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Question Number 67539 by mathmax by abdo last updated on 28/Aug/19
calculate ∫_0 ^∞   (du/(∣u+z∣^2 ))  if z =r e^(iθ)    and −π<θ<π
calculate0duu+z2ifz=reiθandπ<θ<π
Commented by ~ À ® @ 237 ~ last updated on 29/Aug/19
∣u+z∣^2 =(u+z)(u+z^_ )   I=(1/(z−z^_ )) ∫_0 ^∞ ((1/(u+z^_ )) −(1/(u+z ))) du =(1/(2iIm(z)))[ln(((u+z^_ )/(u+z)))]_0 ^∞ =(1/(2irsinθ)) [ln1−ln(e^(−i2θ)  )]=(θ/(rsinθ))
u+z2=(u+z)(u+z_)I=1zz_0(1u+z_1u+z)du=12iIm(z)[ln(u+z_u+z)]0=12irsinθ[ln1ln(ei2θ)]=θrsinθ
Commented by mathmax by abdo last updated on 29/Aug/19
thank you sir.
thankyousir.
Commented by mathmax by abdo last updated on 29/Aug/19
let I =∫_0 ^∞   (du/(∣u+z∣^2 )) ⇒ I =∫_0 ^∞   (du/((u+z)(u+z^− )))  =(1/(z−z^− ))∫_0 ^∞ ((1/(u+z^− ))−(1/(u+z)))du =(1/(2i(rsinθ))) [ln(((u+z^− )/(u+z)))]_0 ^(+∞)   =(1/(2ir sinθ))×(−ln((z^− /z)) ) we have −ln((z^− /z))=ln((z/z^− ))  (z/z^− ) =(z^2 /(∣z∣^2 )) =((r^2  e^(2iθ) )/r^2 ) =e^(2iθ)  ⇒ln((z/z^− )) =2iθ ⇒  I =(1/(2ir sinθ))×(2iθ) ⇒I =(θ/(r sinθ)) .
letI=0duu+z2I=0du(u+z)(u+z)=1zz0(1u+z1u+z)du=12i(rsinθ)[ln(u+zu+z)]0+=12irsinθ×(ln(zz))wehaveln(zz)=ln(zz)zz=z2z2=r2e2iθr2=e2iθln(zz)=2iθI=12irsinθ×(2iθ)I=θrsinθ.

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