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Question Number 221095 by SdC355 last updated on 24/May/25
Complex integral 1. ∫_(−∞) ^( +∞) (dz/((z^2 +1)^ν ))=?? 2. ∫_(−∞) ^(+∞) (e^(iπt) /(t^2 +1)) dt=?? 3. ∮_( C) (1/z) dz=?? , C;x^2 +y^2 =1
$$\mathrm{Complex}\:\mathrm{integral} \\ $$$$\mathrm{1}.\:\int_{−\infty} ^{\:+\infty} \:\:\:\frac{\mathrm{d}{z}}{\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{\nu} }=?? \\ $$$$\mathrm{2}.\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{\boldsymbol{{i}}\pi{t}} }{{t}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{t}=?? \\ $$$$\mathrm{3}.\:\oint_{\:{C}} \:\frac{\mathrm{1}}{{z}}\:\mathrm{d}{z}=??\:,\:{C};{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1} \\ $$
Answered by MrGaster last updated on 24/May/25
For(3):∮_( C) (1/z)=∫_0 ^(2π) (1/e^(iθ) )∙ie^(iθ) dθ =i∫_0 ^(2π) dθ =2πi
$$\mathrm{For}\left(\mathrm{3}\right):\oint_{\:{C}} \:\frac{\mathrm{1}}{{z}}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{\mathrm{1}}{{e}^{{i}\theta} }\centerdot{ie}^{{i}\theta} {d}\theta \\ $$$$={i}\int_{\mathrm{0}} ^{\mathrm{2}\pi} {d}\theta \\ $$$$=\mathrm{2}\pi{i} \\ $$
Answered by MrGaster last updated on 24/May/25
For(2):∫_(−∞) ^(+∞) (e^(iπt) /(t^2 +1))dt=2πi∙Res_(t=i) ((e^(iπt) /(t^2 +1))) =2πi∙(lim_(t→i) (t−i)∙(e^(iπt) /((t−i)(t+i)))) =2πi∙(e^(−π) /(2i)) =πe^(−π)
$$\mathrm{For}\left(\mathrm{2}\right):\int_{−\infty} ^{+\infty} \frac{{e}^{{i}\pi{t}} }{{t}^{\mathrm{2}} +\mathrm{1}}{dt}=\mathrm{2}\pi{i}\centerdot\mathrm{Res}_{{t}={i}} \left(\frac{{e}^{{i}\pi{t}} }{{t}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$=\mathrm{2}\pi{i}\centerdot\left(\underset{{t}\rightarrow{i}} {\mathrm{lim}}\left({t}−{i}\right)\centerdot\frac{{e}^{{i}\pi{t}} }{\left({t}−{i}\right)\left({t}+{i}\right)}\right) \\ $$$$=\mathrm{2}\pi{i}\centerdot\frac{{e}^{−\pi} }{\mathrm{2}{i}} \\ $$$$=\pi{e}^{−\pi} \\ $$
Answered by MrGaster last updated on 24/May/25
For (1): ∫_(−∞) ^( +∞) (dz/((z^2 +1)^ν ))=2∫_0 ^∞ (dz/((z^2 +1))) →^(z=tan θ) 2∫_0 ^(π/2) ((sec^2 θ dθ)/((sec^2 θ)^ν )) =2∫_0 ^(π/2) cos^(2ν−2) θ dθ =2∙(1/2)B((1/2),ν−(1/2)) =((Γ((1/2))Γ(ν−(1/2)))/(Γ(ν))) =(((√π)Γ(ν−(1/2)))/(Γ(ν)))
$$\mathrm{For}\:\left(\mathrm{1}\right):\:\int_{−\infty} ^{\:+\infty} \:\:\:\frac{\mathrm{d}{z}}{\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{\nu} }=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{dz}}{\left({z}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\overset{{z}=\mathrm{tan}\:\theta} {\rightarrow}\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sec}^{\mathrm{2}} \theta\:{d}\theta}{\left(\mathrm{sec}^{\mathrm{2}} \theta\right)^{\nu} } \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}\nu−\mathrm{2}} \theta\:{d}\theta \\ $$$$=\mathrm{2}\centerdot\frac{\mathrm{1}}{\mathrm{2}}{B}\left(\frac{\mathrm{1}}{\mathrm{2}},\nu−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\nu−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\nu\right)} \\ $$$$=\frac{\sqrt{\pi}\Gamma\left(\nu−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\nu\right)} \\ $$

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