# Help with complex integration

Show 40 post(s) from this thread on one page
Page 1 of 3 123 Last
• June 11th 2010, 06:21 AM
Opaque02
Help with complex integration
Hey all

I've got the integral

{int}_0^infinity ((x^2/(1+x^6)) dx [sorry for this, I'm not sure how to use the math thing here]

The question is asking us to evaluate it using complex integrals. All I have with me is a textbook that has one example of this which I don't understand, and lecture notes, which have an example, but to do it, it says to look at the example in the book, which doesn't help me :(.

Anyways, if anyone could help sometime, I would greatly appreciate it, as I have my complex analysis exam in a few days and apparently I need to know how to do this :)

Thanks in advanced :D
• June 11th 2010, 06:25 AM
Ackbeet
So you need to compute $\int_{0}^{\infty}\frac{x^{2}}{1+x^{6}}\,dx$, right? (You can click on the integral to see what code I used to generate the output.)

I would say the first step would be to decide a contour to use. Are you familiar with residue theory?
• June 11th 2010, 06:26 AM
Ackbeet
Incidentally, these posts really ought to be in the Analysis, Topology and Differential Geometry section.
• June 11th 2010, 06:44 AM
chisigma
An very similar integral has been solved in...

http://www.mathhelpforum.com/math-he...-integral.html

Here is...

$\int_{-\infty}^{+\infty} \frac{x^{2}}{1+x^{6}}\cdot dx = 2 \pi i \sum_{k} r_{k}$ (1)

... where the $r_{k}$ are the residues of $f(z) = \frac{z^{2}}{1+z^{6}}$ with positive imaginary part that are at $z=e^{i (2k+1) \frac{\pi}{6}}$, $k=0,1,2$. All the computations are the same and, of course, because the limits of integration are $0$ and $\infty$, the integral (1) must be divided by 2...

Kind regards

$\chi$ $\sigma$
• June 11th 2010, 07:19 AM
Opaque02
Quote:

Originally Posted by Ackbeet
So you need to compute $\int_{0}^{\infty}\frac{x^{2}}{1+x^{6}}\,dx$, right? (You can click on the integral to see what code I used to generate the output.)

I would say the first step would be to decide a contour to use. Are you familiar with residue theory?

Vaguely, but once again, my lecturer refers only to the book, which I don't understand. Could you perhaps give me a quick run through on how to do it?

Also thanks for the math code :D.

Many thanks
• June 11th 2010, 07:22 AM
Ackbeet
Like I said, the first step is to choose a contour. Here are some guidelines:

1. Pick a contour that includes the range of values over which the original integral runs.

2. Pick a simple, closed contour.

3. Pay attention to where the poles of the integrand are located relative to your closed contour. Sometimes you can simplify things by not including any poles, or by only including poles where it is easy to compute residues, etc.

4. Pick a contour that has a good deal of symmetry. That will usually simplify expressions.

So, what's your candidate?
• June 11th 2010, 07:29 AM
Opaque02
Quote:

Originally Posted by Ackbeet
Like I said, the first step is to choose a contour. Here are some guidelines:

1. Pick a contour that includes the range of values over which the original integral runs.

2. Pick a simple, closed contour.

3. Pay attention to where the poles of the integrand are located relative to your closed contour. Sometimes you can simplify things by not including any poles, or by only including poles where it is easy to compute residues, etc.

4. Pick a contour that has a good deal of symmetry. That will usually simplify expressions.

So, what's your candidate?

Hmm. I'm still confused about how all the stuff like contours work. So, I can basically pick any contour, and I will still get the same answer?

So, for this function, would I choose a nice closed contour, like a circle or something along those lines? Or is there something nicer that will be easier to understand?
• June 11th 2010, 07:41 AM
Ackbeet
If you make no mistakes, there's loads of freedom in choosing contours. Let me outline the general idea of the solution. You pick a contour such as I have described. Let's say that the contour has various "legs of the journey", or is partitioned into the following contours: $C_{1}, C_{2}, C_{3},\dots,C_{n}$, where one of them, $C_{j}$, corresponds to the original integral, and $\cup_{k=1}^{n}C_{k}=C$. You've got an original integrand $f(x)$. Now, let's suppose further that you know what the value of
$\oint_{C}f(z)\,dz$ is. What you do is set

$\int_{C_{1}}f(z)\,dz+\dots+\int_{C_{j}}f(z)\,dz+\d ots+\int_{C_{n}}f(z)\,dz=\oint_{C}f(z)\,dz$.

The hope is that all the $\int_{C_{k}}f(z)\,dz$, except the one you're interested in, are very easy to do. You might be able to eliminate them on the basis of an ML estimate, for example. Then, you simply shove everything except the integral in which you're interested over to the other side:

$\int_{C_{j}}f(z)\,dz=\oint_{C}f(z)\,dz-\sum_{k=1, k\not=j}^{n}\int_{C_{k}}f(z)\,dz$.

Everything on the RHS is either known or easy to compute. Voila! You're done. That's the high-level view of contour integration as it's used to evaluate real-valued integrals. Make sense?
• June 11th 2010, 07:44 AM
Opaque02
Quote:

Originally Posted by Ackbeet
If you make no mistakes, there's loads of freedom in choosing contours. Let me outline the general idea of the solution. You pick a contour such as I have described. Let's say that the contour has various "legs of the journey", or is partitioned into the following contours: $C_{1}, C_{2}, C_{3},\dots,C_{n}$, where one of them, $C_{j}$, corresponds to the original integral, and $\cup_{k=1}^{n}C_{k}=C$. You've got an original integrand $f(x)$. Now, let's suppose further that you know what the value of
$\oint_{C}f(z)\,dz$ is. What you do is set

$\int_{C_{1}}f(z)\,dz+\dots+\int_{C_{j}}f(z)\,dz+\d ots+\int_{C_{n}}f(z)\,dz=\oint_{C}f(z)\,dz$.

The hope is that all the $\int_{C_{k}}f(z)\,dz$, except the one you're interested in, are very easy to do. You might be able to eliminate them on the basis of an ML estimate, for example. Then, you simply shove everything except the integral in which you're interested over to the other side:

$\int_{C_{j}}f(z)\,dz=\oint_{C}f(z)\,dz-\sum_{k=1, k\not=j}^{n}\int_{C_{k}}f(z)\,dz$.

Everything on the RHS is either known or easy to compute. Voila! You're done. That's the high-level view of contour integration as it's used to evaluate real-valued integrals. Make sense?

Wow. Yeah, that makes so much more sense. So, what are some examples of some contours that we could choose? Would we choose something nice and simple, like a circle, or would we try for something different? This suddenly is making a lot more sense than it did 10 minutes ago :)
• June 11th 2010, 07:49 AM
Ackbeet
First thing I would do is locate the poles of your integrand. So where are those?
• June 11th 2010, 07:55 AM
Opaque02
Quote:

Originally Posted by Ackbeet
First thing I would do is locate the poles of your integrand. So where are those?

So, that's when $x^6+1=0$? So, the sixth roots of -1. This gives e^(i(-\pi/2+2n\pi)), with n=0, 1, 2, 3, 4 and 5? Is that what you mean?

On a side note, how do I put the e^(i(-\pi/2+2n\pi)) into math code? Whenever I tried, it just put the first bracket in the power, not anything else :(
• June 11th 2010, 08:02 AM
Ackbeet
That's the idea. I'm not sure you quite have the poles. [To include more than just one thing in an exponent, use curly braces: compare the code e^{2x} with e^2x, and you'll see what I mean (Use the Preview Post feature: it's wonderful!).] The poles are $e^{i(-\pi/2+n\pi/3)}$, I believe, for $n=0,1,\dots,5$. Now here's a rule about contours: you can't pick a contour that runs right over a pole. You have to avoid any poles in your contour (otherwise, you'd divide by zero). So, what do you think? Does any contour jump out at you as a candidate?
• June 11th 2010, 08:04 AM
Ackbeet
Sorry: when I said "avoid poles in the contour", I meant that no poles can be in the path of the contour. The closed region of the entire contour $C$ can contain poles, no problem.
• June 11th 2010, 08:07 AM
Opaque02
Quote:

Originally Posted by Ackbeet
That's the idea. I'm not sure you quite have the poles. [To include more than just one thing in an exponent, use curly braces: compare the code e^{2x} with e^2x, and you'll see what I mean (Use the Preview Post feature: it's wonderful!).] The poles are $e^{i(-\pi/2+n\pi/3)}$, I believe, for $n=0,1,\dots,5$. Now here's a rule about contours: you can't pick a contour that runs right over a pole. You have to avoid any poles in your contour (otherwise, you'd divide by zero). So, what do you think? Does any contour jump out at you as a candidate?

Ah, ok. I see what i did. Or, rather, didn't do. I needed to put it to the power of $1/6$, since we wanted the sixth pole. Ok, so I get $e^{i(-\pi/12+n\pi/3)}$. Would I use a circle as my contour, since this gives a nice, closed, symmetrical shape going around the poles? I think you're managing to teach me something I haven't been able to understand for 7 weeks now :D
• June 11th 2010, 08:11 AM
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 $\oint_{C}f(z)\,dz=0$. 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.
Show 40 post(s) from this thread on one page
Page 1 of 3 123 Last