Question Number 164991 by mkam last updated on 24/Jan/22
$$\boldsymbol{{Solve}}\:\boldsymbol{{by}}\:\boldsymbol{{resideo}}\:\boldsymbol{{theorem}}\:\int_{−\infty} ^{\:\infty} \:\frac{\boldsymbol{{z}}^{\mathrm{2}} }{\boldsymbol{{z}}^{\mathrm{4}} +\mathrm{1}}\:\boldsymbol{{dz}} \\ $$
Commented by mkam last updated on 24/Jan/22
$$????? \\ $$
Answered by aleks041103 last updated on 26/Jan/22
![Define the region Ω_r :={z∣Im(z)≥0 and ∣z∣≤r}⊂C where r∈R^+ . Therefore the boundary is: ∂Ω_r =Γ_r ∪Λ_r , where Γ_r :={z∣Im(z)>0 and ∣z∣=r} and Λ_r :={z∣z∈R and z∈[−r,r]}. Then: ∮_( ∂Ω_r ) (z^2 /(z^4 +1))dz=∫_Γ_r (z^2 /(z^4 +1))dz + ∫_Λ_r (z^2 /(z^4 +1))dz= =∫_0 ^π ((r^2 e^(2it) )/(r^4 e^(4it) +1))d(re^(it) )+∫_(−r) ^( r) (z^2 /(z^4 +1))dz if r→∞: lim_(r→∞) ∫_Γ_r (z^2 /(z^4 +1))dz=∫_0 ^π (lim_(r→∞) ((ir^3 e^(3it) )/(r^4 e^(4it) +1)))dt=0 ⇒lim_(r→∞) ∮_( ∂Ω_r ) (z^2 /(z^4 +1))dz=∫_(−∞) ^( ∞) (z^2 /(z^4 +1))dz From the residue theorem: lim_(r→∞) ∮_( ∂Ω_r ) (z^2 /(z^4 +1))dz=2πiΣ_j Res((z^2 /(z^4 +1)),z_j ∈Ω_∞ ) Ω_∞ :={z∣Im(z)≥0} ⇒∫_(−∞) ^( ∞) (z^2 /(z^4 +1))dz=2πiΣ_j Res((z^2 /(z^4 +1)),Im(z_j )≥0) The poles of (z^2 /(z^4 +1)) are: z^4 +1=0⇒z=e^(iπ/4) ,e^(3iπ/4) ,e^(5iπ/4) ,e^(7iπ/4) or z=(1/( (√2)))+(1/( (√2)))i,−(1/( (√2)))+(1/( (√2)))i,−(1/( (√2)))−(1/( (√2)))i,(1/( (√2)))−(1/( (√2)))i Only the poles at z=e^(iπ/4) and z=e^(3iπ/4) lie inside the region Ω_∞ . ⇒∫_(−∞) ^( ∞) (z^2 /(z^4 +1))dz=2πi(Res(e^(iπ/4) )+Res(e^(3iπ/4) )) (z^2 /(z^4 +1))=(z^2 /((z−e^(iπ/4) )(z−e^(3iπ/4) )(z−e^(5iπ/4) )(z−e^(7iπ/4) ))) Therefore the poles are simple ⇒Res(e^(iπ/4) )=lim_(z→e^(iπ/4) ) (((z−e^(iπ/4) )z^2 )/(z^4 +1))=^(l′hopital) =lim_(z→e^(iπ/4) ) ((3z^2 −2ze^(iπ/4) )/(4z^3 ))=(1/4)e^(−iπ/4) Res(e^(3iπ/4) )=lim_(z→e^(3iπ/4) ) (((z−e^(3iπ/4) )z^2 )/(z^4 +1))= =lim_(z→e^(3iπ/4) ) ((3z^2 −2ze^(3iπ/4) )/(4z^3 ))=(1/4)e^(−3iπ/4) ∫_(−∞) ^( ∞) (z^2 /(z^4 +1))dz=2πi((1/4)e^(−iπ/4) +(1/4)e^(−3iπ/4) )= =((πi)/2)((1/( (√2)))−(i/( (√2)))+(−(1/( (√2))))−(i/( (√2))))= =((πi)/2)(−((2i)/( (√2))))= =(π/( (√2)))](https://www.tinkutara.com/question/Q165040.png)
$${Define}\:{the}\:{region} \\ $$$$\:\Omega_{{r}} :=\left\{{z}\mid{Im}\left({z}\right)\geqslant\mathrm{0}\:{and}\:\mid{z}\mid\leqslant{r}\right\}\subset\mathbb{C}\: \\ $$$${where}\:{r}\in\mathbb{R}^{+} . \\ $$$${Therefore}\:{the}\:{boundary}\:{is}: \\ $$$$\partial\Omega_{{r}} =\Gamma_{{r}} \cup\Lambda_{{r}} ,\:{where}\: \\ $$$$\Gamma_{{r}} :=\left\{{z}\mid{Im}\left({z}\right)>\mathrm{0}\:{and}\:\mid{z}\mid={r}\right\} \\ $$$${and}\:\Lambda_{{r}} :=\left\{{z}\mid{z}\in\mathbb{R}\:{and}\:{z}\in\left[−{r},{r}\right]\right\}. \\ $$$${Then}: \\ $$$$\underset{\:\partial\Omega_{{r}} } {\oint}\frac{{z}^{\mathrm{2}} }{{z}^{\mathrm{4}} +\mathrm{1}}{dz}=\int_{\Gamma_{{r}} } \frac{{z}^{\mathrm{2}} }{{z}^{\mathrm{4}} +\mathrm{1}}{dz}\:+\:\int_{\Lambda_{{r}} } \frac{{z}^{\mathrm{2}} }{{z}^{\mathrm{4}} +\mathrm{1}}{dz}= \\ $$$$=\int_{\mathrm{0}} ^{\pi} \frac{{r}^{\mathrm{2}} {e}^{\mathrm{2}{it}} }{{r}^{\mathrm{4}} {e}^{\mathrm{4}{it}} +\mathrm{1}}{d}\left({re}^{{it}} \right)+\int_{−{r}} ^{\:{r}} \frac{{z}^{\mathrm{2}} }{{z}^{\mathrm{4}} +\mathrm{1}}{dz} \\ $$$${if}\:{r}\rightarrow\infty: \\ $$$$\underset{{r}\rightarrow\infty} {{lim}}\int_{\Gamma_{{r}} } \frac{{z}^{\mathrm{2}} }{{z}^{\mathrm{4}} +\mathrm{1}}{dz}=\int_{\mathrm{0}} ^{\pi} \left(\underset{{r}\rightarrow\infty} {{lim}}\frac{{ir}^{\mathrm{3}} {e}^{\mathrm{3}{it}} }{{r}^{\mathrm{4}} {e}^{\mathrm{4}{it}} +\mathrm{1}}\right){dt}=\mathrm{0} \\ $$$$\Rightarrow\underset{{r}\rightarrow\infty} {{lim}}\underset{\:\partial\Omega_{{r}} } {\oint}\frac{{z}^{\mathrm{2}} }{{z}^{\mathrm{4}} +\mathrm{1}}{dz}=\int_{−\infty} ^{\:\infty} \frac{{z}^{\mathrm{2}} }{{z}^{\mathrm{4}} +\mathrm{1}}{dz} \\ $$$${From}\:{the}\:{residue}\:{theorem}: \\ $$$$\underset{{r}\rightarrow\infty} {{lim}}\underset{\:\partial\Omega_{{r}} } {\oint}\frac{{z}^{\mathrm{2}} }{{z}^{\mathrm{4}} +\mathrm{1}}{dz}=\mathrm{2}\pi{i}\underset{{j}} {\sum}{Res}\left(\frac{{z}^{\mathrm{2}} }{{z}^{\mathrm{4}} +\mathrm{1}},{z}_{{j}} \in\Omega_{\infty} \right) \\ $$$$\Omega_{\infty} :=\left\{{z}\mid{Im}\left({z}\right)\geqslant\mathrm{0}\right\} \\ $$$$\Rightarrow\int_{−\infty} ^{\:\infty} \frac{{z}^{\mathrm{2}} }{{z}^{\mathrm{4}} +\mathrm{1}}{dz}=\mathrm{2}\pi{i}\underset{{j}} {\sum}{Res}\left(\frac{{z}^{\mathrm{2}} }{{z}^{\mathrm{4}} +\mathrm{1}},{Im}\left({z}_{{j}} \right)\geqslant\mathrm{0}\right) \\ $$$${The}\:{poles}\:{of}\:\frac{{z}^{\mathrm{2}} }{{z}^{\mathrm{4}} +\mathrm{1}}\:{are}: \\ $$$${z}^{\mathrm{4}} +\mathrm{1}=\mathrm{0}\Rightarrow{z}={e}^{{i}\pi/\mathrm{4}} ,{e}^{\mathrm{3}{i}\pi/\mathrm{4}} ,{e}^{\mathrm{5}{i}\pi/\mathrm{4}} ,{e}^{\mathrm{7}{i}\pi/\mathrm{4}} \\ $$$${or}\:{z}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{i},−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{i},−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{i},\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{i} \\ $$$${Only}\:{the}\:{poles}\:{at} \\ $$$${z}={e}^{{i}\pi/\mathrm{4}} \:{and}\:{z}={e}^{\mathrm{3}{i}\pi/\mathrm{4}} \:{lie}\:{inside}\:{the}\:{region} \\ $$$$\Omega_{\infty} . \\ $$$$\Rightarrow\int_{−\infty} ^{\:\infty} \frac{{z}^{\mathrm{2}} }{{z}^{\mathrm{4}} +\mathrm{1}}{dz}=\mathrm{2}\pi{i}\left({Res}\left({e}^{{i}\pi/\mathrm{4}} \right)+{Res}\left({e}^{\mathrm{3}{i}\pi/\mathrm{4}} \right)\right) \\ $$$$\frac{{z}^{\mathrm{2}} }{{z}^{\mathrm{4}} +\mathrm{1}}=\frac{{z}^{\mathrm{2}} }{\left({z}−{e}^{{i}\pi/\mathrm{4}} \right)\left({z}−{e}^{\mathrm{3}{i}\pi/\mathrm{4}} \right)\left({z}−{e}^{\mathrm{5}{i}\pi/\mathrm{4}} \right)\left({z}−{e}^{\mathrm{7}{i}\pi/\mathrm{4}} \right)} \\ $$$${Therefore}\:{the}\:{poles}\:{are}\:{simple} \\ $$$$\Rightarrow{Res}\left({e}^{{i}\pi/\mathrm{4}} \right)=\underset{{z}\rightarrow{e}^{{i}\pi/\mathrm{4}} } {{lim}}\frac{\left({z}−{e}^{{i}\pi/\mathrm{4}} \right){z}^{\mathrm{2}} }{{z}^{\mathrm{4}} +\mathrm{1}}\overset{{l}'{hopital}} {=} \\ $$$$=\underset{{z}\rightarrow{e}^{{i}\pi/\mathrm{4}} } {{lim}}\frac{\mathrm{3}{z}^{\mathrm{2}} −\mathrm{2}{ze}^{{i}\pi/\mathrm{4}} }{\mathrm{4}{z}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{4}}{e}^{−{i}\pi/\mathrm{4}} \\ $$$${Res}\left({e}^{\mathrm{3}{i}\pi/\mathrm{4}} \right)=\underset{{z}\rightarrow{e}^{\mathrm{3}{i}\pi/\mathrm{4}} } {{lim}}\frac{\left({z}−{e}^{\mathrm{3}{i}\pi/\mathrm{4}} \right){z}^{\mathrm{2}} }{{z}^{\mathrm{4}} +\mathrm{1}}= \\ $$$$=\underset{{z}\rightarrow{e}^{\mathrm{3}{i}\pi/\mathrm{4}} } {{lim}}\frac{\mathrm{3}{z}^{\mathrm{2}} −\mathrm{2}{ze}^{\mathrm{3}{i}\pi/\mathrm{4}} }{\mathrm{4}{z}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{4}}{e}^{−\mathrm{3}{i}\pi/\mathrm{4}} \\ $$$$\int_{−\infty} ^{\:\infty} \frac{{z}^{\mathrm{2}} }{{z}^{\mathrm{4}} +\mathrm{1}}{dz}=\mathrm{2}\pi{i}\left(\frac{\mathrm{1}}{\mathrm{4}}{e}^{−{i}\pi/\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{4}}{e}^{−\mathrm{3}{i}\pi/\mathrm{4}} \right)= \\ $$$$=\frac{\pi{i}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{{i}}{\:\sqrt{\mathrm{2}}}+\left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)−\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)= \\ $$$$=\frac{\pi{i}}{\mathrm{2}}\left(−\frac{\mathrm{2}{i}}{\:\sqrt{\mathrm{2}}}\right)= \\ $$$$=\frac{\pi}{\:\sqrt{\mathrm{2}}} \\ $$
Commented by Tawa11 last updated on 25/Jan/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by Ar Brandon last updated on 26/Jan/22
![I=∫_(−∞) ^(+∞) (z^2 /(z^4 +1))dz=∫_0 ^(+∞) (((z^2 +1)+(z^2 −1))/(z^4 +1))dz =∫_0 ^(+∞) ((z^2 +1)/(z^4 +1))dz+∫_0 ^(+∞) ((z^2 −1)/(z^4 +1))dz =∫_0 ^(+∞) ((1+(1/z^2 ))/(z^2 +(1/z^2 )))dz+∫_0 ^(+∞) ((1−(1/z^2 ))/(z^2 +(1/z^2 )))dz =∫_0 ^(+∞) ((1+(1/z^2 ))/((z−(1/z))^2 +2))dz+∫_0 ^(+∞) ((1−(1/z^2 ))/((z+(1/z))^2 −2))dz =(1/( (√2)))[arctan(((z^2 −1)/( (√2)z)))]_0 ^(+∞) −(1/( 2(√2)))[ln∣((z^2 +(√2)z+1)/(z^2 −(√2)z+1))∣]_0 ^(+∞) =(1/( (√2)))((π/2)+(π/2))=(π/( (√2)))](https://www.tinkutara.com/question/Q165133.png)
$${I}=\int_{−\infty} ^{+\infty} \frac{{z}^{\mathrm{2}} }{{z}^{\mathrm{4}} +\mathrm{1}}{dz}=\int_{\mathrm{0}} ^{+\infty} \frac{\left({z}^{\mathrm{2}} +\mathrm{1}\right)+\left({z}^{\mathrm{2}} −\mathrm{1}\right)}{{z}^{\mathrm{4}} +\mathrm{1}}{dz} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{+\infty} \frac{{z}^{\mathrm{2}} +\mathrm{1}}{{z}^{\mathrm{4}} +\mathrm{1}}{dz}+\int_{\mathrm{0}} ^{+\infty} \frac{{z}^{\mathrm{2}} −\mathrm{1}}{{z}^{\mathrm{4}} +\mathrm{1}}{dz} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{1}+\frac{\mathrm{1}}{{z}^{\mathrm{2}} }}{{z}^{\mathrm{2}} +\frac{\mathrm{1}}{{z}^{\mathrm{2}} }}{dz}+\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{1}−\frac{\mathrm{1}}{{z}^{\mathrm{2}} }}{{z}^{\mathrm{2}} +\frac{\mathrm{1}}{{z}^{\mathrm{2}} }}{dz} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{1}+\frac{\mathrm{1}}{{z}^{\mathrm{2}} }}{\left({z}−\frac{\mathrm{1}}{{z}}\right)^{\mathrm{2}} +\mathrm{2}}{dz}+\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{1}−\frac{\mathrm{1}}{{z}^{\mathrm{2}} }}{\left({z}+\frac{\mathrm{1}}{{z}}\right)^{\mathrm{2}} −\mathrm{2}}{dz} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[\mathrm{arctan}\left(\frac{{z}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{\mathrm{2}}{z}}\right)\right]_{\mathrm{0}} ^{+\infty} −\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{\mathrm{2}}}\left[\mathrm{ln}\mid\frac{{z}^{\mathrm{2}} +\sqrt{\mathrm{2}}{z}+\mathrm{1}}{{z}^{\mathrm{2}} −\sqrt{\mathrm{2}}{z}+\mathrm{1}}\mid\right]_{\mathrm{0}} ^{+\infty} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right)=\frac{\pi}{\:\sqrt{\mathrm{2}}} \\ $$