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Term Test 2 / Re: TT2 Problem 4
« on: November 24, 2018, 09:39:16 AM »
(a) $x^{2} + 1 = 0$
so $x = i$ $(x = -i$ not in range of graph )
By Residue theorem
$\int f\left(z\right)dz = 2\pi i \sum Res(f, i)= (\dfrac{1}{\sqrt{z}(z + i)}, i) \times 2\pi i = \dfrac{1}{\sqrt{i}(2i)}\times 2\pi i = \dfrac{\pi}{\sqrt{i}} $
(b) $|\int_{R} \dfrac{dz}{\sqrt{z}(z^{2} + i)}| \leq \pi R \dfrac{1}{\sqrt{R}(R^{2} - i)}\rightarrow 0$ When $R\rightarrow \infty$
$|\int_{\varepsilon} \dfrac{dz}{\sqrt{z}(z^{2} + i)}| \leq \pi \varepsilon \dfrac{1}{\sqrt{\varepsilon}(i - \varepsilon^{2})} \rightarrow 0$ When $\varepsilon\rightarrow 0$
so, we have $\int ^{0}_{-\infty} f\left(z\right)dz + \int ^{\infty}_{0} f\left(z\right)dz = \dfrac{\pi}{\sqrt{i}} $
(c) $\int ^{0}_{-\infty} f\left(z\right)dz + \int ^{\infty}_{0} f\left(z\right)dz = 2I = \dfrac{\pi}{\sqrt{i}} $
$I = \dfrac{\pi}{2 \sqrt{i}} $
so $x = i$ $(x = -i$ not in range of graph )
By Residue theorem
$\int f\left(z\right)dz = 2\pi i \sum Res(f, i)= (\dfrac{1}{\sqrt{z}(z + i)}, i) \times 2\pi i = \dfrac{1}{\sqrt{i}(2i)}\times 2\pi i = \dfrac{\pi}{\sqrt{i}} $
(b) $|\int_{R} \dfrac{dz}{\sqrt{z}(z^{2} + i)}| \leq \pi R \dfrac{1}{\sqrt{R}(R^{2} - i)}\rightarrow 0$ When $R\rightarrow \infty$
$|\int_{\varepsilon} \dfrac{dz}{\sqrt{z}(z^{2} + i)}| \leq \pi \varepsilon \dfrac{1}{\sqrt{\varepsilon}(i - \varepsilon^{2})} \rightarrow 0$ When $\varepsilon\rightarrow 0$
so, we have $\int ^{0}_{-\infty} f\left(z\right)dz + \int ^{\infty}_{0} f\left(z\right)dz = \dfrac{\pi}{\sqrt{i}} $
(c) $\int ^{0}_{-\infty} f\left(z\right)dz + \int ^{\infty}_{0} f\left(z\right)dz = 2I = \dfrac{\pi}{\sqrt{i}} $
$I = \dfrac{\pi}{2 \sqrt{i}} $