### Author Topic: TT2A Problem 4  (Read 1833 times)

#### Victor Ivrii

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##### TT2A Problem 4
« on: November 24, 2018, 05:15:23 AM »
Calculate an improper integral
$$I=\int_0^\infty \frac{\sqrt{x}\,dx}{(x^2+2x+2)}.$$

Hint:

(a) Calculate
$$J_{R,\varepsilon} = \int_{\Gamma_{R,\varepsilon}} f(z)\,dz, \qquad f(z)=\frac{\sqrt{z}}{(z^2+2z+2)}$$
where we have chosen the branch of $\sqrt{z}$ such that it is analytic inside  $\Gamma$ and is real-valued for $z=x+i0$ with $x>0$. $\Gamma=\Gamma_{R,\varepsilon}$ is the contour on the figure below:

(b) Prove that $\int_{\gamma_R} \frac{\sqrt{z}\,dz}{(z^2+1)}\to 0$ as $R\to \infty$ and $\int_{\gamma_\varepsilon} \frac{\sqrt{z}\,dz}{(z^2+1)}\to 0$ as $\varepsilon\to 0^+$ where $\gamma_R$ and $\gamma_\varepsilon$ are large and small circles on the picture. This will give you a value of
$$\int_{\infty}^0 f(x-i0)\,dx + \int_0^{\infty} f(x+i0)\,dx \tag{*}$$
where $f(x\pm i0)=\lim _{\delta\to 0^+} f(x+i\delta)$.

(c) Express both integrals using $I$.
« Last Edit: November 29, 2018, 08:26:30 AM by Victor Ivrii »

#### Heng Kan

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##### Re: TT2A Problem 4
« Reply #1 on: November 25, 2018, 12:17:50 PM »
There are four scanned pictures.  The first one is for part a, the second and third one are for part b, the last one is for part c.

#### Victor Ivrii

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##### Re: TT2A Problem 4
« Reply #2 on: November 29, 2018, 08:24:35 AM »
$\newcommand{\Res}{\operatorname{Res}}$
Correct. Since it is not well readable (the answer is the same):

(a) As $R> 1$ there are  two singularities inside $\Gamma_{R,\varepsilon}$, namely,  simple poles at
$z= -1+i=\sqrt{2}e^{3\pi i/4}$ and $z= -1+i=\sqrt{2}e^{5\pi i/4}$ due to branch selection. The residues are
\begin{align*}
&\Res (\frac{\sqrt{z}}{(z^2+2z+2)}, -1+i)= \frac{\sqrt{z}}{{2z+2}}\Bigl|_{-1+i}=\frac{1}{2i}2^{1/4}e^{3\pi i/8}\\
\text{and}\\
&\Res (\frac{\sqrt{z}}{(z^2+2z+2)}, -1-i)= \frac{\sqrt{z}}{{2z+2}}\Bigl|_{z=-1-i}=-\frac{1}{2i}2^{1/4}e^{5\pi i/8}=
\frac{1}{2i}2^{1/4}e^{-3\pi i/8}
\end{align*}
due to the branch selection and therefore
$$J= 2\pi i \times 2^{1/4}\frac{1}{2i}\bigl[e^{3i\pi/8}+ e^{-3i\pi/8}\bigr]= 2^{5/4}\pi \cos(3\pi/8) .$$

(b) \begin{gather*}
|\int_{\gamma_R}  \frac{\sqrt{z}\, dz}{(z^2+1)}|\le \frac{  2\pi R \sqrt{R}}{(R-1)^2}\to 0\qquad \text{as\ \ }R\to \infty\\
\text{and}\\
|\int_{\gamma_\varepsilon}  \frac{\sqrt{z}\,dz}{(z^2+1)}|\le \frac{ 2\pi \varepsilon \sqrt{\varepsilon}}{(1-\varepsilon)^2}\to 0\qquad \text{as\ \ }\varepsilon \to 0
\end{gather*}

(c)  In (*) the second integral is $I$ and the first one is
$$\int_{\infty}^0 \frac{-\sqrt{x}\,dx}{(x^2+1)} =I$$
after change of variables. Thus
$$2I =2^{5/4}\pi \sin(3\pi/8) \implies I=2^{1/4}\pi \cos(3\pi/8).$$

« Last Edit: November 29, 2018, 08:26:49 AM by Victor Ivrii »