- #1
transmini
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Homework Statement
Compute the Integral: ##\int_{-\infty}^\infty \space \frac{e^{-2ix}}{x^2+4}dx##
Homework Equations
##\int_C \space f(z) = 2\pi i \sum \space res \space f(z)##
The Attempt at a Solution
At first I tried doing this using a bounded integral but couldn't seem to get the right answer from there. I could get it using Jordan's Lemma on a semi circular curve traced in the lower half plane. I'm just curious why the original approach didn't work.
$$\int_{-R}^R \space \frac{e^{-2ix}}{x^2+4}dx + \int_{C_R} \space \frac{e^{-2iz}}{z^2+4}dz = 2\pi i (\lim_{z\to 2i} (z-2i)\frac{e^{-2iz}}{z^2+4}) = 2\pi i (\lim_{z\to 2i} \frac{e^{-2iz}}{z+2i})$$
$$|\int_{C_R} \space \frac{e^{-2iz}}{z^2+4}dz| \leq \int_{C_R} \space |\frac{e^{-2iz}}{z^2+4}|dz \leq \int_{C_R} \space \frac{|e^{-2iz}|}{|z+2i||z-2i|}dz \leq \int_{C_R} \space \frac{1}{(R-2i)(R+2i)}dz$$
$$|\int_{C_R} \space \frac{e^{-2iz}}{z^2+4}dz| \leq \frac{1}{(R-2i)(R+2i)} \int_{C_R} \space dz = \frac{1}{(R-2i)(R+2i)} \pi R = \frac{\pi R}{(R-2i)(R+2i)}$$
which goes to 0 as R approaches infinity. The last inequality in the 2nd line comes from using the form of the triangle inequality saying ##|a+b| \geq |a|-|b|##.
Then as ## R\to\infty##,
$$\int_{-\infty}^\infty \space \frac{e^{-2ix}}{x^2+4}dx + \int_{C_\infty} \space \frac{e^{-2iz}}{z^2+4}dz = 2\pi i (\lim_{z\to 2i} \frac{e^{-2iz}}{z+2i})$$
$$\int_{-\infty}^\infty \space \frac{e^{-2ix}}{x^2+4}dx + 0 = 2\pi i \frac{e^{-2i*2i}}{2i+2i} = 2\pi i \frac{e^4}{4i}$$
$$\int_{-\infty}^\infty \space \frac{e^{-2ix}}{x^2+4}dx = \frac{\pi e^4}{2}$$
where using Jordan's Lemma in the negatively traced curve gives ##\frac{\pi}{2e^4}##. Any ideas as to why the bounded integral method does not work, or where I made a mistake in the derivation at? Thanks ahead of time.