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Homework Statement
Hi everyone. I'm currently taking a graduate math physics course and complex integrals are beating the crap out of me. Some of my questions may be relatively basic. Forgive me, I'm trying to teach myself and am regretting not taking a course on complex analysis as an undergrad. I have several problems. I'll list them all at once and mark the corresponding work for each separate problem below.
Thank you in advance. Any help is greatly appreciated!
1. Solve the following integral twice. Once by closing the contour in the Upper Half Plane, and once by closing in the Lower Half Plane:
[tex]I=\int_{-\infty}^{\infty} \frac{1}{x^3 + i} dx[/tex]
2. Solve: [tex]I=\int_0^{\infty} \frac{1}{1+x^3} dx[/tex]
HINT for #2: Use the following contour in the complex plane: 1. The real axis from 0 to R. 2. The arc, [itex]\theta=0..\frac{2\pi}{3}[/itex]. 3. The line from the end of that arc back to the origin. Take the limit as R approaches infinity.
3. Show [tex]I=\int_{-\infty}^{\infty}\frac{\sin(x)}{x}dx=\pi[/tex] NOT by computing a principle value, but instead by letting [itex]x-0\,\rightarrow\, x-(0+i\epsilon)\,}[/itex] and then compting the integral and taking the limit as epsilon approaches 0.
Homework Equations
Work to come, still editing post.
The Attempt at a Solution
1.
Work with the complex integral:
[tex]I_c=\oint\frac{dz}{z^3+i}[/tex]
Close the contour in the UHP:
[tex]I_c=\lim_{R\rightarrow\infty}\int_{-R}^{R}\frac{dx}{x^3 + i}+\lim_{R\rightarrow\infty}\int_{C_R}\frac{dz}{z^3+i}[/tex]
where [itex]C_R[/itex] is the half circle of radius R in the UHP.
Since [itex]z^3=R^3\exp{3i\theta}[/itex], the second integral goes to zero in the limit. Then, using the residue theorem:
[tex]I_c=2\pi i \Sigma (Residues)=\lim_{R\rightarrow\infty}\int_{-R}^{R}\frac{dx}{x^3 + i}[/tex]
If my work is correct up until here, then I think my only problem is with computing the residues. Up until now I have only needed to compute residues for functions with denominators that easily factored, making the residue computation easy. Here, I'm lost.
I know there are zeros when z=-i. The zeroes are all along the unit circle at [itex] \theta=\frac{\pi}{6},\frac{\pi}{2}, \frac{5\pi}{6},...[/itex] (Is this correct?)
How do I compute find the residues at these points in the easiest way? Find the Laurent series? I'm sorry if I'm missing something stupid. I am just pressed for time and could be missing a multitude of things.
2.
Dick, Way too tired tonight and left this problem in my office! I'll post this work tomorrow morning! Sorry, I know you need something to do.
3.
So, let [itex]x-0\,\rightarrow\, x-(0+i\epsilon)\,}[/itex] :
Therefore:
[tex]I=\lim_{\epsilon\rightarrow0}\int_{-\infty}^{\infty}\frac{\sin(x)}{x-i\epsilon}dx[/tex]
[tex]I=\frac{1}{2i}\lim_{\epsilon\rightarrow0}\int_{-\infty}^{\infty}\frac{e^{ix}dx}{x-i\epsilon}-\frac{1}{2i}\lim_{\epsilon\rightarrow0}\int_{-\infty}^{\infty}\frac{e^{-ix}dx}{x-i\epsilon}[/tex]
Now if I let epsilon go to zero now, I have an integral that we discussed in class. I know the first integral is equal to [itex]i\pi[/itex] and the second [itex]-i\pi[/itex].
This simplifies to the correct result, but technically I used a principle value to compute the exponential integrals, which was against the rules of the problem. Is there another easy way to find those integrals?
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