## Residue Calculus Problem

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Mathmagic
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### Residue Calculus Problem

I'm working on a maths assignment right now, and have run into a bit of an issue with the following question:

Show that if [imath]0<b<a[/imath]
$\int_{-\infty}^\infty \frac{\cos{x}}{(x^2+a^2)(x^2+b^2)}\,dx=\frac{\pi}{a^2-b^2}(\frac{e^{-b}}{b}-\frac{e^{-a}}{a})$

I started out by expressing [imath]\cos{x}[/imath] in terms of complex exponentials. We also know that the integrated function has simple poles at [imath]\pm ai[/imath] and [imath]\pm bi[/imath]. I created a contour from -R to R along the real axis, and going about a circle of radius R in the positive sense from R to -R. Thus, there are only two poles in the interior of the contour ([imath]ai[/imath] and [imath]bi[/imath]), and finding the residues at poles yields:
$Res(f,ai)=\frac{e^{-a}+e^a}{-4i(a^2-b^2)a}$
$Res(f,bi)=\frac{e^{-b}+e^b}{4i(a^2-b^2)b}$
Using the Cauchy Integral Formula, we know that the original integral is equal to the sum of the residues multiplied by [imath]2\pi i[/imath] minus the line integral along the semi-circle contour (not including the real axis).

The first part with the residues I found was equal to
$\frac{\pi}{a^2-b^2}(\frac{e^{-b}+e^b}{2b}-\frac{e^{-a}+e^a}{2a})$
And I showed that the contour integral went to zero as R went to infinity (which is what the value of the bounds need to be) by saying f is bounded by the cosine (between 1 and -1), so the max value that f gets to with R is approximately 1/R4. I then said that the max value that the integral goes to is the length of the contour (which is simply [imath]\pi R[/imath] multiplied by the max value of f, which is [imath]\frac{\pi}{R^3}[/imath], which goes to zero as R goes to infinity.

Obviously, the result I arrived at is not what is required, and this is why I come to you for a bit of help. Did I go wrong in approximating/bounding the contour integral, or did I screw up my residues?
Last edited by Mathmagic on Fri Feb 13, 2009 7:28 am UTC, edited 1 time in total.
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eta oin shrdlu
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### Re: Residue Calculus Problem

What is the value of cos(iR) for large real R? (Hint: it's not bounded on [-1,1].)

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### Re: Residue Calculus Problem

Yup. I haven't gone through all the details, but if you're trying to bound cosine between 1 and -1 you've forgotten that we're not in Kansas anymore when we look at complex values. Maybe you can do something with a symmetry argument if you turn the contour on its side? You'd have to do something tricky to avoid integrating over a pole, though.
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Mathmagic
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### Re: Residue Calculus Problem

Okay thanks guys. So did I find the residues correctly? And what would be the best way to bound the contour integral? We've just started doing these complex line integrals and I've never really done this bounding thing before. The "general" rule that my prof gave was that it was bounded by the length of the contour multiplied by the max absolute value of the function being integrated. Is this the right approach for this question?

Oh and thanks for pointing out the cosine bound problem. I totally forgot I was dealing with cos(z) and not cos(x). cos(Ri) is just cosh(R), which isn't bounded at all. It behaves like an exponential, whose power series completely swamps the R4 term in the denominator, so I'm guessing I didn't bound it correctly in the first place.
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eta oin shrdlu
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### Re: Residue Calculus Problem

You correctly found the residues for your choice of f(z). But as you've discovered, the integral you have chosen does not converge over the large semicircle, so you can't immediately use this result to find the integral you're interested in. Can you find a related function which does have the convergence properties you want, and use its integral to find the integral of f?

cspirou
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### Re: Residue Calculus Problem

Ever heard of Jordan's lemma? It's only for curves in the upper half of the complex plane. Well on of those terms can be interpreted as being in the lower plane and goes to infinity.

It's better to do cos(x)=Re[e^(ix)] instead of breaking up into exponentials. Then just take the real part in the end. As a bonus you also get the answer for Sin(x) as well.

Mathmagic
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### Re: Residue Calculus Problem

eta oin shrdlu wrote:Can you find a related function which does have the convergence properties you want, and use its integral to find the integral of f?
cspirou wrote:It's better to do cos(x)=Re[e^(ix)] instead of breaking up into exponentials.

Thanks a lot guys. I think it should work if I choose [imath]f(z)[/imath] as:
$f(z)=\frac{e^{iz}}{(z^2+a^2)(z^2+b^2)}$
Since [imath]e^{iz}[/imath] behaves as cosine does along the real axis, and the polynomial in [imath]z^2[/imath] behaves as [imath]x^2[/imath] does along the real axis.

I'll try this and see what I come up with.

EDIT: Looks like it did the trick! Thanks again everyone for your help.
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Mathmagic
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### Re: Residue Calculus Problem

I'm working on the final review assignment for this same course and I'm trying to figure out what my prof did in the following steps:

Find the residue at [imath]z=0[/imath] of [imath]f(z)=\frac{1}{z^2\sinh{z}}[/imath]

Solution:

Since $\sinh{z}=z+\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots$ it follows that $\frac{1}{z^2\sinh{z}}=\frac{1}{z^2(z+\frac{z^3}{3!}+\cdots)}=\frac{1}{z^3}\frac{1}{1+\frac{z^2}{3!}+\cdots}=\frac{1}{z^3}(1-\frac{z^2}{3!}+\cdots)$
----

How did he go from [imath]\frac{1}{1+\frac{z^2}{3!}+\cdots}[/imath] to [imath](1-\frac{z^2}{3!}+\cdots)[/imath]?

I tried using the "normal" route and finding [imath]C_{-1}[/imath] of the Laurent series using the method found here for a pole of order 3, and I think it worked (i.e. my math worked out, but I'm not sure if I did it correctly). Clearly, that method is not ideal for this problem. What am I missing here?
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skeptical scientist
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### Re: Residue Calculus Problem

The first thing to realize is that the ellipses don't represent the same terms in both places. I'm assuming that for the rest of the problem, it didn't really matter what the ellipses were, and just having the z2 term is enough? Anyways, you can work it out from the formula 1/(1-x)=1+x+x2+x3+...: 1/(1+z3/3!+...)=1-(z3/3!+...)+(z3/3!+...)2-(z3/3!+...)3+..., and of course that's the same as 1+z3/3!+higher order terms.

Last edited by skeptical scientist on Sat Apr 25, 2009 4:51 am UTC, edited 1 time in total.
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Mathmagic
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### Re: Residue Calculus Problem

skeptical scientist wrote:The first thing to realize is that the ellipses don't represent the same terms in both places. I'm assuming that for the rest of the problem, it didn't really matter what the ellipses were, and just having the z2 term is enough?
I realize that. I interpreted the first series to be the series for sinh, and the second series to be the series for sinh but divided by z and with a (-1)n term thrown in there. If this is not the case, it may have been what was throwing me off. That's a nifty trick with the geometric series expansion though. Thank you!
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Fafnir43
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### Re: Residue Calculus Problem

By the way, handy LaTeX tip: if you use \left( and \right) instead of ( and ) the brackets will scale to fit whatever's inside them, so you can get this: $\left(\frac{a}{b}\right)$ rather than this: $(\frac{a}{b})$ by using \left(\frac{a}{b}\right) rather than (\frac{a}{b}). Took me a while to notice that!
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Mathmagic
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### Re: Residue Calculus Problem

I just took the final exam today, and it went well except for one snag I caught when trying to find the following integral:

$\int_{-\infty}^{\infty} \frac{x\sin{kx}}{x^2+1} \,dx.$

I calculated the residues correctly, but when it came to using Cauchy's integral formula, I couldn't find a contour that caused the complex line integral to converge. I reasoned it like this:

The value of the line integral is bound by the length of the path multiplied by the maximum value of the absolute value of f(x). For a semicircle extending from R to -R in the positive region of the complex plane, this gives:

$\pi{R}\frac{R\sin{Rk}}{R^2+1}$
The limit of this function as R approaches infinity is bound on [imath][-\pi,\pi][/imath]

Did I choose the wrong contour or did I just bound it incorrectly?
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Nitrodon
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### Re: Residue Calculus Problem

The bound is incorrect, since sin(z) is not bounded in the complex plane.

Replace sin(kx) with the imaginary part of e^(ikx). I suspect this will make the problem easier, but I haven't bothered working it out.

odulwa
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### Re: Residue Calculus Problem

To get your last problem, you have to use jordan's lemma as someone pointed out.

$\int_{-\infty}^{\infty} \frac{x\sin{kx}}{x^2+1} \,dx.=Im\left(\int_{C_R\cup[-\infty,\infty]} \frac{ze^{ikz}}{z^2+1} \,dz.\right)$

If the integral along the half circle is zero as R goes to infinity.
To bound an integral on this form on a positive half circle in the complex plane, by jordan's lemma:

$\left| \int_{C_R} \frac{ze^{ikz}}{z^2+1} \,dz.\right|<\frac{max(|f(z)|)\pi}{k}$

and since [imath]max(|f(z)|)[/imath] is zero as R goes to infinity, you're good to go.

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### Re: Residue Calculus Problem

Mathmagic wrote:I just took the final exam today, and it went well except for one snag I caught when trying to find the following integral:

$\int_{-\infty}^{\infty} \frac{x\sin{kx}}{x^2+1} \,dx.$

I calculated the residues correctly, but when it came to using Cauchy's integral formula, I couldn't find a contour that caused the complex line integral to converge. I reasoned it like this:

The value of the line integral is bound by the length of the path multiplied by the maximum value of the absolute value of f(x). For a semicircle extending from R to -R in the positive region of the complex plane, this gives:

$\pi{R}\frac{R\sin{Rk}}{R^2+1}$
The limit of this function as R approaches infinity is bound on [imath][-\pi,\pi][/imath]

Did I choose the wrong contour or did I just bound it incorrectly?

You can write sin(kx) as im(exp(ikx)), and note that x*exp(ikx) is (1/i)(d/dk)exp(ikx). this gives you an integral which can be evaluated with an infinite semicircle in the upper-half of the complex plane.
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LaserGuy
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### Re: Residue Calculus Problem

Mathmagic wrote:I'm working on the final review assignment for this same course and I'm trying to figure out what my prof did in the following steps:

Find the residue at [imath]z=0[/imath] of [imath]f(z)=\frac{1}{z^2\sinh{z}}[/imath]

Solution:

Since $\sinh{z}=z+\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots$ it follows that $\frac{1}{z^2\sinh{z}}=\frac{1}{z^2(z+\frac{z^3}{3!}+\cdots)}=\frac{1}{z^3}\frac{1}{1+\frac{z^2}{3!}+\cdots}=\frac{1}{z^3}(1-\frac{z^2}{3!}+\cdots)$
----

How did he go from [imath]\frac{1}{1+\frac{z^2}{3!}+\cdots}[/imath] to [imath](1-\frac{z^2}{3!}+\cdots)[/imath]?

I tried using the "normal" route and finding [imath]C_{-1}[/imath] of the Laurent series using the method found here for a pole of order 3, and I think it worked (i.e. my math worked out, but I'm not sure if I did it correctly). Clearly, that method is not ideal for this problem. What am I missing here?

If you're interested, in this case, to go from one series to the other, just do long division of polynomials and it should work out.

Mathmagic
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### Re: Residue Calculus Problem

Thanks for the help guys. I just realized that this same issue tripped me up on the problem I mentioned in the OP. D'oh! I guess now I'll remember this for good.
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