calculate-0-du-u-z-2-if-z-r-e-i-and-pi-lt-lt-pi-

Question Number 67539 by mathmax by abdo last updated on 28/Aug/19

c a l c u l a t e \int_{0}^{\infty} \frac{d u}{∣ u + z ∣^{2}} i f z = r e^{i θ} a n d - π < θ < π

Commented by ~ À ® @ 237 ~ last updated on 29/Aug/19

∣ u + z ∣^{2} = (u + z) (u + \overline{z})

I = \frac{1}{z - \overline{z}} \int_{0}^{\infty} (\frac{1}{u + \overline{z}} - \frac{1}{u + z}) d u = \frac{1}{2 i I m (z)} {[l n (\frac{u + \overline{z}}{u + z})]}_{0}^{\infty} = \frac{1}{2 i r s i n θ} [l n 1 - l n (e^{- i 2 θ})] = \frac{θ}{r s i n θ}

Commented by mathmax by abdo last updated on 29/Aug/19

t h a n k y o u s i r .

Commented by mathmax by abdo last updated on 29/Aug/19

l e t I = \int_{0}^{\infty} \frac{d u}{∣ u + z ∣^{2}} \Rightarrow I = \int_{0}^{\infty} \frac{d u}{(u + z) (u + \bar{z})}

= \frac{1}{z - \bar{z}} \int_{0}^{\infty} (\frac{1}{u + \bar{z}} - \frac{1}{u + z}) d u = \frac{1}{2 i (r s i n θ)} {[l n (\frac{u + \bar{z}}{u + z})]}_{0}^{+ \infty}

= \frac{1}{2 i r s i n θ} \times (- l n (\frac{\bar{z}}{z})) w e h a v e - l n (\frac{\bar{z}}{z}) = l n (\frac{z}{\bar{z}})

\frac{z}{\bar{z}} = \frac{z^{2}}{∣ z ∣^{2}} = \frac{r^{2} e^{2 i θ}}{r^{2}} = e^{2 i θ} \Rightarrow l n (\frac{z}{\bar{z}}) = 2 i θ \Rightarrow

I = \frac{1}{2 i r s i n θ} \times (2 i θ) \Rightarrow I = \frac{θ}{r s i n θ} .