathrm{d}{y}= \\ $$$$=\left\{\overset{−} {{x}}=−{x}\right\}=\mathrm{2}\underset{\mathrm{0}} {\overset{{t}} {\int}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{sin}\left({x}\right)}{{x}}\mathrm{d}{x}\mathrm{d}{y}=\pi\underset{\mathrm{0}} {\overset{{t}} {\int}}\mathrm{d}{y}=\pi{t} \\ $$$$ \\ $$$$\underset{\mathrm{0}} {\overset{{t}} {\int}}\underset{−\infty} {\overset{\infty} {\int}}\frac{\mathrm{sin}\left({xy}\right)}{{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{d}{x}\mathrm{d}{y}=\underset{\mathrm{0}} {\overset{{t}} {\int}}\underset{−\infty} {\overset{\infty} {\int}}\left[\frac{\mathrm{sin}\left({xy}\right)}{{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}−\frac{\mathrm{sin}\left({x}\mathrm{0}\right)}{{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}\right]\mathrm{d}{x}\mathrm{d}{y}= \\ $$$$\underset{\mathrm{0}} {\overset{{t}} {\int}}\underset{−\infty} {\overset{\infty} {\int}}\left[\frac{\mathrm{sin}\left({xz}\right)}{{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}\right]_{{z}\rightarrow\mathrm{0}^{+} } ^{{z}\rightarrow{y}^{−} } \mathrm{d}{x}\mathrm{d}{y}= \\ $$$$\underset{\mathrm{0}} {\overset{{t}} {\int}}\underset{−\infty} {\overset{\infty} {\int}}\underset{\mathrm{0}} {\overset{{y}} {\int}}\frac{\mathrm{cos}\left({xz}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z}\mathrm{d}{x}\mathrm{d}{y}=\underset{\mathrm{0}} {\overset{{t}} {\int}}\underset{\mathrm{0}} {\overset{{y}} {\int}}\underset{−\infty} {\overset{\infty} {\int}}\frac{\mathrm{cos}\left({xz}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{x}\mathrm{d}{z}\mathrm{d}{y} \\ $$$$\underset{−\infty} {\overset{\infty} {\int}}\frac{\mathrm{cos}\left({xt}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{x}=\pi−\pi{t}+\underset{\mathrm{0}} {\overset{{t}} {\int}}\underset{\mathrm{0}} {\overset{{y}} {\int}}\underset{−\infty} {\overset{\infty} {\int}}\frac{\mathrm{cos}\left({xz}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{x}\mathrm{d}{z}\mathrm{d}{y} \\ $$$${J}\left({t}\right)=\underset{\mathrm{0}} {\overset{{t}} {\int}}\underset{\mathrm{0}} {\overset{{y}} {\int}}\underset{−\infty} {\overset{\infty} {\int}}\frac{\mathrm{cos}\left({xz}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{x}\mathrm{d}{z}\mathrm{d}{y} \\ $$$${J}''\left({t}\right)−{J}\left({t}\right)=−\pi{t}+\pi \\ $$$${J}''\left({t}\right)−{J}'\left({t}\right)+{J}'\left({t}\right)−{J}\left({t}\right)=\left[{J}'\left({t}\right)−{J}\left({t}\right)\right]'+{J}'\left({t}\right)−{J}\left({t}\right)=−\pi{t}+\pi \\ $$$${K}\left({t}\right)={J}'\left({t}\right)−{J}\left({t}\right) \\ $$$${K}'\left({t}\right)+{K}\left({t}\right)=−\pi{t}+\pi \\ $$$${L}\left({t}\right)={e}^{{t}} {K}\left({t}\right) \\ $$$${K}\left({t}\right)={e}^{−{t}} {L}\left({t}\right) \\ $$$${K}'\left({t}\right)=−{e}^{−{t}} {L}\left({t}\right)+{e}^{−{t}} {L}'\left({t}\right) \\ $$$${e}^{−{t}} {L}'\left({t}\right)=−\pi{t}+\pi \\ $$$${L}'\left({t}\right)=\pi\left({e}^{{t}} −{te}^{{t}} \right) \\ $$$$\int{L}'\left({t}\right)\mathrm{d}{t}={L}\left({t}\right)=\pi\int\left({e}^{{t}} −{te}^{{t}} \right)\mathrm{d}{t}=\pi\left({e}^{{t}} −{te}^{{t}} +{e}^{{t}} \right)+{A}=\pi\left(\mathrm{2}{e}^{{t}} −{te}^{{t}} \right)+{A} \\ $$$${K}\left({t}\right)=\pi\left(\mathrm{2}−{t}\right)+{Ae}^{−{t}} ={J}'\left({t}\right)−{J}\left({t}\right) \\ $$$${M}\left({t}\right)={e}^{−{t}} {J}\left({t}\right) \\ $$$${J}\left({t}\right)={e}^{{t}} {M}\left({t}\right) \\ $$$${J}'\left({t}\right)={e}^{{t}} {M}\left({t}\right)+{e}^{{t}} {M}'\left({t}\right) \\ $$$${e}^{{t}} {M}'\left({t}\right)=\pi\left(\mathrm{2}−{t}\right)+{Ae}^{−{t}} \\ $$$${M}'\left({t}\right)=\pi\left(\mathrm{2}{e}^{−{t}} −{te}^{−{t}} \right)+{Ae}^{−\mathrm{2}{t}} \\ $$$$\int{M}'\left({t}\right)\mathrm{d}{t}={M}\left({t}\right)=\pi\int\left(\mathrm{2}{e}^{−{t}} −{te}^{−{t}} \right)\mathrm{d}{t}+{Ae}^{−\mathrm{2}{t}} \\ $$$$=\pi\left(−\mathrm{2}{e}^{−{t}} +{te}^{−{t}} +{e}^{−{t}} \right)+{Ae}^{−\mathrm{2}{t}} +{B} \\ $$$$=\pi\left({te}^{−{t}} −{e}^{−{t}} \right)+{Ae}^{−\mathrm{2}{t}} +{B}={M}\left({t}\right) \\ $$$${J}\left({t}\right)=\pi\left({t}−\mathrm{1}\right)+{Ae}^{−{t}} +{Be}^{{t}} =\underset{\mathrm{0}} {\overset{{t}} {\int}}\underset{\mathrm{0}} {\overset{{y}} {\int}}\underset{−\infty} {\overset{\infty} {\int}}\frac{\mathrm{cos}\left({xz}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{x}\mathrm{d}{z}\mathrm{d}{y} \\ $$$${J}\left(\mathrm{0}\right)=\mathrm{0}\Rightarrow−\pi+{A}+{B}=\mathrm{0} \\ $$$${J}'\left({t}\right)=\underset{\mathrm{0}} {\overset{{t}} {\int}}\underset{−\infty} {\overset{\infty} {\int}}\frac{\mathrm{cos}\left({xy}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{x}\mathrm{d}{y}=\pi−{Ae}^{−{t}} +{Be}^{{t}} \\ $$$${J}'\left(\mathrm{0}\right)=\mathrm{0}\Rightarrow\pi−{A}+{B}=\mathrm{0} \\ $$$${A}=\pi \\ $$$${B}=\mathrm{0} \\ $$$${J}'\left({t}\right)=\underset{\mathrm{0}} {\overset{{t}} {\int}}\underset{−\infty} {\overset{\infty} {\int}}\frac{\mathrm{cos}\left({xy}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{x}\mathrm{d}{y}=\pi−\pi{e}^{−{t}} \\ $$$${J}''\left({t}\right)=\underset{−\infty} {\overset{\infty} {\int}}\frac{\mathrm{cos}\left({xt}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{x}=\pi{e}^{−{t}} \\ $$$${J}''\left(\mathrm{1}\right)=\underset{−\infty} {\overset{\infty} {\int}}\frac{\mathrm{cos}\left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{x}=\frac{\pi}{{e}} \\ $$