Toronto Math Forum

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
$$

$$
=2iI
$$

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

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

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

-54π