Problem: Contour integral, order of a pole

[degs2k4]

Homework Statement

The Attempt at a Solution

PART1)

Series of Taylor for e^(3z) + 2 (based on series for e^z)

$$\sum_{n=0}^{\infty} \frac{(3z)^{n}}{n!}= (1+2) + 3z + \frac{(3z)^2}{2!} + \frac{(3z)^3}{3!} + ...$$

Series of Taylor for 3e^(z) (based on series for e^z)

$$\sum_{n=0}^{\infty} \frac{3z^{n}}{n!}= 3 + 3z + \frac{z^2}{2!} + \frac{3z^3}{3!} + ...$$

Subtracting each term of the above 2 series, and dividing by z^3 we get the Laurent series:

$$0 + 0 + \frac{6}{2!z} + \frac{24}{3!} + ...$$

Since the laurent series (negative powers) stops at n=1, the pole is of order 1.

the residue af f(z) at z=0 is 0, since z=0 is outside the contour C.

Questions:
1) Is this ok? if not, why?
2) Is there any other way of knowing a pole order in this function without using Laurent Series ? (eg by limits of derivates?) If not, we must know as much (Taylor)Laurent series as we can in order to solve those problems ?
3) If the both the numerator/denominator of the function were polynomial and z = a a pole of it, we only need to check the power of (z-a) in order to know the order of the pole?PART2)

The integral of the contour C is 0, as no singularities can be found inside C.

$$\oint_C = 0 = \int_A + \int_B + \int_C + \int_D$$

I want to demonstrate that \int_D tends to 0 as R tends to infinite. For doing so I tried the following parametrization and check

the limit when R tends to 0, but couldn't reach something showing that the limit is 0.

$$z = Re^{it}, t=0...\pi$$
$$dz = iRe^{it} dt$$

I also want to demonstrate that \int_B tends to 0 as R tends to infinite, but can't see how to do it neither. Visually I can see

that when R tends to 0 the integrals over A and C become so big that the integral over B is negligible (zero).

Questions:
How could I demonstrate that those two integrals tend to to 0 (which I think they do) ?

Thanks in advance...

 

[HallsofIvy]

You seem to have dropped the "i" in both $e^{iz}$ and $e^{3iz}$. That will affect the signs.

 

[jackmell]

How could I demonstrate that those two integrals tend to to 0 (which I think they do) ?

Thanks in advance...

Try and just work on one thing at a time, and even one thing at a time, break it up into pieces. For example, the integral over D, using something similar to what we did on that other one, can you show just this part (use inequalities):

$$\lim_{R\to\infty} \int_D \frac{e^{3iz}}{z^3}dz=0,\quad z=Re^{it},\quad 0\leq t\leq \pi$$

 

[degs2k4]

You seem to have dropped the "i" in both $e^{iz}$ and $e^{3iz}$. That will affect the signs.

Thanks, that's true. Fixing it:

Series of Taylor for e^(3z) + 2 (based on series for e^z)

$$2 + \sum_{n=0}^{\infty} \frac{(3iz)^{n}}{n!}= (1+2) + 3iz - \frac{9z^2}{2!} - \frac{27iz^3}{3!} + ...$$

Series of Taylor for 3e^(z) (based on series for e^z)

$$3 \sum_{n=0}^{\infty} \frac{(iz)^{n}}{n!}= 3 + 3iz - \frac{3z^2}{2!} - \frac{3iz^3}{3!} + ...$$

Subtracting each term of the above 2 series, and dividing by z^3 we get the Laurent series:

$$0 + 0 - \frac{6}{2!z} -24i + ...$$


Since the laurent series (negative powers) stops at n=1, the pole is of order 1.

the residue af f(z) at z=0 is 0, since z=0 is outside the contour C.

Is it this part fully ok now?

Try and just work on one thing at a time, and even one thing at a time, break it up into pieces. For example, the integral over D, using something similar to what we did on that other one, can you show just this part (use inequalities):

$$\lim_{R\to\infty} \int_D \frac{e^{3iz}}{z^3}dz=0,\quad z=Re^{it},\quad 0\leq t\leq \pi$$

Thanks again for your suggestion. OK, taking the first part:

$$\lim_{R\to\infty} \int_D \frac{e^{3iz}}{z^3}dz=0,\quad z=Re^{it},\quad 0\leq t\leq \pi$$

I think we can do this:

$$\left| \int_{0}^{\pi} \frac{e^{3i(Re^{it})}iRe^{it} }{R^3e^{3it}} dt \right|\leq \frac{1}{R^3} \int_{0}^{\pi} \left| e^{3iR(cost +isint)} \right| dt = \frac{1}{R^3} \int_{0}^{\pi} \left| e^{3iRcost - 3Rsint)} \right| dt \leq \frac{1}{R^3} \int_{0}^{\pi} e^{-3R(sint)} dt \leq \frac{1}{R^3} \int_{0}^{\pi} e^{-3R(myline)} dt$$

I choose myline as y=0 so that sint > myline according to this pic:

And then,

$$\frac{1}{R^3} \int_{0}^{\pi} e^{-3R(myline)} dt = \frac{1}{R^3} \int_{0}^{\pi} e^{-3R(0)} dt = \frac{1}{R^3} \int_{0}^{\pi} dt = \frac{\pi}{R^{3}}$$

and we can say that when R tends to infinite, it becomes 0. Is it ok like this?

 

[vela]

Subtracting each term of the above 2 series, and dividing by z^3 we get the Laurent series:

$$0 + 0 - \frac{6}{2!z} -24i + ...$$


Since the laurent series (negative powers) stops at n=1, the pole is of order 1.

the residue af f(z) at z=0 is 0, since z=0 is outside the contour C.

Is it this part fully ok now?

No, the residue is just the coefficient of the 1/z term. The contour only matters when you evaluate an integral: if the pole is inside the contour, its residue contributes to the integral.

 

[degs2k4]

No, the residue is just the coefficient of the 1/z term. The contour only matters when you evaluate an integral: if the pole is inside the contour, its residue contributes to the integral.

I see! Thanks!

 

[jackmell]

I think we can do this:

$$\left| \int_{0}^{\pi} \frac{e^{3i(Re^{it})}iRe^{it} }{R^3e^{3it}} dt \right|\leq \frac{1}{R^3} \int_{0}^{\pi} \left| e^{3iR(cost +isint)} \right| dt$$
$$= \frac{1}{R^3} \int_{0}^{\pi} \left| e^{3iRcost - 3Rsint)} \right| dt \leq \frac{1}{R^3} \int_{0}^{\pi} e^{-3R(sint)} dt \leq \frac{1}{R^3} \int_{0}^{\pi} e^{-3R(myline)} dt$$

That should be:

$$\lim_{R\to\infty}\frac{1}{R^2}\int_0^{\pi} e^{-3R\sin(t)}dt$$

But it's not necessary to use that myline approach here. Don't recall if it was in the other problem. Just note that in the range $0\leq t\leq \pi$, we have $\left|e^{-3R\sin(t)}\right|<=1$ so that whole integral is less than the constant $\pi$ and so:

$$\lim_{R\to\infty}\frac{1}{R^2}\int_0^{\pi} e^{-3R\sin(t)}dt\leq \lim_{R\to\infty}\frac{1}{R^2}\pi=0$$.

Now you can do the same with the expression $e^{iz}$, and $2$ and conclude the integral of the entire expression goes to zero as well.

Also, it's really annoying when you post something that's wider than a page and it affects the entire thread causing us to keep using the arrow buttons to see it. Try to break things up with multiple lines so it width is no more than a page width.

 

[degs2k4]

That should be:

$$\lim_{R\to\infty}\frac{1}{R^2}\int_0^{\pi} e^{-3R\sin(t)}dt$$

But it's not necessary to use that myline approach here. Don't recall if it was in the other problem. Just note that in the range $0\leq t\leq \pi$, we have $\left|e^{-3R\sin(t)}\right|<=1$ so that whole integral is less than the constant $\pi$ and so:

$$\lim_{R\to\infty}\frac{1}{R^2}\int_0^{\pi} e^{-3R\sin(t)}dt\leq \lim_{R\to\infty}\frac{1}{R^2}\pi=0$$.

Now you can do the same with the expression $e^{iz}$, and $2$ and conclude the integral of the entire expression goes to zero as well.

Also, it's really annoying when you post something that's wider than a page and it affects the entire thread causing us to keep using the arrow buttons to see it. Try to break things up with multiple lines so it width is no more than a page width.

Thanks jackmell! All your comments are very helpful!

Ok, I think I won't have any problem regarding integral over line D. I guess that we can apply exactly the same method with the integral over B, right ? the integral bounds and the parametrization look almost exactly the same (except using ro instead of R, and the direction being taken clockwise)

 

[jackmell]

Thanks jackmell! All your comments are very helpful!

Ok, I think I won't have any problem regarding integral over line D. I guess that we can apply exactly the same method with the integral over B, right ? the integral bounds and the parametrization look almost exactly the same (except using ro instead of R, and the direction being taken clockwise)

I haven't worked it out yet. Not sure how to in fact. Maybe others can help too. But looks to me we need a different approach.

Note if you try and just look at the part:

$$\frac{e^{3iz}}{z^3}$$,

as $\rho\to 0$, that expression tend to infinity because of the rho in the denominator. However it may be that:

$$\lim_{\rho\to 0} \int_{\pi}^0 \frac{e^{3iz}-3e^{iz}+2}{z^3}dz\overset{?}{\to}\frac{0}{0}$$

which means you'll need to take the limit of the entire expression but I'm not sure. Maybe need to try the myline approach on it then take the limit. And one of the most important things I can suggest to you is to just try things and learn to recognize that often, the things you initially try don't work but sometimes lead you to the route that does. :)

 

[Count Iblis]

Note that part B) will not go to zero, it will be minus -1/2 times 2 pi i times times the residue. This is just the formula mentioned at the start of the problem. What then happens is that you decided to exclue the pole from the contour, but you now get its resudue back multipoled by minus -1/2. If you bring this to the other side, you see that you got hapf the residue of the pole.

Had you included the pole, the contour integral would have picked up the contribution from the residue. However, you would now have had half the residue which you would have had to move to the other side. So, either way, you end up with half the residue.


So, you're essentially rederiving Plemelj's formula: if you are lazy you can just let the contour run through a pole and define the integral as a suitably defined principal part, instead of bothering about making a small detour and letting the radius go to zero. You the residue then counts for half. In general if the contour interects the pole and makes an angle on the pole, then you need to replace 2 pi i times the residue by the i times the angle times the residue.

 

[jackmell]

Note that part B) will not go to zero, it will be minus -1/2 times 2 pi i times times the residue. This is just the formula mentioned at the start of the problem. What then happens is that you decided to exclue the pole from the contour, but you now get its resudue back multipoled by minus -1/2. If you bring this to the other side, you see that you got hapf the residue of the pole.

Had you included the pole, the contour integral would have picked up the contribution from the residue. However, you would now have had half the residue which you would have had to move to the other side. So, either way, you end up with half the residue.


So, you're essentially rederiving Plemelj's formula: if you are lazy you can just let the contour run through a pole and define the integral as a suitably defined principal part, instead of bothering about making a small detour and letting the radius go to zero. You the residue then counts for half. In general if the contour interects the pole and makes an angle on the pole, then you need to replace 2 pi i times the residue by the i times the angle times the residue.

Thank you. :)

 

[degs2k4]

I'm sorry, but I'm afraid I don't understand what you mean...

Note that part B) will not go to zero, it will be minus -1/2 times 2 pi i times times the residue. This is just the formula mentioned at the start of the problem. What then happens is that you decided to exclue the pole from the contour, but you now get its resudue back multipoled by minus -1/2. If you bring this to the other side, you see that you got hapf the residue of the pole.

I don't understand why it "gets the residue back multipoled by -1/2".

According to the formula,:
$$\theta_{1}=0$$
$$\theta_{2}=\pi$$
$$0 \leq \theta_{2} - \theta_{1} \leq 2\pi$$
$$0 \leq \pi \leq 2\pi$$

$$\int_{B} f(z)dz = \lim_{\rho\to 0} \int_{B} f(z)dz = i (\theta_{2}-\theta_{1}) Res(f(0)) = i \pi Res(f(0))$$

 

[Count Iblis]

theta1 should be pi and theta2 should be zero. Because you start the small circle at theta1 = pi and it ends at theta2 = 0.

So, you get minus pi i times the residue, which brought to the other side of the equation yields that the sum of the other parts of the contour integral is pi i times the residue. Part A and C together while letting rho shrink to zero is how you can define "moving through the pole". You compute that by adding the half circle B connecting A and C so that you get a closed contour, the integral over which is zero, and then evaluating B separately in the limit of rho to zero.

So, you see that if the contour "moves through a pole", as defined above, in a straght line, you get a contribution of pi i times the residue instead of 2 pi i, so the residue counts for 1/2 relative to what it would be if the pole were inside the contour. This a special case of Plemej's formula.

 

[degs2k4]

theta1 should be pi and theta2 should be zero. Because you start the small circle at theta1 = pi and it ends at theta2 = 0.

So, you get minus pi i times the residue, which brought to the other side of the equation yields that the sum of the other parts of the contour integral is pi i times the residue. Part A and C together while letting rho shrink to zero is how you can define "moving through the pole". You compute that by adding the half circle B connecting A and C so that you get a closed contour, the integral over which is zero, and then evaluating B separately in the limit of rho to zero.

So, you see that if the contour "moves through a pole", as defined above, in a straght line, you get a contribution of pi i times the residue instead of 2 pi i, so the residue counts for 1/2 relative to what it would be if the pole were inside the contour. This a special case of Plemej's formula.

Thanks for the detailed explanation! I think I understood it now. Despite the fact that the pole is outside the contour, since the contour moves to a pole there is a half contribution from the residue. I'll do it this way and post again the problem results in this thread as soon as I am done.

Thanks again everyone for your time and patience!

 

[degs2k4]

Stuck, again...
What I have so far:

The asked integral is:

$$\int_{0}^{\infty} \frac{(sinx)^{3}}{x^3} dx$$We want to solve it by complex integration:$$\int_{0}^{\infty} \frac{(sinz)^{3}}{z^3} dz = \frac{1}{2} \int_{-\infty}^{\infty} \frac{(sinz)^{3}}{z^3} dz =$$

$$= \frac{1}{2} \lim_{R\to\infty, \rho\to0} [ \int_{-R}^{-\rho} \frac{(sinz)^{3}}{z^3} dz + \int_{\rho}^{R} \frac{(sinz)^{3}}{z^3} dz ] =$$

$$= \lim_{R\to\infty, \rho\to0} \int_{\rho}^{R} \frac{(sinz)^{3}}{z^3} dz$$

Since the num is sinus, we take the imaginary part of the complex function (e^iz = cosz + i sinz)

$$Im ( \int_{0}^{\infty} \frac{(sinz)^{3}}{z^3} dz ) = \lim_{R\to\infty, \rho\to0} \int_{\rho}^{R} \frac{(e^z)^{3}}{z^3} dz$$

Back into our contour integral:

$$\oint_C = 0 = \int_A + \int_B + \int_C + \int_D$$

when rho -> 0 and R-> infty, and since the residue is -3 (coefficient from Laurent), we know the integral for the B and D:

$$0 = \int_{-\infty}^{0} f(z) dz - (\pi i 3) + \int_{0}^{\infty} f(z) dz + 0$$

So we can have:

$$\pi i 3 = \int_{-\infty}^{0} f(z) dz + \int_{0}^{\infty} f(z) dz = 2 \int_{0}^{\infty} f(z) dz$$

$$\frac{\pi i 3}{2} = \int_{0}^{\infty} f(z) dz$$now, here comes the part where I am stuck in. We need to represent f(z) by means of A = (sin(z))^3/z^3:$$Im( \int_{\rho}^{R} \frac{e^{iz3} + 3e^{iz} + 2}{z^3} dz ) = \int_{\rho}^{R} \frac{sinz^{3}}{z^3} dz + \int_{\rho}^{R} \frac{3sinz+ 2}{z^3} dz = \int_{\rho}^{R} A dz+ \int_{\rho}^{R} \frac{3sinz+ 2}{z^3} dz$$

How can I evaluate the other part of he integral ?

 

[Count Iblis]

You need to correctly represent sin^3(z) in terms of complex exponentials. This is something you should do at the start. The way the problem is set up invites you to think about this at the end, but if you had stumbled on this integral yourself and you were thinking about using contour integration, you really would have had to do this first.

One way is to manipulate the standard trignometric identities to express sin^3(z) in terms of sin(z) and sin(3z). Another way is to write:

sin(z) = [exp(iz) + exp(-iz)]/(2 i)

and then take the third power of both sides.

 

[degs2k4]

One way is to manipulate the standard trignometric identities to express sin^3(z) in terms of sin(z) and sin(3z).

I'm sorry about my lack of knowledge, but I don't remember how to do that...

Another way is to write:

sin(z) = [exp(iz) + exp(-iz)]/(2 i)

and then take the third power of both sides.

In [exp(iz) + exp(-iz)]/(2 i), replacing exp(zi) = cosz + i sinz, the right part of that expression becomes [cos (z)]/ (2i) which is not sin z right?

I am sorry, I'm quite confused with this...

 

[jackmell]

Know what, that's kinda' hard, for me anyway. Remember you are evaluating the integral:

$$\int_{A+C} \frac{e^{3iz}-3e^{iz}+2}{z^3}dz$$

Do this then: break up that integral into two parts: one goes from $-\infty$ to zero, the other from zero to $$\infty$$ but make a simple change of variables in the first one to get it in the form of zero to $\infty$ so now you have two integrals from zero to $\infty$. Now combine them and compare the numerator of that expression to the expansion:

$$\left(\sin(z)\right)^3=\left(\frac{e^{iz}-e^{-iz}}{2i}\right)^3$$

You can expand that cube right? Hope that's not insulting.

 

[degs2k4]

ok, the first part about what you said is:

$$\int_{-\infty}^{\infty} \frac{e^{3iz} - 3e^{iz} + 2}{z^{3}} dz =$$

$$\int_{-\infty}^{0} \frac{e^{3iz} - 3e^{iz} + 2}{z^{3}} dz + \int_{0}^{\infty} \frac{e^{3iz} - 3e^{iz} + 2}{z^{3}} dz =$$

$$\int_{0}^{\infty} \frac{-e^{3iz} + 3e^{iz} - 2}{z^{3}} dz + \int_{0}^{\infty} \frac{e^{3iz} - 3e^{iz} + 2}{z^{3}} dz =$$

$$\int_{0}^{\infty} \frac{-e^{3iz} + 3e^{iz} + e^{3iz} - 3e^{iz}}{z^{3}} dz$$

and if we expand the cube:

$$(e^{iz} - e^{-iz})^{3} = e^{3iz} - e^{-3iz} + 3e^{-iz} -3e^{iz}$$

Which it seems the nemuerators are quite close one to each other, but the signs of the powers are different...

 

[vela]

You didn't do the change of variables correctly going from the second line to the third line. Try again.

 

[jackmell]

You have the conversion for

$$\int_{-\infty}^0 \frac{e^{3iz}-3e^{iz}+2}{z^3}dz$$

almost correct. I assume you let $z=-r$ or something like that then $e^{3iz}$ would go to $e^{-3ir}$ correct? Same with $e^{iz}$. Then finish making the remaining substitutions including switching the limits of integration.

 

[degs2k4]

You have the conversion for

$$\int_{-\infty}^0 \frac{e^{3iz}-3e^{iz}+2}{z^3}dz$$

almost correct. I assume you let $z=-r$ or something like that then $e^{3iz}$ would go to $e^{-3ir}$ correct? Same with $e^{iz}$. Then finish making the remaining substitutions including switching the limits of integration.

ok,

$$\int_{-\infty}^0 \frac{e^{3iz}-3e^{iz}+2}{z^3}dz =$$

$$\int_{-\infty}^0 \frac{e^{-3ir}-3e^{-ir}+2}{-r^3} (-1)dr =$$

$$\int_{-\infty}^0 \frac{-e^{-3ir}+3e^{-ir}-2}{-r^3} dr =$$

$$\int_{0}^{\infty} \frac{-e^{-3ir}+3e^{-ir}-2}{r^3} dr$$

Now, joining with the other part of the integral, we finally get:

$$\int_{-\infty}^{0} \frac{e^{3iz} - 3e^{iz} + 2}{z^{3}} dz + \int_{0}^{\infty} \frac{e^{3iz} - 3e^{iz} + 2}{z^{3}} dz =$$


$$\int_{0}^{\infty} \frac{-e^{-3ir}+3e^{-ir}-2}{r^3} dr + \int_{0}^{\infty} \frac{e^{3iz} - 3e^{iz} + 2}{z^{3}} dz$$


Now the problem is that we have 2 integrals that look almost exactly to the numerator of the cube expression when joined, but one has a variable change z=-r. If we make the variable change in the second integral in order to join the integrals, it no longer looks like the numerator of the cube...?!

 

[jackmell]

ok,

$$\int_{-\infty}^0 \frac{e^{3iz}-3e^{iz}+2}{z^3}dz =$$

$$\int_{-\infty}^0 \frac{e^{-3ir}-3e^{-ir}+2}{-r^3} (-1)dr =$$

$$\int_{-\infty}^0 \frac{-e^{-3ir}+3e^{-ir}-2}{-r^3} dr =$$

$$\int_{0}^{\infty} \frac{-e^{-3ir}+3e^{-ir}-2}{r^3} dr$$

Now, joining with the other part of the integral, we finally get:

$$\int_{-\infty}^{0} \frac{e^{3iz} - 3e^{iz} + 2}{z^{3}} dz + \int_{0}^{\infty} \frac{e^{3iz} - 3e^{iz} + 2}{z^{3}} dz =$$


$$\int_{0}^{\infty} \frac{-e^{-3ir}+3e^{-ir}-2}{r^3} dr + \int_{0}^{\infty} \frac{e^{3iz} - 3e^{iz} + 2}{z^{3}} dz$$


Now the problem is that we have 2 integrals that look almost exactly to the numerator of the cube expression when joined, but one has a variable change z=-r. If we make the variable change in the second integral in order to join the integrals, it no longer looks like the numerator of the cube...?!

It doesn't matter what the variable of integration looks like ok. Just make them all r's or whatever so that the integral over the real axis is:

$$\int_0^{\infty}\frac{e^{3ir}-e^{-3ir}-3e^{ir}+3e^{-ir}}{r^3}dr=\frac{8}{i}\int_0^{\infty} \frac{sin^3 (z)}{z^3}dz$$

I think that' right. Please go over it to verify (don't forget the 2i cubed in the denominator of the sin expansion).

 

[degs2k4]

It doesn't matter what the variable of integration looks like ok. Just make them all r's or whatever so that the integral over the real axis is:

$$\int_0^{\infty}\frac{e^{3ir}-e^{-3ir}-3e^{ir}+3e^{-ir}}{r^3}dr=\frac{8}{i}\int_0^{\infty} \frac{sin^3 (z)}{z^3}dz$$

I think that' right. Please go over it to verify (don't forget the 2i cubed in the denominator of the sin expansion).

I have been thinking about that 8/i, and still have no idea of where it comes from... I mean, if you power 2i by the cubic power it becomes -8 right?

 

[jackmell]

You've got to try and get the basic complex algebra straight degs.

$$\frac{\sin^3(z)}{z^3}=\frac{e^{3iz}-e^{-3iz}-3e^{iz}+3e^{-iz}}{-8i z^3}$$

$$\frac{8}{i}\frac{\sin^3(z)}{z^3}=\frac{e^{3iz}-e^{-3iz}-3e^{iz}+3e^{-iz}}{ z^3}$$

I just multiplied by -8i then multiplied -8i by i/i to give me 8/i or you could have left it at -8i.

 

[degs2k4]

I just multiplied by -8i then multiplied -8i by i/i to give me 8/i or you could have left it at -8i.

Oh I see! Thanks, I knew -8i but could not the 8/i. Now I think I have all clear...

So the solution should be:

$$\int_0^{\infty} \frac{sin^3 (x)}{x^3}dx = \frac{1}{-8i} \int_0^{\infty}\frac{e^{3ir}-e^{-3ir}-3e^{ir}+3e^{-ir}}{r^3}dr =$$
$$= \frac{1}{-8i} \frac{\pi i 3}{2} = \frac{3\pi}{-16}$$

By the way, I just have a last question regarding contour integrals problems: in this problem we used a great formula which tells us that the integral when rho->0 (the pole at z=0) is (pi*i*Residue). This formula also appears in Serge Lang's Complex Analysis Book.

However, according to another book, if rho->a and a is not zero, for example, in the following image:

we have a different formula that tells us that the integral when rho-> a is pi*Residue, so we can do the following:

$$\int_C = \int_{I} + \int_{S_{-a}(E)} + \int_{II} + \int_{S_{a}(E)} + \int_{III} + \int_{Arc} =$$

$$= \int_{-R}^{0} + \int_{S_{-a}(E)} + \int_{S_{a}(E)} + \int_{0}^{R} + \int_{Arc} =$$

when E tends to a and -a, and R tends to infty we get:

$$= \int_{-\infty}^{0} + \pi Res(z=-a) + \pi Res(z=a) + \int_{0}^{\infty} + 0$$

The question is in the second step: is it ok to ignore II and join in two integrals ? (the one from -R to 0 and the one of 0 to R) Or we should use the three integrals I, II, III separately?

Thanks again!

 

[jackmell]

Oh I see! Thanks, I knew -8i but could not the 8/i. Now I think I have all clear...

So the solution should be:

$$\int_0^{\infty} \frac{sin^3 (x)}{x^3}dx = \frac{1}{-8i} \int_0^{\infty}\frac{e^{3ir}-e^{-3ir}-3e^{ir}+3e^{-ir}}{r^3}dr =$$
$$= \frac{1}{-8i} \frac{\pi i 3}{2} = \frac{3\pi}{-16}$$

That's not correct. Now, putting all the contours together, I get:

$$\frac{8}{i}\int_{0}^{\infty} \frac{\sin^3(x)}{x^3}-\pi i r=0$$

since the pole is simple and we're going in the reverse direction and above, you guys computed the residue was -3 so then:

$$\frac{8}{i}\int_{0}^{\infty} \frac{\sin^3(x)}{x^3}=3\pi i$$

or:
$$\int_{0}^{\infty} \frac{\sin^3(x)}{x^3}=\frac{3\pi}{8}$$

By the way, I just have a last question regarding contour integrals problems: in this problem we used a great formula which tells us that the integral when rho->0 (the pole at z=0) is (pi*i*Residue). This formula also appears in Serge Lang's Complex Analysis Book.

However, according to another book, if rho->a and a is not zero, for example, in the following image:

we have a different formula that tells us that the integral when rho-> a is pi*Residue

The integral around a simple pole as the radius goes to zero is simply $\theta i r$
where r is the residue and $\theta$ is the arc length of the contour and is positive or negative depending if it's in the positive or reverse orientation.

The question is in the second step: is it ok to ignore II and join in two integrals ? (the one from -R to 0 and the one of 0 to R) Or we should use the three integrals I, II, III separately?

Thanks again!

No. You have to go from -R to -a, then -a to a, then a to R.

 

[degs2k4]

Thanks for your reply and your patience, jackmell.

That's not correct. Now, putting all the contours together, I get:

$$\frac{8}{i}\int_{0}^{\infty} \frac{\sin^3(x)}{x^3}-\pi i r=0$$

since the pole is simple and we're going in the reverse direction and above, you guys computed the residue was -3 so then:

$$\frac{8}{i}\int_{0}^{\infty} \frac{\sin^3(x)}{x^3}=3\pi i$$

or:
$$\int_{0}^{\infty} \frac{\sin^3(x)}{x^3}=\frac{3\pi}{8}$$

I think there's a little mistake in there: I think there should be a 16 in the denominator instead of a 8. Going step by step:

We had initially:

$$0 = \int_{-R}^{-\rho} + \int_{-\rho}^{\rho} + \int_{\rho}^{R} + \int_{Arc} =$$

now, when R->infty, and rho->0

$$\int_{-\infty}^{0} + (-\pi i (-3)) + \int_{0}^{\infty} + 0$$

So we get:

$$\int_{0}^{\infty} = - \frac{\pi i 3}{2}$$

And now just do the final step to get the integral for the cubic sin:

$$\frac{8}{i}\int_{0}^{\infty} \frac{\sin^3(x)}{x^3}=- \frac{\pi i 3}{2} i$$

So the final result should be:

$$\int_{0}^{\infty} \frac{\sin^3(x)}{x^3}= \frac{3 \pi}{16}$$

Is this step-by-step development ok?

You have to go from -R to -a, then -a to a, then a to R.

Why? In the real line I can see 5 diferent integrals...

 

[jackmell]

Thanks for your reply and your patience, jackmell.



I think there's a little mistake in there: I think there should be a 16 in the denominator instead of a 8. Going step by step:

We had initially:

$$0 = \int_{-R}^{-\rho} + \int_{-\rho}^{\rho} + \int_{\rho}^{R} + \int_{Arc} =$$

now, when R->infty, and rho->0

$$\int_{-\infty}^{0} + (-\pi i (-3)) + \int_{0}^{\infty} + 0$$

So we get:

$$\int_{0}^{\infty} = - \frac{\pi i 3}{2}$$

That's not correct. We showed above that these two integrals combined into a single integral from 0 to infinity so no 2. Sides, I checked it against Mathematica and it's 3pi/8.

Why? In the real line I can see 5 diferent integrals.

Ok, you're right, the other two half-circle contours also.

 

[degs2k4]

That's not correct. We showed above that these two integrals combined into a single integral from 0 to infinity so no 2. Sides, I checked it against Mathematica and it's 3pi/8.

Oh! You're completely right, I was mistakenly thinking about the fact of joining the two half integrals as well...

Ok, you're right, the other two half-circle contours also.

Thanks! OK, so this should be right:

$$\int_C = \int_{-R}^{-a-\rho} + \int_{S_{-a}} + \int_{-a+\rho}^{0} + \int_{0}^{a-\rho} + \int_{S_{a}} + \int_{a+\rho}^{R} + \int_{Arc} =$$

when R->\infty, and \rho->0:

$$\int_C = \int_{-\infty}^{-a} + (-\pi i Res(-a)) + \int_{-a}^{0} + \int_{0}^{a} + (-\pi i Res(a)) + \int_{a}^{R} + 0 =$$

So finally:

$$= \int_{-\infty}^{0} + (-\pi i Res(-a)) + \int_{0}^{\infty} + (-\pi i Res(a))$$

You've all wiped out *almost* all my doubts about those integrals! Thank you very much.