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Homework Statement
Show that:
For a = 0
[tex] \int_{0}^{\infty} \frac{cos{ax}+x sin{ax}}{1+x^2} dx = \frac{\pi}{2} [/tex]
For a > 0
[tex] \int_{0}^{\infty} \frac{cos{ax}+x sin{ax}}{1+x^2} dx = \pi e^{-a} [/tex]
For a < 0
[tex] \int_{0}^{\infty} \frac{cos{ax}+x sin{ax}}{1+x^2} dx = 0 [/tex]
Homework Equations
Residue Theorem:
[tex]\oint_{C} f(z)dz = 2 \pi i \sum_{C \ni z_{i}} R(z_{i}) [/tex]
The Attempt at a Solution
The first part is rather trivial. With a=0 it just reduces to the arctan infinity, which is pi over two.
For the second two parts I'm having trouble closing the contour for the second term, xsin(ax). At first I ignored the xsin(ax) term for a>0, and it immediately gave the answer:
[tex] \int_{0}^{\infty} \frac{cos{ax}+x sin{ax}}{1+x^2} dx = \int_{0}^{\infty} \frac{cos{ax}}{1+x^2}dx + \int_{0}^{\infty} \frac{x sin{ax}}{1+x^2} dx [/tex]
Focusing just on the cosine term:
[tex] \int_{0}^{\infty} \frac{cos{ax}}{1+x^2}dx = Real[ \int_{0}^{\infty} \frac{e^{iaz}}{1+z^2} dz ] [/tex]
Closing the contour for this:
[tex] \oint_{C} \frac{e^{iaz}}{1+z^2} dz = Real[ \int_{0}^{\infty} \frac{e^{iaz}}{1+z^2} dz ] + \int_{Arc} \frac{cos{az}}{1+z^2} dz [/tex]
For the Arc Integral:
[tex] z = R e^{i \theta} [/tex]
[tex] dz = R i e^{i \theta} d\theta [/tex]
[tex] \int_{Arc} \frac{cos{az}}{1+z^2} dz = lim_{R \rightarrow \infty} \int_{0}^{2 \pi} \frac{e^{iaRe^{i \theta}}}{1+(Re^{i \theta})^2} (R i e^{i \theta} d\theta) [/tex]
[tex] \int_{Arc} \frac{cos{az}}{1+z^2} dz \approx lim_{R \rightarrow \infty} \frac{1}{R} = 0 [/tex]
So the arc contributes nothing to the contour integral.
Which would mean:
[tex] \oint_{C} \frac{e^{iaz}}{1+z^2} dz = Real[ \int_{0}^{\infty} \frac{e^{iaz}}{1+z^2} dz ] = 2 \pi i R(i) [/tex]
[tex] \frac{e^{iaz}}{1+z^2} = \frac{e^{iaz}}{2i} (\frac{1}{z-i} + \frac{1}{z+i} ) [/tex]
So
[tex] R(i) = \frac{e^{-a}}{2i} [/tex]
[tex] \oint_{C} \frac{e^{iaz}}{1+z^2} dz = Real[ \int_{0}^{\infty} \frac{e^{iaz}}{1+z^2} dz ] = 2 \pi i \frac{e^{-a}}{2i} = \pi e^{-a} [/tex]
Which leads me to believe that the xsin(ax) term contributes nothing to the actual contour integral when a > 0. However that doesn't really help me close the contour for the sin term. Whenever I try to use the same method on the second term, I don't get a real solution.
[tex] \int_{0}^{\infty} \frac{x sin{ax}}{1+x^2} dx = \oint_{C} \frac{ze^{iaz}}{1+z^2} dz + \int_{Arc} \frac{ze^{iaz}}{1+z^2}= Imaginary[ \int_{0}^{\infty} \frac{ze^{iaz}}{1+z^2} dz ] + \int_{Arc} \frac{ze^{iaz}}{1+z^2}[/tex]
For the arc:
[tex] z = R e^{i \theta} [/tex]
[tex] dz = R i e^{i \theta} d\theta [/tex]
[tex] \int_{Arc} \frac{ze^{iaz}}{1+z^2} = lim_{R \rightarrow \infty} \int_{0}^{2 \pi} \frac{(Re^{i \theta}) e^{iaRe^{i \theta}}}{1+(Re^{i \theta})^2} (R i e^{i \theta} d\theta) [/tex]
But in this case the limit does not reduce the integral to zero and I don't know what to do. Is there another method I can use to solve for the sine term, or did I make a mistake somewhere? What is the next step to solve this contour integral?
Oops, I messed up the bounds when transforming it from a real to a complex function. So in actuality the value of the cosine integral is pi e^-a all over two. So the sine term must have the same value. This makes sense, so that when a < 0 the two terms cancel out to give zero, but I'm still having trouble closing the contour for the sine term
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