MAT334-2018F > Reading Week Bonus--sample problems for TT2

Term Test 2 sample P4M

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Victor Ivrii:
Calculate an improper integral
$$I=\int_0^\infty \frac{\sin(x)}{x}\,dx.$$

Hint: (a) Calculate
$$J_{R,\varepsilon} = \int_{\Gamma_{R,\varepsilon}} f(z)\,dz, \qquad f(z):=\frac{e^{iz}}{z}$$
where we have chosen the branch  of $\log(z)$ such that they are analytic on the upper half-plane $\{z\colon \Im z>0\}$ and is real-valued for $z=x>0$. $\Gamma_{R,\varepsilon}$ is the contour on the figure below (one can also replace the large semi-circle $\gamma_R$ by the broken line from $R$ to $R+Ri$ to $-R+Ri$ to $-R$)

(b) Prove that $\int_{\gamma_R} f(z)\,dz\to 0$ as $R\to \infty$ and $\int_{\gamma_\varepsilon} \bigl[f(z)-\frac{1}{z}]\,dz\to 0$ as $\varepsilon\to 0^+$; calculate $\int_{\gamma_\varepsilon}\frac{1}{z}\,dz$.

This will give you a value of

\lim_{\varepsilon\to+0}\Bigl[\int_{-\infty}^{-\varepsilon} f(z)\,dz +  \int_\varepsilon ^{\infty} f(z)\,dz\Bigr].
\label{4-1}

(c) Changing the variable in the first integral express (\ref{4-1}) using $I$.

Fangqi Lu:
There's something wrong with the format of the Latex file, it turns out to be some messy codes when I tried to copy and paste here, therefore I just decided to upload the pdf file, hope it works :)

Victor Ivrii:
It is difficult to make sense from what you uploaded, since in the first display formula you do not show the integrand, and some conclusions seem to be either unwarranted or wrong

It would be useful in the discussion to have it typed, rather than uploaded. Upload your faulty LaTeX file, however I can tell you  looking on your pdf, that the first line is already borked and starts from dollar or double dollar, while it should not (it is just a text, excluding c1 ... which should be $c_1$, ...

Jeffery Mcbride:

\begin{equation*}
I( b) \ =\int ^{\infty }_{0}\frac{e^{-bx} sin( x)}{x} \ dx\ for\ b\ \geq 0.\\
\\
I( 0) \ =\ \int ^{\infty }_{0}\frac{sin( x)}{x} \ ( This\ is\ the\ integral\ of\ interest) .\\
\\
I'( b) \ =\ -\ \int ^{\infty }_{0} \ e^{-bx} sinx\ dx\\
\\
=\ -\ \frac{1}{1\ +\ b^{2}}\\
\end{equation*}

Integrate this with respect to b and we get:

\begin{equation*}
I( b) \ =\ \pi /2\ -\ arctan( b) ,\ setting\ b\ =\ 0,\ gives\ us\ \pi /2\\
so,\\
\\
\int ^{\infty }_{0}\frac{sin( x)}{x} \ \ =\ \pi /2\ \\
\end{equation*}

Fangqi Lu:
First, we divide the graph into several segments, $C_1$, $C_2$, $C_3$, $C_4$ such that:

$$\int_C=\int_{C_1}+\int_{C_2}+\int_{C_3}+\int_{C_4}$$
You need to indicate, what is an integrand, and be consistent with notations of contours.

Since c is close and analytic we know $\int_c=0$ by Cauchy theorem. Let radius at $C_1$ be $\epsilon$ and radius of the whole semi-circle be $R$ We want to let the radius of the smaller semi-circle to be infinitively small and the radius of the larger circle to be infinitively large  so, let $\epsilon \to 0$ and $R\to\infty$
$$\therefore \int^R_\epsilon \Rightarrow \int^\infty_0\\ \because e^{iz}=\cos z+i\sin z$$

First we calculate$\int\frac{e^{iz}}{z}dz=3+4i$         How? By a residue theorem? You need to do it
$$\therefore \int\frac{\cos z}{z}dz=3 \int \frac{\sin z}{z}dz=4$$

$C_3$:
$$\int\frac{e^{iz}}{z}dz$$
$$r(t)=Re^{it} \quad 0\leq t\leq \pi\\ \therefore r'(t)=iRe^{it}, f(z)=\frac{e^{iz}}{z}\\ \therefore f(r(t)) =\frac{e^{ir(t)}}{r(t)}$$

$$=\frac{e^{iRe^{it}}}{Re^{it}}$$

$$=\frac{e^{iR(\cos t+i \sin t)}}{Re^{it}}$$
$$=\frac{e^{-R\sin t+iR\cos t}}{Re^{it}}$$
$$\therefore |\int f(z)\mathrm{d}z|$$
$$=|\int f(r(t))\cdot r'(t)\mathrm{d}t|$$
$$=|\int^\pi_0\frac{e^{-R\sin t+iR\cos t}}{Re^{it}}iRe^{it}dt|$$
$$=|i\int^\pi_0e^{-R\sin t}\cdot e^{iR\cos t}dt|$$
$$\leq \int ^\pi_0 e^{-R\sin t}dt$$
$$Since\,R\rightarrow\infty\,and\,\,in\,the\,interval\, (0, \pi),\, \sin t\,is\,alway\,positive, so\,-R \sin t\rightarrow\infty,\,and\,e^{-\infty}=0\,so\,when\,R\rightarrow-\infty,\,e^{-R\sin t}\rightarrow0$$
$$\therefore |\int f(z)\mathrm{d}z| \leq 0$$
$$\therefore \int_{c3}=0$$

$C_1$:

$$\int_{c1}\frac{e^{iz}}{z}dz=\int_{c1}\frac{1}{z}+f(z)dz, Let\, |f(z)|\leq M(bounded),\,where\,M\,is\,a\,constant$$

$$\leq \int \frac{1}{z}dz+\int M dz$$

$$\leq \int \frac{1}{z}dz+\underbrace{M\pi\varepsilon}_{\underset{0}{\Downarrow}}$$

$$r(t)=\varepsilon e^{it}\,\, \pi \leq t\leq 0$$

$$r'(t)=i\varepsilon e^{it}$$

$$f(z)=\frac{1}{z}$$

$$\therefore f(r(t))=\frac{1}{\varepsilon e^{it}}=\frac{1}{\epsilon}e^{-it}=\int^0_\pi f(r(t))\cdot r'(t)dt=-\pi i$$

Since $C_2$ and $C_4$ are symmetric, we can transform $C_4$ into $C_2$
$C_2$ + $C_4$:

$$\int_{c2+c4}=\int^{-\epsilon}_{-R}\frac{e^{ix}}{z}dz+\int^{R}_\epsilon \frac{e^{iz}}{z}dz$$

These segments are on real line, therefore z=x+yi=x

$$\therefore =\int^{-\epsilon}_{-R}\frac{e^{ix}}{x}dx+\int^R_\epsilon \frac{e^{ix}}{x}dx$$

$$let \quad u=-x$$

$$\therefore du=-dx$$

$$\therefore dx=-du$$

$$=\int^\epsilon_R-\frac{e^{i(-u)}}{-u}du+\int^R_\epsilon \frac{e^{ix}}{x}dx$$

$$=\int^R_\epsilon\frac{e^{i(-u)}}{-u}du+\int^R_\epsilon \frac{e^{ix}}{x}dx$$

$$=\int^R_\epsilon\frac{e^{-ix}}{-x}dx+\int^R_\epsilon\frac{e^{ix}}{x}dx$$

$$=\int^R_\epsilon\frac{e^{ix}-e^{-ix}}{x}dx$$

$$=\int^R_\epsilon\frac{2i\sin x}{x}dx$$

$$Since\, \epsilon \to 0\, and\, R\to\infty \therefore\,=2i\int ^\infty_0\frac{\sin x}{ x}dx$$

$$=2iI$$

$$\int_c=\int_{c1}+\int_{c2}+\int_{c3}+\int_{c4}$$

$$0=-\pi i+0+2iI$$

$$\therefore I=\frac{\pi}{2}$$