Question-148064

Question Number 148064 by tabata last updated on 25/Jul/21

Commented by tabata last updated on 25/Jul/21

$${help}\:{me}\:{sir}\:{in}\:{complex}\:{number} \\ $$

Answered by Olaf_Thorendsen last updated on 25/Jul/21

Q6. I = ∫_0 ^π (dθ/(2−cosθ)) Let t = tan(θ/2) I = ∫_0 ^∞ (1/(2−((1−t^2 )/(1+t^2 )))).((2dt)/(1+t^2 )) I = 2∫_0 ^∞ (dt/(2(1+t^2 )−(1−t^2 ))) I = 2∫_0 ^∞ (dt/(3t^2 +1)) = (2/( (√3)))∫_0 ^∞ (((√3)dt)/(3t^2 +1)) I = (2/( (√3)))[arctan((√3)t)]_0 ^∞ = (π/( (√3)))

$$\mathrm{Q6}. \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\pi} \frac{{d}\theta}{\mathrm{2}−\mathrm{cos}\theta} \\ $$$$\mathrm{Let}\:{t}\:=\:\mathrm{tan}\frac{\theta}{\mathrm{2}} \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{2}−\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}.\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\mathrm{I}\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\mathrm{2}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)−\left(\mathrm{1}−{t}^{\mathrm{2}} \right)} \\ $$$$\mathrm{I}\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\mathrm{3}{t}^{\mathrm{2}} +\mathrm{1}}\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\int_{\mathrm{0}} ^{\infty} \frac{\sqrt{\mathrm{3}}{dt}}{\mathrm{3}{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left[\mathrm{arctan}\left(\sqrt{\mathrm{3}}{t}\right)\right]_{\mathrm{0}} ^{\infty} \:=\:\frac{\pi}{\:\sqrt{\mathrm{3}}} \\ $$

Answered by Olaf_Thorendsen last updated on 25/Jul/21

J = ∫_0 ^∞ ((cos(2x))/(x^2 +1)) dx = (1/2)∫_(−∞) ^(+∞) ((cos(2x))/(x^2 +1)) dx J = (1/2)∫_(−∞) ^(+∞) (e^(2ix) /(x^2 +1)) dx J = (1/2).( 2iπ.Res((e^(2ix) /(x^2 +1)),+i)) J = iπ.lim_(x→+i) ((e^(2ix) /(x+i))) = iπ.(e^(−2) /(2i)) = (π/(2e^2 ))

$$\mathrm{J}\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{cos}\left(\mathrm{2}{x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{\mathrm{cos}\left(\mathrm{2}{x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx} \\ $$$$\mathrm{J}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{{e}^{\mathrm{2}{ix}} }{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx} \\ $$$$\mathrm{J}\:=\:\frac{\mathrm{1}}{\mathrm{2}}.\left(\:\mathrm{2}{i}\pi.\mathrm{Res}\left(\frac{{e}^{\mathrm{2}{ix}} }{{x}^{\mathrm{2}} +\mathrm{1}},+{i}\right)\right) \\ $$$$\mathrm{J}\:=\:{i}\pi.\underset{{x}\rightarrow+{i}} {\mathrm{lim}}\left(\frac{{e}^{\mathrm{2}{ix}} }{{x}+{i}}\right)\:=\:{i}\pi.\frac{{e}^{−\mathrm{2}} }{\mathrm{2}{i}}\:=\:\frac{\pi}{\mathrm{2}{e}^{\mathrm{2}} } \\ $$

Answered by qaz last updated on 25/Jul/21

∫_0 ^∞ ((L{cos (ax)})/(x^2 +1))dx=∫_0 ^∞ (s/((x^2 +1)(x^2 +s^2 )))dx=(π/(2(s+1))) ∫_0 ^∞ ((cos (2x))/(x^2 +1))dx=L^(−1) {(π/(2(s+1)))}(a=2)=(π/(2e^2 ))

$$\int_{\mathrm{0}} ^{\infty} \frac{\mathscr{L}\left\{\mathrm{cos}\:\left(\mathrm{ax}\right)\right\}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{s}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{s}^{\mathrm{2}} \right)}\mathrm{dx}=\frac{\pi}{\mathrm{2}\left(\mathrm{s}+\mathrm{1}\right)} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{cos}\:\left(\mathrm{2x}\right)}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx}=\mathscr{L}^{−\mathrm{1}} \left\{\frac{\pi}{\mathrm{2}\left(\mathrm{s}+\mathrm{1}\right)}\right\}\left(\mathrm{a}=\mathrm{2}\right)=\frac{\pi}{\mathrm{2e}^{\mathrm{2}} } \\ $$

Answered by mathmax by abdo last updated on 25/Jul/21

I=∫_0 ^∞ ((cos(2x))/(x^2 +1))dx ⇒2I=∫_(−∞) ^(+∞) (e^(2ix) /(x^2 +1))dx let ϕ(z)=(e^(2iz) /(z^2 +1)) ϕ(z)=(e^(2iz) /((z−i)(z+i))) ⇒∫_(−∞) ^(+∞) ϕ(z)dz=2iπRes(ϕ,i) =2iπ×(e^(2i(i)) /(2i)) =πe^(−2) ⇒I =(π/(2e^2 ))★

$$\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\:\Rightarrow\mathrm{2I}=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{e}^{\mathrm{2ix}} }{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\:\mathrm{let}\:\varphi\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{2iz}} }{\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\varphi\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{2iz}} }{\left(\mathrm{z}−\mathrm{i}\right)\left(\mathrm{z}+\mathrm{i}\right)}\:\Rightarrow\int_{−\infty} ^{+\infty} \varphi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi\mathrm{Res}\left(\varphi,\mathrm{i}\right) \\ $$$$=\mathrm{2i}\pi×\frac{\mathrm{e}^{\mathrm{2i}\left(\mathrm{i}\right)} }{\mathrm{2i}}\:=\pi\mathrm{e}^{−\mathrm{2}} \:\Rightarrow\mathrm{I}\:=\frac{\pi}{\mathrm{2e}^{\mathrm{2}} }\bigstar \\ $$

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