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Question-111104

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Question Number 111104 by Lordose last updated on 02/Sep/20
Commented by mathdave last updated on 02/Sep/20
the question is not integradable
$${the}\:{question}\:{is}\:{not}\:{integradable} \\ $$
Commented by Dwaipayan Shikari last updated on 02/Sep/20
∫(−1)^x dx=∫e^(xlog(−1)) dx=∫e^(iπx) dx=(1/(iπ))e^(iπx) +C
$$\int\left(−\mathrm{1}\right)^{{x}} {dx}=\int{e}^{{xlog}\left(−\mathrm{1}\right)} {dx}=\int{e}^{{i}\pi{x}} {dx}=\frac{\mathrm{1}}{{i}\pi}{e}^{{i}\pi{x}} +{C} \\ $$
Answered by abdomsup last updated on 02/Sep/20
this integral can be solved...
$${this}\:{integral}\:{can}\:{be}\:{solved}… \\ $$
Commented by mathdave last updated on 02/Sep/20
then solve it since it is integradable
$${then}\:{solve}\:{it}\:{since}\:{it}\:{is}\:{integradable}\: \\ $$
Answered by mathmax by abdo last updated on 03/Sep/20
I =∫_(−∞) ^(+∞) (((−1)^x^2 )/((x^2 +x+1)^2 )) dx ⇒ I =∫_(−∞) ^(+∞) (e^(iπx^2 ) /((x^2 +x+1)^2 ))dx let ϕ(z) =(e^(iπz^2 ) /((z^2 +z+1)^2 )) poles of ϕ? z^2 +z +1 =0→Δ =1−4 =−3 ⇒z_1 =((−1+i(√3))/2) =e^((i2π)/3) z_2 =((−1−i(√3))/2) =e^(−((i2π)/3)) ⇒ϕ(z) =(e^(iπz^2 ) /((z−e^((i2π)/3) )^2 (z−e^(−((i2π)/3)) )^2 )) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,e^((i2π)/3) ) Res(ϕ ,e^((i2π)/3) ) =lim_(z→e^((i2π)/3) ) (1/((2−1)!)){ (z−e^((i2π)/3) )^2 ϕ(z)}^((1)) =lim_(z→e^((i2π)/3) ) {(e^(iπz^2 ) /((z−e^(−((i2π)/3)) )^2 ))}^((1)) =lim_(z→e^((i2π)/3) ) ((2iπze^(iπz^2 ) (z−e^(−((i2π)/3)) )^2 −2(z−e^(−((i2π)/3)) )e^(iπz^2 ) )/((z−e^(−((i2π)/3)) )^4 )) =lim_(z→e^((i2π)/3) ) ((2iπz e^(iπz^2 ) (z−e^(−((i2π)/3)) )−2e^(iπz^2 ) )/((z−e^(−((i2π)/3)) )^3 )) =lim_(z→e^((i2π)/3) ) ((2e^(iπz^2 ) {iπz (z−e^(−((i2π)/3)) )−1})/((z−e^(−((i2π)/3)) )^3 )) =((2e^(iπe^(i((4π)/3)) ) {iπ e^((i2π)/3) (2isin(((2π)/3))−1})/((2isin(((2π)/3)))^3 )) =((2e^(iπ{−e^((iπ)/3) }) {−π(√3)e^((i2π)/3) −1})/(−8i(((√3)/2))^3 )) =2×((e^(−iπ{(1/2)+((i(√3))/2)}) {π(√3)(−(1/2)+((i(√3))/2)+1})/(3i(√3))) =(2/(3i(√3))) × (−i)e^((π(√3))/2) {((π(√3))/2) +((3π)/2)i} =−(1/(3(√3))) e^((π(√3))/2) {π(√3)+3iπ} ⇒ ∫_(−∞) ^(+∞ ) ϕ(z)dz =2iπ(−(1/(3(√3))))e^((π(√3))/2) {π(√3)+3iπ} =I
$$\mathrm{I}\:=\int_{−\infty} ^{+\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{x}^{\mathrm{2}} } }{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{dx}\:\Rightarrow\:\mathrm{I}\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{e}^{\mathrm{i}\pi\mathrm{x}^{\mathrm{2}} } }{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx}\:\mathrm{let} \\ $$$$\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{e}^{\mathrm{i}\pi\mathrm{z}^{\mathrm{2}} } }{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{z}+\mathrm{1}\right)^{\mathrm{2}} }\:\:\mathrm{poles}\:\mathrm{of}\:\varphi? \\ $$$$\mathrm{z}^{\mathrm{2}} \:+\mathrm{z}\:+\mathrm{1}\:=\mathrm{0}\rightarrow\Delta\:=\mathrm{1}−\mathrm{4}\:=−\mathrm{3}\:\Rightarrow\mathrm{z}_{\mathrm{1}} =\frac{−\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \\ $$$$\mathrm{z}_{\mathrm{2}} =\frac{−\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\mathrm{e}^{−\frac{\mathrm{i2}\pi}{\mathrm{3}}} \:\:\:\:\Rightarrow\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{e}^{\mathrm{i}\pi\mathrm{z}^{\mathrm{2}} } }{\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} } \\ $$$$\mathrm{residus}\:\mathrm{theorem}\:\mathrm{give}\:\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\varphi,\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right) \\ $$$$\mathrm{Res}\left(\varphi\:,\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} } \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\:\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \varphi\left(\mathrm{z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} } \:\:\left\{\frac{\mathrm{e}^{\mathrm{i}\pi\mathrm{z}^{\mathrm{2}} } }{\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} } \:\:\:\:\frac{\mathrm{2i}\pi\mathrm{ze}^{\mathrm{i}\pi\mathrm{z}^{\mathrm{2}} } \left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)\mathrm{e}^{\mathrm{i}\pi\mathrm{z}^{\mathrm{2}} } }{\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)^{\mathrm{4}} } \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} } \:\:\:\:\frac{\mathrm{2i}\pi\mathrm{z}\:\mathrm{e}^{\mathrm{i}\pi\mathrm{z}^{\mathrm{2}} } \left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)−\mathrm{2e}^{\mathrm{i}\pi\mathrm{z}^{\mathrm{2}} } }{\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)^{\mathrm{3}} } \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} } \:\:\:\:\frac{\mathrm{2e}^{\mathrm{i}\pi\mathrm{z}^{\mathrm{2}} } \left\{\mathrm{i}\pi\mathrm{z}\:\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)−\mathrm{1}\right\}}{\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{2e}^{\mathrm{i}\pi\mathrm{e}^{\mathrm{i}\frac{\mathrm{4}\pi}{\mathrm{3}}} } \left\{\mathrm{i}\pi\:\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \left(\mathrm{2isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)−\mathrm{1}\right\}\right.}{\left(\mathrm{2isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)^{\mathrm{3}} }\:=\frac{\mathrm{2e}^{\mathrm{i}\pi\left\{−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right\}} \left\{−\pi\sqrt{\mathrm{3}}\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} −\mathrm{1}\right\}}{−\mathrm{8i}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{3}} } \\ $$$$=\mathrm{2}×\frac{\mathrm{e}^{−\mathrm{i}\pi\left\{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right\}} \left\{\pi\sqrt{\mathrm{3}}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}+\mathrm{1}\right\}\right.}{\mathrm{3i}\sqrt{\mathrm{3}}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3i}\sqrt{\mathrm{3}}}\:×\:\left(−\mathrm{i}\right)\mathrm{e}^{\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{2}}} \left\{\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:+\frac{\mathrm{3}\pi}{\mathrm{2}}\mathrm{i}\right\}\:=−\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}}\:\mathrm{e}^{\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{2}}} \left\{\pi\sqrt{\mathrm{3}}+\mathrm{3i}\pi\right\}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty\:} \varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left(−\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}}\right)\mathrm{e}^{\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{2}}} \left\{\pi\sqrt{\mathrm{3}}+\mathrm{3i}\pi\right\}\:=\mathrm{I} \\ $$

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