# Thread: Help with complex integration

1. I mean "analytic integrand inside the region enclosed by the contour".

2. Originally Posted by Ackbeet
Don't forget that you have to include the positive real axis in your contour, because that corresponds to the limits of your original integral. You may or may not want to include poles inside your contour. If you include no poles, then assuming an analytic integrand, you have $\displaystyle \oint_{C}f(z)\,dz=0$. That's easy to compute! On the other hand, if all you have are simple poles (like in this case), you can use residue theory easily enough.
Ah, OK. So, if we wanted the integral over all space, we would choose a full circle, but because its only from 0 to infinity, we only choose the semi circle in the upper half plane? So, for the residues, don't we need to find the series expansion for that, then find the coefficient of the -1 power term, and multiply that by $\displaystyle 2i\pi$? Because that's another place I have trouble with, the series expansions . Blah I hate my lecturer for not teaching this well

3. Your choice of contour seems good to me. That is, I would choose the following:

$\displaystyle C_{1}:-R\to R$ on the real axis, (you'll let $\displaystyle R\to\infty$ later) followed by the semicircle of radius $\displaystyle R$ in the upper half plane going back to $\displaystyle -R$ on the real axis, which we'll call $\displaystyle C_{2}$. Picture that? Technically, you're interested in the second half of $\displaystyle C_{1}$ after you've let $\displaystyle R\to\infty$.

I should point out, that, because when using the residue theory you're assumed to have traversed a contour in the clockwise (positive) direction, you'll need to change the sign of $\displaystyle \oint_{C}f(z)\,dz$ to get the correct answer.

So we've got a simple closed contour that avoids all poles directly in its path, and includes the original limits of integration. What next?

4. Originally Posted by Ackbeet
Your choice of contour seems good to me. That is, I would choose the following:

$\displaystyle C_{1}:-R\to R$ on the real axis, (you'll let $\displaystyle R\to\infty$ later) followed by the semicircle of radius $\displaystyle R$ in the upper half plane going back to $\displaystyle -R$ on the real axis, which we'll call $\displaystyle C_{2}$. Picture that? Technically, you're interested in the second half of $\displaystyle C_{1}$ after you've let $\displaystyle R\to\infty$.

I should point out, that, because when using the residue theory you're assumed to have traversed a contour in the clockwise (positive) direction, you'll need to change the sign of $\displaystyle \oint_{C}f(z)\,dz$ to get the correct answer.

So we've got a simple closed contour that avoids all poles directly in its path, and includes the original limits of integration. What next?
I would say next we find the residues. Is there an easier way to do it rather than Laurent expansion? I really should learn it, but one thing at a time

5. Well, all residues are concerned with the Laurent expansion. However, there are cute ways of computing them if you know the order of the pole. All our poles are simple, which are the easiest. Rule 1 on page 196 of Gamelin states the following:

If $\displaystyle f(z)$ has a simple pole at $\displaystyle z_{0}$, then

$\displaystyle \text{Res}[f(z),z_{0}]=\lim_{z\to z_{0}}(z-z_{0})\,f(z)$.

Try that on for size. At which poles are you going to compute the residues?

6. Originally Posted by Ackbeet
Well, all residues are concerned with the Laurent expansion. However, there are cute ways of computing them if you know the order of the pole. All our poles are simple, which are the easiest. Rule 1 on page 196 of Gamelin states the following:

If $\displaystyle f(z)$ has a simple pole at $\displaystyle z_{0}$, then

$\displaystyle \text{Res}[f(z),z_{0}]=\lim_{z\to z_{0}}(z-z_{0})\,f(z)$.

Try that on for size. At which poles are you going to compute the residues?
Well, we want the ones that are in the positive imaginary direction, so we would want $\displaystyle c_0=e^{i\pi/6}, c_1=i, c_2=e^{5i\pi/6}$

7. Better to say "upper half plane" than "positive imaginary direction" - less confusing. Those are indeed the poles at which to compute the residues. Once you have the residues, the residue theorem gives you the value of $\displaystyle \oint_{C}f(z)\,dz$, modulo a minus sign for traversing the boundary the "wrong" way. You're left with two pieces: the $\displaystyle C_{1}$ and $\displaystyle C_{2}$ integrals. What do you think about them?

8. Originally Posted by Ackbeet
Better to say "upper half plane" than "positive imaginary direction" - less confusing. Those are indeed the poles at which to compute the residues. Once you have the residues, the residue theorem gives you the value of $\displaystyle \oint_{C}f(z)\,dz$, modulo a minus sign for traversing the boundary the "wrong" way. You're left with two pieces: the $\displaystyle C_{1}$ and $\displaystyle C_{2}$ integrals. What do you think about them?
Ah, true about the upper half plane part. Think I even used that in a previous post. Getting tired, almost 3AM here. So, I get the residue to be $\displaystyle \pi/3$. So, for $\displaystyle C_1$ and $\displaystyle C_2$, I remember something about using the ML notation thing. Is that what we have to do for it?

9. Your computation seems correct. In fact, it's more correct than I would have done. True confessions: I just looked it up, and the positive orientation of a contour is counterclockwise (the interior of the region is to your left as you traverse in the positive direction). So forget what I said about the minus sign.

The ML estimate looks like this: suppose $\displaystyle \gamma$ is a piecewise smooth curve. If $\displaystyle h(z)$ is a continuous function on $\displaystyle \gamma$, then

$\displaystyle \left|\int_{\gamma}h(z)\,dz\right|\le\int_{\gamma} |h(z)|\,|dz|$.

Further, if $\displaystyle \gamma$ has length $\displaystyle L$, and $\displaystyle |h(z)|\le M$ on $\displaystyle \gamma$, then

$\displaystyle \left|\int_{\gamma}h(z)\,dz\right|\le ML$.

That's the theorem on page 105 of Gamelin. This estimate might be useful for your $\displaystyle C_{2}$, but not for $\displaystyle C_{1}$. Why, might you ask, is it not useful for $\displaystyle C_{1}$? Because you want to know its exact value! However, if you can prove that the integral over the $\displaystyle C_{2}$ is zero using the ML estimate, then you're essentially done. How could you use the ML estimate to show that the $\displaystyle C_{2}$ integral is zero (as I think you'll find it is, after you've taken the limit as $\displaystyle R\to\infty$)?

10. Originally Posted by Ackbeet
Your computation seems correct. In fact, it's more correct than I would have done. True confessions: I just looked it up, and the positive orientation of a contour is counterclockwise (the interior of the region is to your left as you traverse in the positive direction). So forget what I said about the minus sign.

The ML estimate looks like this: suppose $\displaystyle \gamma$ is a piecewise smooth curve. If $\displaystyle h(z)$ is a continuous function on $\displaystyle \gamma$, then

$\displaystyle \left|\int_{\gamma}h(z)\,dz\right|\le\int_{\gamma} |h(z)|\,|dz|$.

Further, if $\displaystyle \gamma$ has length $\displaystyle L$, and $\displaystyle |h(z)|\le M$ on $\displaystyle \gamma$, then

$\displaystyle \left|\int_{\gamma}h(z)\,dz\right|\le ML$.

That's the theorem on page 105 of Gamelin. This estimate might be useful for your $\displaystyle C_{2}$, but not for $\displaystyle C_{1}$. Why, might you ask, is it not useful for $\displaystyle C_{1}$? Because you want to know its exact value! However, if you can prove that the integral over the $\displaystyle C_{2}$ is zero using the ML estimate, then you're essentially done. How could you use the ML estimate to show that the $\displaystyle C_{2}$ integral is zero (as I think you'll find it is, after you've taken the limit as $\displaystyle R\to\infty$)?
So we have

$\displaystyle \left|\int_{C_{2}}(x^2/{x^6+1})\,dx\right|\le {ML}$

We know that L is just the length of the semi circle we have, so $\displaystyle R\pi$, and M is something which I don't remember, but know how to calculate. We get $\displaystyle M=R^2/{R^6+1}$. Since, as $\displaystyle lim_{R\to\infty} {M}$, we get the denominator going to 0 much faster than the numerator, so we that that integral term equal 0. So, we just need to find the $\displaystyle C_{1}$, and then we can find the final thing . Would we do the ML for the $\displaystyle C_{1}$ too?

11. You should be a little more careful with your parentheses, especially when adding things together in denominators. *ahem* Your basic idea for the ML estimate is correct. However, without the parentheses, it's actually incorrect. The basic idea is correct, however. The integrand beats out the (correct) value of L you gave, so the ML is zero in the limit. Hence the integral over $\displaystyle C_{2}$ is zero. As for $\displaystyle C_{1}$, I would simply recognize that

$\displaystyle 2\int_{0}^{\infty}\frac{x^{2}}{1+x^{6}}\,dx=\int_{ C_{1}}\frac{x^{2}}{1+x^{6}}\,dx$,

since the integrand is even.

12. Curious to see what your final answer was. What did you get?

Regards.

13. I got pi/6. Also, is it just me or is there no more Math button? Maybe I just fail and can't see it

14. $\displaystyle \pi/6$ is the correct answer. Yes, the math button is gone for now. I don't know if the admins will bring it back or not. In the meantime, you can manually add the [tex] tags around LaTeX code.

15. Ahh, good point. Thanks for all the help

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