Question Number 39165 by math khazana by abdo last updated on 03/Jul/18
$${calculate}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}\:} \:+{t}^{\mathrm{6}} } \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 03/Jul/18
Commented by tanmay.chaudhury50@gmail.com last updated on 03/Jul/18
$$\:\:\:\:\mathrm{1}.\mathrm{5}×\mathrm{1}>\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} +{t}^{\mathrm{6}} }>\mathrm{3}×\left(\mathrm{0}.\mathrm{5}\right)^{\mathrm{2}} \\ $$$$\mathrm{1}.\mathrm{5}>\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} +{t}^{\mathrm{6}} }>\mathrm{0}.\mathrm{75} \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 04/Jul/18
![let I =∫_0 ^∞ (dt/(1+t^4 +t^6 )) 2I = ∫_(−∞) ^(+∞) (dt/(1+t^4 +t^6 )) let ϕ(z)= (1/(1+z^4 +z^8 )) poles of ϕ? 1+z^4 +z^8 =0 ⇔ 1+z^4 +(z^4 )^2 =0 ⇔ ((1−(z^4 )^3 )/(1−z^4 )) =0 and z^4 ≠1 ⇔ z^(12) =1 and z^4 ≠1 z^4 =1 ⇒4θ=2kπ ⇒θ_k =((kπ)/2) and k∈[[0,3]] z_k =e^(i((kπ)/2)) let z=re^(iα) so z^(12) =1 ⇒r=1 and 12α =2kπ ⇒ α_k =((kπ)/6) with k∈[[0,11]]⇒ z_k =e^(i((kπ)/6)) and k∈[[0,5]] z_0 =1 (non pole of ϕ) z_1 =e^(i(π/6)) z_1 ^4 =e^(i((2π)/3)) ≠1 ⇒z_1 pole of ϕ z_2 =e^(i(π/3)) (z_2 ^4 ≠1 ⇒z_2 pole of ϕ) z_3 =e^(i(π/2)) (z_3 ^4 =1 so z_3 non pole of ϕ) z_4 =e^(i((2π)/3)) ( z_4 ^4 ≠1 so z_4 pole of ϕ) z_5 =e^(i ((5π)/6)) (z_5 ^4 ≠1 so z_5 is pole of ϕ) z_6 = e^(iπ) ( z_6 ^4 =1 non pole of ϕ) z_7 =e^(i((7π)/6)) =−e^(i(π/6)) (pole of ϕ) z_8 =e^(i((8π)/6)) = e^(i((4π)/3)) (pole of ϕ) z_9 =e^(i((9π)/6)) =e^(i((3π)/2)) (non pole of ϕ) z_(10) =e^(i((10π)/6)) = e^(i((5π)/3)) (pole ofϕ) z_(11) = e^(i((11π)/6)) =−e^(−i(π/6)) (pole of ϕ) be continued...](https://www.tinkutara.com/question/Q39294.png)
$${let}\:\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} \:+{t}^{\mathrm{6}} } \\ $$$$\mathrm{2}{I}\:\:=\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} \:+{t}^{\mathrm{6}} }\:{let}\:\varphi\left({z}\right)=\:\frac{\mathrm{1}}{\mathrm{1}+{z}^{\mathrm{4}} \:+{z}^{\mathrm{8}} } \\ $$$${poles}\:{of}\:\varphi? \\ $$$$\mathrm{1}+{z}^{\mathrm{4}} \:+{z}^{\mathrm{8}} \:=\mathrm{0}\:\Leftrightarrow\:\mathrm{1}+{z}^{\mathrm{4}} \:+\left({z}^{\mathrm{4}} \right)^{\mathrm{2}} =\mathrm{0}\: \\ $$$$\Leftrightarrow\:\frac{\mathrm{1}−\left({z}^{\mathrm{4}} \right)^{\mathrm{3}} }{\mathrm{1}−{z}^{\mathrm{4}} }\:=\mathrm{0}\:\:{and}\:{z}^{\mathrm{4}} \neq\mathrm{1}\:\Leftrightarrow\:{z}^{\mathrm{12}} \:=\mathrm{1}\:{and}\:{z}^{\mathrm{4}} \neq\mathrm{1} \\ $$$${z}^{\mathrm{4}} =\mathrm{1}\:\Rightarrow\mathrm{4}\theta=\mathrm{2}{k}\pi\:\Rightarrow\theta_{{k}} =\frac{{k}\pi}{\mathrm{2}}\:\:{and}\:{k}\in\left[\left[\mathrm{0},\mathrm{3}\right]\right] \\ $$$${z}_{{k}} ={e}^{{i}\frac{{k}\pi}{\mathrm{2}}} \:\:\:{let}\:\:{z}={re}^{{i}\alpha} \:\:\:{so}\:{z}^{\mathrm{12}} =\mathrm{1}\:\Rightarrow{r}=\mathrm{1}\:{and} \\ $$$$\mathrm{12}\alpha\:=\mathrm{2}{k}\pi\:\Rightarrow\:\alpha_{{k}} =\frac{{k}\pi}{\mathrm{6}}\:\:{with}\:{k}\in\left[\left[\mathrm{0},\mathrm{11}\right]\right]\Rightarrow \\ $$$${z}_{{k}} ={e}^{{i}\frac{{k}\pi}{\mathrm{6}}} \:\:\:{and}\:{k}\in\left[\left[\mathrm{0},\mathrm{5}\right]\right] \\ $$$${z}_{\mathrm{0}} =\mathrm{1}\:\left({non}\:{pole}\:{of}\:\varphi\right) \\ $$$${z}_{\mathrm{1}} ={e}^{{i}\frac{\pi}{\mathrm{6}}} \:\:\:\:{z}_{\mathrm{1}} ^{\mathrm{4}} \:={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:\:\neq\mathrm{1}\:\Rightarrow{z}_{\mathrm{1}} {pole}\:{of}\:\varphi \\ $$$${z}_{\mathrm{2}} ={e}^{{i}\frac{\pi}{\mathrm{3}}} \:\:\:\:\left({z}_{\mathrm{2}} ^{\mathrm{4}} \:\neq\mathrm{1}\:\Rightarrow{z}_{\mathrm{2}} {pole}\:{of}\:\varphi\right) \\ $$$${z}_{\mathrm{3}} ={e}^{{i}\frac{\pi}{\mathrm{2}}} \:\left({z}_{\mathrm{3}} ^{\mathrm{4}} =\mathrm{1}\:{so}\:{z}_{\mathrm{3}} \:{non}\:{pole}\:{of}\:\varphi\right) \\ $$$${z}_{\mathrm{4}} \:={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:\left(\:{z}_{\mathrm{4}} ^{\mathrm{4}} \:\neq\mathrm{1}\:{so}\:{z}_{\mathrm{4}} {pole}\:{of}\:\varphi\right) \\ $$$${z}_{\mathrm{5}} \:={e}^{{i}\:\frac{\mathrm{5}\pi}{\mathrm{6}}} \:\:\:\left({z}_{\mathrm{5}} ^{\mathrm{4}} \neq\mathrm{1}\:{so}\:{z}_{\mathrm{5}} {is}\:{pole}\:{of}\:\varphi\right) \\ $$$${z}_{\mathrm{6}} \:=\:{e}^{{i}\pi} \left(\:{z}_{\mathrm{6}} ^{\mathrm{4}} \:=\mathrm{1}\:{non}\:{pole}\:{of}\:\varphi\right) \\ $$$${z}_{\mathrm{7}} ={e}^{{i}\frac{\mathrm{7}\pi}{\mathrm{6}}} \:=−{e}^{{i}\frac{\pi}{\mathrm{6}}} \:\:\left({pole}\:{of}\:\varphi\right) \\ $$$${z}_{\mathrm{8}} ={e}^{{i}\frac{\mathrm{8}\pi}{\mathrm{6}}} \:=\:{e}^{{i}\frac{\mathrm{4}\pi}{\mathrm{3}}} \:\:\:\left({pole}\:{of}\:\varphi\right) \\ $$$${z}_{\mathrm{9}} ={e}^{{i}\frac{\mathrm{9}\pi}{\mathrm{6}}} \:={e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{2}}} \:\left({non}\:{pole}\:{of}\:\varphi\right) \\ $$$${z}_{\mathrm{10}} ={e}^{{i}\frac{\mathrm{10}\pi}{\mathrm{6}}} \:=\:{e}^{{i}\frac{\mathrm{5}\pi}{\mathrm{3}}} \left({pole}\:{of}\varphi\right) \\ $$$${z}_{\mathrm{11}} =\:{e}^{{i}\frac{\mathrm{11}\pi}{\mathrm{6}}} \:=−{e}^{−{i}\frac{\pi}{\mathrm{6}}} \:\:\left({pole}\:{of}\:\varphi\right)\:\:{be}\:{continued}… \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 04/Jul/18
$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:+{Res}\left(\varphi,{z}_{\mathrm{2}} \right)\:+{Res}\left(\varphi,{z}_{\mathrm{4}} \right)\:\right. \\ $$$$+{Res}\left(\varphi,{z}_{\mathrm{5}} \right)\:+{Res}\left(\varphi,\:{z}_{\mathrm{7}} \right)\:+{Res}\left(\bar {\varphi},{z}_{\mathrm{8}} \right)\:+{Res}\left(\varphi,{z}_{\mathrm{10}} \right) \\ $$$$\left.+{Res}\left(\varphi,{z}_{\mathrm{11}} \right)\:\right\} \\ $$$${Res}\left(\varphi,{z}_{{k}} \right)\:=\:\:\frac{\mathrm{1}}{\mathrm{4}{z}_{{k}} ^{\mathrm{3}} \:+\:\mathrm{6}{z}_{{k}} ^{\mathrm{5}} }\:\Rightarrow \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:\:=\:\frac{\mathrm{1}}{\mathrm{4}\left({e}^{{i}\frac{\pi}{\mathrm{6}}} \right)^{\mathrm{3}} \:+\mathrm{6}\left({e}^{\frac{{i}\pi}{\mathrm{6}}} \right)^{\mathrm{5}} } \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}{i}\:+\mathrm{6}\:{e}^{\frac{{i}\mathrm{5}\pi}{\mathrm{6}}} } \\ $$$${Res}\left(\varphi,{z}_{\mathrm{2}} \right)\:=\:\:\:\frac{\mathrm{1}}{\mathrm{4}\left({e}^{\frac{{i}\pi}{\mathrm{3}}} \right)^{\mathrm{3}} \:+\mathrm{6}\:\left({e}^{\frac{{i}\pi}{\mathrm{3}}} \right)^{\mathrm{5}} }\:=\:\frac{\mathrm{1}}{−\mathrm{4}\:+\mathrm{6}\:{e}^{\frac{{i}\mathrm{5}\pi}{\mathrm{3}}} } \\ $$$${Res}\left(\varphi,{z}_{\mathrm{4}} \right)=\:\frac{\mathrm{1}}{\mathrm{4}\left({e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)^{\mathrm{3}} \:+\mathrm{6}\left({e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)^{\mathrm{5}} }\:=\frac{\mathrm{1}}{\mathrm{4}\:+\mathrm{6}\:{e}^{\frac{{i}\mathrm{10}\pi}{\mathrm{3}}} } \\ $$$$=…. \\ $$