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\centerline{{\it  Class 31} {\bf (Friday April 10)}}
 
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\item[SUMMARY] Applications of Residues to Real Integrals
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\item[CURRENT READING] Brown \& Curchill pages 210--219
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\item[NEXT READING] Brown \& Curchill 204--210
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The beauty of Complex Residue Calculus is that it allows us to evaluate a vast number of contour integrals. In fact, we can show that we can use residues to evaluate associated {\bf real} integrals which would otherwise be very difficult to get exact values for become quite easy as contour integrals.\\

Recall the definition of $z = e^{i \theta} = \cos ( \theta) + i \sin (\theta)$
Therefore, we can write $\cos(\theta)$ and $\sin(\theta)$ in terms of $z$.\\
$\cos \theta = \hspace{3in} \sin \theta = $
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\underline{\bf Exercise}\\
Rewrite the integral ${\displaystyle \int_0^{2 \pi} \frac{d \theta}{3 + 2 \sin \theta}}$ in terms of $z$, using $z = e^{i \theta}$ where $0 \le \theta \le 2 \pi$.
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Evaluate the real integral by evaluating the contour integral.
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\underline{\sc Groupwork}\\
Show that ${\displaystyle \int_0^{2 \pi} \frac{\cos 3 \theta}{5 - 4 \cos \theta} d \theta = \frac{i}{2} \oint_{|z|=1} \frac{z^6+1}{z^3(2z^2-5z+2)} dz }$

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You should be able to compute the value of the contour integral and show it equals ${\displaystyle \frac{\pi}{12}.}$
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{\bf Evaluation of real integrals of the form ${\displaystyle \int_{-\infty}^{\infty} f(x) \ dx}$}\\
We can also evaluate {\bf Improper Integrals} more easily by evaluating associated contour integrals. however, we have to have some conditions on the integrand $f(z)$.

{\bf Boundedness Theorem 1}\\
If $|f(z)| \le {\displaystyle \frac{ M}{R^k}}$ for $z = Re^{i \theta}$ where $k>1$ and $M$ are constants and $C_R$ is the contour consisting of the semi-circle of radius $R$ from $\theta=0$ to $\theta = \pi$, then
${\displaystyle \lim_{R \rightarrow \infty} \int_{C_R} f(z) e^{i n z} \ dz = 0}$
%$$\lim_{R \rightarrow \infty} \int_{C_R} f(z) \ dz = 0$$

{\bf Boundedness Theorem 2}\\
If $|f(z)| \le {\displaystyle \frac{ M}{R^k}}$ for $z = Re^{i \theta}$ where $k>1$, $n>0$  and $M$ are constants and $C_R$ is the contour consisting of the semi-circle of radius $R$ from $\theta=0$ to $\theta = \pi$, then
${\displaystyle \lim_{R \rightarrow \infty} \int_{C_R} f(z) e^{i n z} \ dz = 0}$
The second boundedness theorem is sometimes called {\bf Jordan's Lemma}.\\

You can use these theorems above to say that 
$$\int_{-R}^{R} f(x) dx + \int_{C_R} f(z) \ dz = \oint_C f(z) \ dz$$
 and then take the limit as $R \rightarrow \infty$ to say that 
$$\int_{-\infty}^{\infty} f(x) \ dx = \oint f(z)\ dz = 2 \pi i \sum {\boldmath Res}(f)$$
if the boundedness theorems above apply to $f(z)$ (since then $\int_{C_R} f(z) \ dz \rightarrow 0$ as $R \rightarrow \infty$)

%\underline{\bf Exercise}\\
%Show that ${\displaystyle \int_0^{\infty} \frac{1}{x^2+1} \ dx = \frac{\pi}{2} }$
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\underline{Examples}\\
Show that ${\displaystyle \int_{-\infty}^{\infty} \frac{1}{(x^2+1)^2} \ dx = \frac{\pi}{2} }$
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Show that (for $a>0$) ${\displaystyle \int_0^{\infty} \frac{\cos(ax)}{x^2+1} \ dx = \frac{\pi}{2}e^{-a}}$
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