ReEvaluating real integrals using residue calculus

[loom91]

Hi,

I recently came to know that some real integrals become easier to evaluate using the techniques of residue calculus from complex analysis. I thought that this would be a good tool to pickup apart from the standard techniques of by parts and substitution that we are taught. I have a few questions in this regard.

1)How can I recognise an integral that will be more easily evaluated using residue calculus than ordinary methods?
2)How do I find out if the complex arc part of the contour integral vanishes?
3)Can you give me a few simple problems to practice my rudimentary knowledge?

Thanks.

Molu

 

[data1217]

Hi,

I recently came to know that some real integrals become easier to evaluate using the techniques of residue calculus from complex analysis. I thought that this would be a good tool to pickup apart from the standard techniques of by parts and substitution that we are taught. I have a few questions in this regard.

1)How can I recognise an integral that will be more easily evaluated using residue calculus than ordinary methods?
2)How do I find out if the complex arc part of the contour integral vanishes?
3)Can you give me a few simple problems to practice my rudimentary knowledge?

Thanks.

Molu

1) There are some standard formats/ combinations of functions.
2) Sorry, I am not very clear about this question.
3) Question: $$I = \int_{0}^{\Pi} \frac{1}{{(a+b \cosx)^2}} dx$$
Answer : $$I = \frac {ab^2 \Pi}{(\sqrt{a^2-b^2})^3}$$

 

[loom91]

1) There are some standard formats/ combinations of functions.
2) Sorry, I am not very clear about this question.
3) Question: $$I = \int_{0}^{\Pi} \frac{1}{{(a+b \cosx)^2}} dx$$
Answer : $$I = \frac {ab^2 \Pi}{(\sqrt{a^2-b^2})^3}$$

1)What are these formats?

2)When evaluating the real integral, a contour integral is evaluated using the residue theorem over a contour only a part of which lies in the real line. The desired real integral is then found if the part of the contour lying outside the real line can be shown to vanish. How do I show this or at least recognise that this will be the case?

3)Are you sure that integral is right? You seem to have missed some term, otherwise the integrand is a constant function.

 

[saltydog]

Improper integrals of rational functions in which the denominator is different from zero for all real x and the degree of the denominator is at least 2 higher than the numerator. Like:

$$\int_{-\infty}^{\infty} \frac{dx}{(1+x^2)^2}$$

$$\int_{-\infty}^{\infty}\frac{x^3}{1+x^8}dx$$

The integral over the arc goes to zero since the absolute value of the integrand goes to zero faster than $\pi R$ goes to infinity. This is a consequence of the ML-inequality commonly used in CA to place upper bounds on contour integrals: Since the degree of the denominator is at least 2 higher than numerator, then:

$$|f(x)|<\frac{k}{|z|^2}$$

for sufficiently large R. Now, since you're integrating over a half-circle arc in these problems, the length of the contour is just $\pi R$ thus the integral over this region cannot be larger than:

$$\left|\int_{C} f(z)dz\right|<\frac{k}{R^2} \pi R=\frac{k \pi}{R}$$

since $|z|=R$ on C.

This goes to zero as R goes to infinity.

Rational functions like the above, multiplied by Sin(x) and Cos(x) can also be evaluated by residue integration too like:

$$\int_{-\infty}^{\infty}\frac{Sin(2x)}{x^2+x+1}dx$$

$$\int_{-\infty}^{\infty}\frac{Cos(kx)}{x^2+1}dx$$

 

[loom91]

Improper integrals of rational functions in which the denominator is different from zero for all real x and the degree of the denominator is at least 2 higher than the numerator. Like:

$$\int_{-\infty}^{\infty} \frac{dx}{(1+x^2)^2}$$

$$\int_{-\infty}^{\infty}\frac{x^3}{1+x^8}dx$$

The integral over the arc goes to zero since the absolute value of the integrand goes to zero faster than $\pi R$ goes to infinity. This is a consequence of the ML-inequality commonly used in CA to place upper bounds on contour integrals: Since the degree of the denominator is at least 2 higher than numerator, then:

$$|f(x)|<\frac{k}{|z|^2}$$

for sufficiently large R. Now, since you're integrating over a half-circle arc in these problems, the length of the contour is just $\pi R$ thus the integral over this region cannot be larger than:

$$\left|\int_{C} f(z)dz\right|<\frac{k}{R^2} \pi R=\frac{k \pi}{R}$$

since $|z|=R$ on C.

This goes to zero as R goes to infinity.

Rational functions like the above, multiplied by Sin(x) and Cos(x) can also be evaluated by residue integration too like:

$$\int_{-\infty}^{\infty}\frac{Sin(2x)}{x^2+x+1}dx$$

$$\int_{-\infty}^{\infty}\frac{Cos(kx)}{x^2+1}dx$$

Thanks saltydog for clearing it up. I will now try to evaluate the examples you gave. One more question, can residue calculus be usefull in integrations with finite limits and what would one do with the arc then?

 

[saltydog]

Very good. Residue integration can be used for bounded integrands of sines and cosines. Like this:

$$\int_0^{2\pi} Cos(\theta)d\theta$$

Make the substitution $z=e^{i\theta}$ noting that:

$$Cos(\theta)=1/2(e^{i\theta}+e^{-i\theta})$$

Since $\theta$ makes one complete circuit, then z represents a circle around the (complex) origin thus the contour is closed and the integral becomes:

$$\frac{1}{2i}\oint(1+\frac{1}{z^2})dz$$

The residue at the origin is zero so the integral is zero as expected. How about this one then:

$$\int_0^{2\pi}Cos^2 \theta d\theta$$

That one's not zero. Can you prove it's $\pi$ using residue integration? Usually though this is done for more complex integrals like:

$$\int_0^{2\pi}\frac{Cos(\theta)}{3+Cos(\theta)}d\theta$$

Remember, only count the residues inside the contour.

 

[loom91]

Very good. Residue integration can be used for bounded integrands of sines and cosines. Like this:

$$\int_0^{2\pi} Cos(\theta)d\theta$$

Make the substitution $z=e^{i\theta}$ noting that:

$$Cos(\theta)=1/2(e^{i\theta}+e^{-i\theta})$$

Since $\theta$ makes one complete circuit, then z represents a circle around the (complex) origin thus the contour is closed and the integral becomes:

$$\frac{1}{2i}\oint(1+\frac{1}{z^2})dz$$

The residue at the origin is zero so the integral is zero as expected.

I'm not sure I understand this conversion into contour integral. Could you fill in some of the steps?

Making the substitution $z=e^{i\theta}$, I get

$$\frac{1}{2}\oint(z+\frac{1}{z})dz$$

over the unit circle which seems to have a simple pole at the origin and a residue of 1, giving an integral of [itex]\pi i[/tex] using the residue theorem. What am I doing wrong?

 

[loom91]

How about this one then:

$$\int_0^{2\pi}Cos^2 \theta d\theta$$

That one's not zero. Can you prove it's $\pi$ using residue integration?

First I made the substitution $z=e^{i\theta}$, giving me

$$\oint [\frac {z^4 + 1}{4z^2} + \frac {1}{2}]dz$$ around the unit circle. The contour integral of the constant term vanishes. Next I evaluate the residue at the second-order pole at origin using the formula

$$Res(f, c) = lim_{z\rightarrow c} \frac {d^{n-1}}{dz^{n-1}} [f(z)(z-c)^n]$$, getting 0 (which is surprising, can a residue at a pole be 0?).

Using residue theorem, the integral evaluates to

$$2\pi i * 0 = 0$$

Hard luck again. What went wrong?

 

[shmoe]

What am I doing wrong?

If $z=e^{i\theta}$ then $dz=ie^{i\theta}d\theta$

You can generally deal with any rational function of sin's and cos's, this substitution will then yield a rational function of z. It's just a matter of locating the poles inside the unit circle.

You should also keep in mind transformations that can get you into the correct form, like

$$\int_0^\pi\frac{d\theta}{2+\cos{\theta}}$$

isn't an integral over a full circle with that substitution. How to fix this?

(which is surprising, can a residue at a pole be 0?)

yes it can. what's the residue of $z^{-2}$ at z=0?


Do you have a complex analysis text? Ahlfors has a small section on the main common types. A book aimed more at the engineering crowd will probably be more example and problem rich though.

 

[saltydog]

Thanks Shmoe. I think the differential of z caused the problem for Loom. It's just:

$$dz=ie^{i\theta}d\theta$$

so then:

$$d\theta=\frac{1}{iz}dz$$

all the rest should fall into place. Also, I got most of the examples I gave above from the quintessential engineering math reference book: "Advanced Engineering Mathematics" by Kreyszig. It would certainly not do for math purist but . . . I ain't proud.

 

[loom91]

Oh, I forgot to transform the differential form, how stupid of me.:">

I don't have a complex analysis text, but I will be getting Spivak, which I think has a few chapters on complex analysis. Don't know whether it covers residues though.

 

[loom91]

Very good. Residue integration can be used for bounded integrands of sines and cosines. Like this:

$$\int_0^{2\pi} Cos(\theta)d\theta$$

Make the substitution $z=e^{i\theta}$ noting that:

$$Cos(\theta)=1/2(e^{i\theta}+e^{-i\theta})$$

Since $\theta$ makes one complete circuit, then z represents a circle around the (complex) origin thus the contour is closed and the integral becomes:

$$\frac{1}{2i}\oint(1+\frac{1}{z^2})dz$$

The residue at the origin is zero so the integral is zero as expected.

Got that one.

How about this one then:

$$\int_0^{2\pi}Cos^2 \theta d\theta$$

That one's not zero. Can you prove it's $\pi$ using residue integration?

Problem again, I get 2pi.

After substitution, I have

$$d\theta = \frac {1}{iz} dz$$
$$cos^2\theta = \frac{1}{4}(z^2+\frac{1}{z^2})+\frac {1}{2}$$

Making the integral

$$I = \oint \frac {1}{iz}[\frac {1}{4}(z^2+\frac {1}{z^2})+\frac{1}{2}]dz =-i\oint \frac {z^4+2z^2+1}{4z^3} dz$$

$$f(z)=\frac {z^4+2z^2+1}{4z^3}$$

$$Res(f, 0) = lim_{z\rightarrow 0} \frac {d^2}{dz^2} (\frac{z^4+2z^2+1}{4z^3}z^3) = 1$$

$$I= -i * 2\pi i * 1 = 2\pi$$

What's wrong this time? Thanks for your patience.

 

[loom91]

If $z=e^{i\theta}$ then $dz=ie^{i\theta}d\theta$

You can generally deal with any rational function of sin's and cos's, this substitution will then yield a rational function of z. It's just a matter of locating the poles inside the unit circle.

You should also keep in mind transformations that can get you into the correct form, like

$$\int_0^\pi\frac{d\theta}{2+\cos{\theta}}$$

isn't an integral over a full circle with that substitution. How to fix this?

Substitute $z=e^{i2\theta}$? However this gives a rather poor form of the integrand $cos\theta = \frac{1}{2}(\sqrt{z}+\sqrt{\frac{1}{z}})$, which doesn't seem easy to integrate using residues.

Another possible method is to add a real portion from +1 to -1 to close the contour giving a half circle, in which case the function would have to be separately integrated over the real portion and the result substracted from the total integral to get the actual result. But this may be easier (being a rational function) than the original trigonometric one.

But this one also wouldn't work because the pole would lie on the contour. Since the function will be defined almost everywhere on the real line, can I ignore the pole at the origin when integrating the real part and take an improper integral on both sides of the origin?

 

[shmoe]

there's a problem with your residue calculation. you have:

$$f(z)=\frac{z}{4}+\frac{1}{2z}+\frac{1}{4z^3}$$

so the residue will be 1/2, it's the coefficient of the 1/z term in the Laurent series. Nothing fancy is needed to find the residue here, it's already a Laurent series.

The "multiply by z^3, take second derivative" that you tried to do as you had a pole of order 3 should be "multiply by z^3/2!, take second derivative". There's a division by 1/(n-1)! to take into account the (n-1)! you get when you differentiate n-1 times. Remember how they derived this from the Laurent sereis?

 

[shmoe]

Another possible method is to add a real portion from +1 to -1 to close the contour giving a half circle, in which case the function would have to be separately integrated over the real portion and the result substracted from the total integral to get the actual result. But this may be easier (being a rational function) than the original trigonometric one.

But this one also wouldn't work because the pole would lie on the contour. Since the function will be defined almost everywhere on the real line, can I ignore the pole at the origin when integrating the real part and take an improper integral on both sides of the origin?

You couldn't ignore this pole. If you wanted the upper half circle to be equal to an integral from [-1,1] (+ some residues) you need to have a closed contour. It would be possible to avoid this pole with a small half circle whose radius tends to 0 as you are taking your improper integral approaching this pole, but this is more work than is needed.

Notice:

$$\int_0^\pi\frac{d\theta}{2+\cos{\theta}}=\frac{1}{2}\int_0^{2\pi}\frac{d\theta}{2+\cos{\theta}}$$

and now you will get a full circle to integrate over.

 

[saltydog]

Got that one.



Problem again, I get 2pi.


$$Res(f, 0) = lim_{z\rightarrow 0} \frac {d^2}{dz^2} (\frac{z^4+2z^2+1}{4z^3}z^3) = 1$$

$$I= -i * 2\pi i * 1 = 2\pi$$

What's wrong this time? Thanks for your patience.

You forgot the factoral coefficient in the calculation of a residue of higher order, you know, the :

$$\frac{1}{(m-1)!}$$

part. That would have one-halfed it down to 1/2 or . . . like Shmoe said, it's already in a Laurent series form with coefficient 1/2.

 

[loom91]

You forgot the factoral coefficient in the calculation of a residue of higher order, you know, the :

$$\frac{1}{(m-1)!}$$

part. That would have one-halfed it down to 1/2 or . . . like Shmoe said, it's already in a Laurent series form with coefficient 1/2.

Of course, it was already a Laurent series, and I forgot the factorial. No wonder I'm never good at math. Finally I've got this one. I'll move onto the next. Thanks to all.

 

[loom91]

You couldn't ignore this pole. If you wanted the upper half circle to be equal to an integral from [-1,1] (+ some residues) you need to have a closed contour. It would be possible to avoid this pole with a small half circle whose radius tends to 0 as you are taking your improper integral approaching this pole, but this is more work than is needed.

Notice:

$$\int_0^\pi\frac{d\theta}{2+\cos{\theta}}=\frac{1}{2}\int_0^{2\pi}\frac{d\theta}{2+\cos{\theta}}$$

and now you will get a full circle to integrate over.

Are you using some result or theorem here? I haven't seen a result saying constants in the limits can be taken out of the integral. In fact I can see only a particular case where this would be possible, when the integrand had a linear primitive.

 

[shmoe]

symmetry, $$\cos(\theta)=\cos(2\pi-\theta)$$. The integral over 0,pi will be the same as the integral over pi, 2pi

 

[loom91]

Usually though this is done for more complex integrals like:

$$\int_0^{2\pi}\frac{Cos(\theta)}{3+Cos(\theta)}d\theta$$

Remember, only count the residues inside the contour.

Hmm, I get the answer $$\frac{1}{2} \pi (4-3\sqrt {2})$$. That's not the right answer, is it?

 

[saltydog]

Hello Loom. So you did the z-substitution and got:

$$\frac{1}{i}\oint \frac{z^2+1}{z(z^2+6z+1)}dz$$

right?

So we're integrating around a circle of radius 1. Now, when you factor the denom, only 0 and one other root are poles inside this circle. Calculating the residues of both I get:

$$\mathop\text{Res}\limits_{z=0}\left\{\frac{z^2+1}{z(z^2+6z+1)}\right\}=1$$

$$\mathop\text{Res}\limits_{z=-3+2\sqrt{2}}\left\{\frac{z^2+1}{z(z^2+6z+1)}\right\} =-\frac{3}{2\sqrt{2}}$$

You know, I think Mathematica helps in this regards. You can at least see what the answer is and then work it out by hand towards that result.

Edit: just noticed the answer you reported is correct. I misread it. Nice job.

 

[loom91]

Hello Loom. So you did the z-substitution and got:

$$\frac{1}{i}\oint \frac{z^2+1}{z(z^2+6z+1)}dz$$

right?

So we're integrating around a circle of radius 1. Now, when you factor the denom, only 0 and one other root are poles inside this circle. Calculating the residues of both I get:

$$\mathop\text{Res}\limits_{z=0}\left\{\frac{z^2+1}{z(z^2+6z+1)}\right\}=1$$

$$\mathop\text{Res}\limits_{z=-3+2\sqrt{2}}\left\{\frac{z^2+1}{z(z^2+6z+1)}\right\} =-\frac{3}{2\sqrt{2}}$$

You know, I think Mathematica helps in this regards. You can at least see what the answer is and then work it out by hand towards that result.

Edit: just noticed the answer you reported is correct. I misread it. Nice job.

Yahoo! I really loved the trial version of mathematica, but it was far too expensive for me to afford. I have Matlab 7, but it doesn't seem to able to do anything useful.

 

[loom91]

Have you got some more, or some link to a place where I can get a few more problems on bounded real integrals? I'm not doing the improper ones because we don't have improper integrals in our syllabus, though there was recently an uproar over examples doing (bounded) improper integrals as proper ones in textbooks (they were including singularities as endpoints of the interval of integration).

 

[shmoe]

A decent looking set of general notes:

http://www.math.ust.hk/~maykwok/courses/ma304/05_06/Complex_7.pdf#search=%22evaluating%20real%20integrals%20%20residues%22

You can always make up your own integrals of this type, just take some rational function of sin's and cos's and go from there. MATLAB can at least give you a numerical approximation, so you have a good idea if you are correct, though these should get boring fast.

 

[saltydog]

Have you got some more

Alright, how about this one:

$$\int_0^{2\pi}\frac{1}{(Cos(\theta)+2)^n}d\theta$$

Can it be done using residue integration for any integer n? Can it be done any other way?
What about just for n=1, 2,3, maybe 4, then Mathematical Induction? I don't know. Might work on it though.

 

[mcfc]

If instead of $\oint cos^2 \theta d\theta$
we had
$\oint {1 \over 1 +cos^2 \theta } d\theta$
how would that be solved?