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Question Number 129684 by mnjuly1970 last updated on 17/Jan/21

... advanced calculus... prove that: ∫_0 ^( π) cos(tan(x)−cot(x))dx=(π/e^2 )

$$\:\:\:\:\:\:\:\:\:\:\:\:\:...\:{advanced}\:\:{calculus}... \\ $$$$\:\:{prove}\:\:{that}: \\ $$$$\:\:\int_{\mathrm{0}} ^{\:\pi} {cos}\left({tan}\left({x}\right)−{cot}\left({x}\right)\right){dx}=\frac{\pi}{{e}^{\mathrm{2}} } \\ $$$$ \\ $$

Answered by mindispower last updated on 17/Jan/21

=2∫_0 ^(π/2) cos(tg(x)−cot(x))dx =2∫_0 ^∞ ((cos(t−(1/t)))/(1+t^2 ))dt =Re∫_(−∞) ^∞ (e^(i(t−(1/t))) /(1+t^2 )) t=re^(iε) ,ε∈[0,π[ e^(i(re^(iε) −(e^(−iε) /r))) =e^(i(t−(1/t))) , e^i (re^(iε) −(e^(−iε) /r))→0,r→∞ wehn r→0 r→0^+ e^(i(re^(iε) +i((sin(ε)−cos(ε))/r))) →0 ⇒∫_(−∞) ^(+∞) (e^(i(t−(1/t))) /(1+t^2 ))dt=2iπ.(Res((e^(i(t−(1/t))) /(1+t^2 )),t=i) =2iπRe((e^(i(i−(1/i))) /(2i)))=2iπ.((e^(−2) /(2i)))=(π/e^2 ) 2∫_0 ^∞ ((cos(t−(1/t)))/(1+t^2 ))dt=Re((π/e^2 ))=(π/e^2 )

$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}\left({tg}\left({x}\right)−{cot}\left({x}\right)\right){dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{cos}\left({t}−\frac{\mathrm{1}}{{t}}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$={Re}\int_{−\infty} ^{\infty} \frac{{e}^{{i}\left({t}−\frac{\mathrm{1}}{{t}}\right)} }{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${t}={re}^{{i}\epsilon} ,\epsilon\in\left[\mathrm{0},\pi\left[\right.\right. \\ $$$${e}^{{i}\left({re}^{{i}\epsilon} −\frac{{e}^{−{i}\epsilon} }{{r}}\right)} ={e}^{{i}\left({t}−\frac{\mathrm{1}}{{t}}\right)} , \\ $$$${e}^{{i}} \left({re}^{{i}\epsilon} −\frac{{e}^{−{i}\epsilon} }{{r}}\right)\rightarrow\mathrm{0},{r}\rightarrow\infty \\ $$$${wehn}\:{r}\rightarrow\mathrm{0} \\ $$$${r}\rightarrow\mathrm{0}^{+} \\ $$$${e}^{{i}\left({re}^{{i}\epsilon} +{i}\frac{{sin}\left(\epsilon\right)−{cos}\left(\epsilon\right)}{{r}}\right)} \rightarrow\mathrm{0} \\ $$$$\Rightarrow\int_{−\infty} ^{+\infty} \frac{{e}^{{i}\left({t}−\frac{\mathrm{1}}{{t}}\right)} }{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\mathrm{2}{i}\pi.\left({Res}\left(\frac{{e}^{{i}\left({t}−\frac{\mathrm{1}}{{t}}\right)} }{\mathrm{1}+{t}^{\mathrm{2}} },{t}={i}\right)\right. \\ $$$$=\mathrm{2}{i}\pi{Re}\left(\frac{{e}^{{i}\left({i}−\frac{\mathrm{1}}{{i}}\right)} }{\mathrm{2}{i}}\right)=\mathrm{2}{i}\pi.\left(\frac{{e}^{−\mathrm{2}} }{\mathrm{2}{i}}\right)=\frac{\pi}{{e}^{\mathrm{2}} } \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{cos}\left({t}−\frac{\mathrm{1}}{{t}}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}={Re}\left(\frac{\pi}{{e}^{\mathrm{2}} }\right)=\frac{\pi}{{e}^{\mathrm{2}} } \\ $$$$ \\ $$$$ \\ $$

Answered by mathmax by abdo last updated on 17/Jan/21

I=∫_0 ^(π/2) cos(tanx−(1/(tanx)))dx +∫_(π/2) ^π cos(tanx −(1/(tanx)))dx(→x=(π/2)+t) =2∫_0 ^(π/2) cos(tanx−(1/(tanx)))dx =_(tanx=t) 2∫_0 ^∞ cos(t−(1/t))(dt/(t^2 +1)) =∫_(−∞) ^(+∞) ((cos(t−(1/t)))/(t^2 +1))dt =Re(∫_(−∞) ^(+∞) (e^(i(t−(1/t))) /(t^2 +1))dt) let ϕ(z)=(e^(i(z−(1/z))) /(z^2 +1)) ⇒ϕ(z)=(e^(i(z−(1/z))) /((z−i)(z+i))) residus[theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) =2iπ×(e^(i(i−(1/i))) /(2i)) =π e^(−1−1) =(π/e^2 ) ⇒ ★ I =(π/e^2 )★

$$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}\left(\mathrm{tanx}−\frac{\mathrm{1}}{\mathrm{tanx}}\right)\mathrm{dx}\:+\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \:\mathrm{cos}\left(\mathrm{tanx}\:−\frac{\mathrm{1}}{\mathrm{tanx}}\right)\mathrm{dx}\left(\rightarrow\mathrm{x}=\frac{\pi}{\mathrm{2}}+\mathrm{t}\right) \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\left(\mathrm{tanx}−\frac{\mathrm{1}}{\mathrm{tanx}}\right)\mathrm{dx}\:=_{\mathrm{tanx}=\mathrm{t}} \:\:\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\mathrm{cos}\left(\mathrm{t}−\frac{\mathrm{1}}{\mathrm{t}}\right)\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cos}\left(\mathrm{t}−\frac{\mathrm{1}}{\mathrm{t}}\right)}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dt}\:=\mathrm{Re}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{e}^{\mathrm{i}\left(\mathrm{t}−\frac{\mathrm{1}}{\mathrm{t}}\right)} }{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dt}\right)\:\mathrm{let} \\ $$$$\varphi\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{i}\left(\mathrm{z}−\frac{\mathrm{1}}{\mathrm{z}}\right)} }{\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow\varphi\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{i}\left(\mathrm{z}−\frac{\mathrm{1}}{\mathrm{z}}\right)} }{\left(\mathrm{z}−\mathrm{i}\right)\left(\mathrm{z}+\mathrm{i}\right)}\:\mathrm{residus}\left[\mathrm{theorem}\:\mathrm{give}\right. \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\varphi,\mathrm{i}\right)\:=\mathrm{2i}\pi×\frac{\mathrm{e}^{\mathrm{i}\left(\mathrm{i}−\frac{\mathrm{1}}{\mathrm{i}}\right)} }{\mathrm{2i}}\:=\pi\:\mathrm{e}^{−\mathrm{1}−\mathrm{1}} \:=\frac{\pi}{\mathrm{e}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\bigstar\:\mathrm{I}\:=\frac{\pi}{\mathrm{e}^{\mathrm{2}} }\bigstar \\ $$

Answered by mnjuly1970 last updated on 20/Jan/21

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