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

Term Test 2 sample P4M

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Victor Ivrii:
Jeff
What you are doing is correct but does not make sense. More precisely $\int_{-\infty}^\infty e^{ibz}\cos(z)\,dz$ diverges in the "normal" sense but converges in the sense of distributions, which are more advanced ..

In fact, you can calculate $\int_0^R e^{ix}\cos (x)\, dx$ without taking the limit first. However one needs to justify the differentiation.

Since this problem is more difficult than the corresponding problems in Test 2 or posted alternative, I sketch a correct solution. First, we consider $\int_C \frac{e^{iz}}{z}\,dz=0$ due to Cauchy theorem. Next, integral over large semicircle indeed tends to $0$ because
$$
|\int_C \frac{e^{iz}}{z}\,dz|\le \int_0^{\pi} e^{-\Im z(\theta)}\,d\theta=
2  \int_0^{\pi/2} e^{-R\sin (\theta)}\,d\theta\le
2  \int_0^{\pi/2} e^{-R\theta/2}\,d\theta\le \frac{4}{R}
$$
where we used $\sin (\theta)\ge \theta /2$ as $0<\theta <\pi/2$.

Next, integral over small semicircle equals
$$
\int _{C_4} \frac{e^{iz}-1}{z}\,dz +\int _{C_4} \frac{1}{z}\,dz
$$
where
$$
|\int _{C_4} \frac{e^{iz}-1}{z}\,dz |\le \int _{0}^{\pi} \frac{|e^{iz(\theta)}-1|} d\theta\le  \int_0^\pi \varepsilon\,d\theta \le \pi \varepsilon
$$
tends to $0$ as $\varepsilon \to 0^+$ and $\int _{C_4} \frac{1}{z}\,dz=\int_\pi^0 i d\theta =-\pi i$.

Therefore
$$
\Bigl[\int_{-R}^{-\varepsilon} \frac{e^{ix}}{x}\,dx + \int_{\varepsilon }^{R} \frac{e^{ix}}{x}\,dx\Bigr]\to \pi i\tag{*}\label{eq-X}
$$
as $R\to +\infty$, $\varepsilon \to 0^+$. However it is true for the sum only, limits of each part simply do not exist! $\int_0^* \frac{e^{ix}}{x}\,dx $ diverges at $0$.

Observe that the LHE in (\ref{eq-X}) equals
$$
\Bigl[\int_{R}^{\varepsilon} \frac{e^{-ix}}{-x}\,d(-x) + \int_{\varepsilon }^{R} \frac{e^{ix}}{x}\,dx\Bigr]=
\Bigl[-\int_{\varepsilon}^{R} \frac{e^{-ix}}{x}\,dx + \int_{\varepsilon }^{R} \frac{e^{ix}}{x}\,dx\Bigr]=
\int_{\varepsilon}^R \frac{e^{ix}-e^{-ix}}{x}\,dx= 2i\int_\varepsilon^R \frac{\sin(x)}{x}\,dx
$$
which tends to $2i J $ as $R\to \infty$, $\varepsilon\to 0^+$ and therefore $J= \pi i/2i =\pi/2$.

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