MAT334-2018F > Term Test 2

TT2B Problem 4

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Victor Ivrii:
After Alex Qi correction from Zixuan post it follows $\newcommand{\Res}{\operatorname{Res}}$

(a) As $R>1$ there is just one singularity inside $\Gamma_{R,\varepsilon}$, namely a simple pole at $z=i$. The residue is
$\Res (\frac{\sqrt{z}\log(z)}{(z^2+1)}, i)= \frac{\sqrt{z}\log(i)}{{2z}}|_{z=i}= \frac{\pi i/2 }{2i}e^{i\pi/4}$ due to the branch selection and therefore $J= 2\pi i \times \frac{\pi i/2 }{2i}e^{i\pi/4}= \frac{\pi^2}{2} e^{3i\pi/4}=
\frac{\pi^2}{2\sqrt{2}} (-1-i)$.

(b)
$$
\int_{\gamma_R}  \frac{\sqrt{z}\log(z)\,dz}{(z^2+1)}|\le \frac{  \pi R \sqrt{R}(\ln(R)+\pi)}{(R-1)^2}\to 0\qquad \text{as }\ \ R\to \infty
$$
and
$$
\int_{\gamma_\varepsilon}  \frac{\sqrt{z}\log(z)\,dz}{(z^2+1)}|\le \frac{ \pi \varepsilon \sqrt{\varepsilon} (|\log \varepsilon|+\pi)}{(1-\varepsilon)^2}\to 0\qquad \text{as  }\ \ \varepsilon \to 0
$$
(c) In (*) the second integral is $I$ and the first one is
$$
\int_{-\infty}^0 \frac{e^{i\pi/2}\sqrt{|x|}(\ln |x|+\pi i)\, dx}{(x^2+1)} =
i\int_{\infty}^0 \frac{\sqrt{|x|} (\ln |x|+\pi i) dx}{(x^2+1)}=
i I -\pi K
$$
with
$$
K=\int_0^\infty\frac{\sqrt{|x|}  dx}{(x^2+1)}
$$
after change of variables. Thus
$$
(i+1)I -\pi K= \frac{\pi^2}{2\sqrt{2}} (-1+i).
$$
Since both $I,K$ are real
$$
I =\frac{\pi^2}{2\sqrt{2}}.
$$
As an added value we get $I-\pi K =-\frac{\pi^2}{2\sqrt{2}}$ and $K=\frac{\pi}{\sqrt{2}}$.

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