integration involving fraction

[logistic_guy]

Very good. That was super. Now which of these zeros lie on or in the contour? Think carefully.

thank

$\displaystyle x_1$ is up so it's inside the contour
$\displaystyle x_2$ is left so it's outside the contour
$\displaystyle x_3$ is down so it's outside the contour

it's easy why need to think carefully?

Yes, sorry, @logistic_guy . It is a bit hard to tell which level of answers should be given. I don't know much about the members. Some ask very basic questions, others quite sophisticated ones, and yours swam somewhere between them. Complex integration of real functions is not trivial. And real integration of real functions can be tricky as @mario99 has shown us.

I'm still baffled by how he got rid of the divergence problem of the two critical terms before integration. It meant that he used partial fractions but took a step back and recombined two critical quotients again to avoid infinite values. I very much like that example because it shows that "one method serves all" usually doesn't work for integrations.

I don't want to bother you with another article, but I quote it anyway in case a) some readers might be interested, b) you're brave and give me another shot (I first thought your question was primarily about complex integration), c) someone wants to link it because it lists many of the standard tricks in real integration. And if nothing of it counts: it has examples!
www.physicsforums.com/insights/the-art-of-integration/

Don't give up. Integration can be fun:

thank

i'll read the article right now

 

[mario99]

thank

$\displaystyle x_1$ is up so it's inside the contour
$\displaystyle x_2$ is left so it's outside the contour
$\displaystyle x_3$ is down so it's outside the contour

it's easy why need to think carefully?

I told you to think carefully because I knew that you will tell me $\displaystyle \frac{1}{2} + \text{i}\frac{\sqrt{3}}{2}$ lies inside the bounded region of the contour. But before we decide that we have to know more about $R$ which you did not say any information about it. If $R < 1$, all three points will lie outside the bounded region and we have to use a different contour. Therefore, the only logical choice for us is to assume $R > 1$. Why did I choose $1$ and not other numbers? Because I have calculated the magnitude of the point $\displaystyle \frac{1}{2} + \text{i}\frac{\sqrt{3}}{2}$ which is $\displaystyle \sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} = 1$.

Once we have known the zero, now we would want to choose a method to calculate the residue at it. There are a lot of methods, so which method you have learnt to use to find the residues?

 

[logistic_guy]

I told you to think carefully because I knew that you will tell me $\displaystyle \frac{1}{2} + \text{i}\frac{\sqrt{3}}{2}$ lies inside the bounded region of the contour. But before we decide that we have to know more about $R$ which you did not say any information about it. If $R < 1$, all three points will lie outside the bounded region and we have to use a different contour. Therefore, the only logical choice for us is to assume $R > 1$. Why did I choose $1$ and not other numbers? Because I have calculated the magnitude of the point $\displaystyle \frac{1}{2} + \text{i}\frac{\sqrt{3}}{2}$ which is $\displaystyle \sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} = 1$.

Once we have known the zero, now we would want to choose a method to calculate the residue at it. There are a lot of methods, so which method you have learnt to use to find the residues?

i'll be honest with you. my goal of this question is partial fraction decomposition. i study your solution and i understand it

you're asking me now beyond my understanding

 

[mario99]

i'll be honest with you. my goal of this question is partial fraction decomposition. i study your solution and i understand it

you're asking me now beyond my understanding

My bad! I thought you understand complex analysis, especially after surprising me by finding the complex roots. What I understand from your comment is that you are not interested to solve the problem with complex analysis?

 

[logistic_guy]

My bad! I thought you understand complex analysis, especially after surprising me by finding the complex roots. What I understand from your comment is that you are not interested to solve the problem with complex analysis?

i'm interested but i don't understand residues

 

[fresh_42]

i'm interested but i don't understand residues

This is not an easy subject. It requires writing functions as power series
$$f(z)=\sum_{n=-\infty }^\infty a_n (z-z_0)^n$$and investigate the coefficients $a_n.$ I admit that the article I linked in post #15 isn't an easy read but it summarizes the concept. It is finally an integration in the complex number plane. Your integrand
$$\dfrac{1}{x^3+1}=\dfrac{1}{x+1}\cdot \dfrac{1}{x^2-x+1}=\dfrac{1}{x+1}\cdot \dfrac{1}{x-x_1} \cdot \dfrac{1}{x-x_2}$$with $x_{1,2}=\dfrac{1}{2}\pm \dfrac{i\sqrt{3}}{2}$ has three locations (one real number, $-1$, two complex numbers $x_1$ and $x_2$) where it is not defined and we have to deal with these so-called poles. We already ran into divergencies by real integration, now we have two more of them. The residue theorem and Cauchy's integral formula give us a method but it involves things like counting how often we circle around such a pole and to compute "residues" and finally a limit consideration since the integral includes an infinity $\int_0^\infty .$

 

[mario99]

i'm interested but i don't understand residues

I will explain three methods to find the residues and you decide which one is easier for you to understand.

Let $\displaystyle z_1 = e^{\frac{i\pi}{3}} = \frac{1}{2} + \text{i}\frac{\sqrt{3}}{2} = \sqrt[3]{-1} \ \ \ \ \$ (Our main zero.)

First method.

It is exactly what professor fresh_42 wrote in post #46. We have a Laurent series $\displaystyle f(z) = \sum_{k=-\infty}^{\infty}a_k(z-z_0)^k$

And we want to use it to expand our function $\displaystyle \frac{1}{z^3 + 1} = \left(\frac{1}{z + 1}\right)\left(\frac{1}{z - z_1}\right)\left(\frac{1}{z - z_2}\right)$

where $z_2 = e^{\frac{i5\pi}{3}}$.

Since $\displaystyle \frac{1}{z + 1}$ is analytic at $\displaystyle z_1$, it has a Taylor expansion equal to:

$\displaystyle \frac{1}{1 + \sqrt[3]{-1}} - \frac{z - \sqrt[3]{-1}}{(1 + \sqrt[3]{-1})^2} + \frac{(z - \sqrt[3]{-1})^2}{(1 + \sqrt[3]{-1})^3} - . . . .$

Also since $\displaystyle \frac{1}{z - z_2}$ is analytic at $\displaystyle z_1$, it has a Taylor expansion equal to:

$\displaystyle -\frac{\text{i}}{\sqrt{3}} + \frac{1}{3}\left(z - \sqrt[3]{-1}\right) + \frac{\text{i}(z - \sqrt[3]{-1})^2}{3\sqrt{3}} - . . . .$

We are only interested in the first term from the expansions, so our Laurent expansion is:

$\displaystyle \frac{1}{z^3 + 1} =\left(\frac{1}{1 + \sqrt[3]{-1}}\right)\left(\frac{1}{z + z_1}\right)\left(-\frac{\text{i}}{\sqrt{3}}\right) + ....$


The residue of $\displaystyle \frac{1}{z^3 + 1}$ at $\displaystyle z_1$ is the coefficient of $\displaystyle \frac{1}{z + z_1}$ in the Laurent expansion which is $\displaystyle \left(\frac{1}{1 + \sqrt[3]{-1}}\right)\left(-\frac{\text{i}}{\sqrt{3}}\right) = \ \ \ ?$

It is your job to simplify it.

Second method.


Let $\displaystyle z_0 = e^{\frac{i\pi}{3}} = \frac{1}{2} + \text{i}\frac{\sqrt{3}}{2} = \sqrt[3]{-1}$

And

Let $\displaystyle \frac{1}{z^3 + 1} = \frac{p(z)}{q(z)}$

This means:

$\displaystyle p(z) = 1$
$\displaystyle q(z) = z^3 + 1$
$\displaystyle q'(z) = 3z^2$

If the condition below is satisfied:

$\displaystyle p(z_0) \neq 0$
$\displaystyle q(z_0) = 0$
$\displaystyle q'(z_0) \neq 0$

Then,

The residue of $\displaystyle \frac{1}{z^3 + 1}$ at $\displaystyle z_0$ is $\displaystyle \frac{p(z_0)}{q'(z_0)}$

Try it

Third method.

I think that this is the easiest one. Use this formula:

The residue of $\displaystyle \frac{1}{z^3 + 1}$ at $\displaystyle z_0 = \lim_{z\rightarrow z_0}\frac{1}{(k - 1)!}\frac{d^{k-1}}{dz^{k-1}}[(z - z_0)f(z)]$

where $\displaystyle k$ is the order of the pole
$\displaystyle z_0$ is our zero inside the contour
$\displaystyle \frac{d^{k-1}}{dz^{k-1}}$ is the kth derivative
$\displaystyle f(z)$ is our main function


Examples of the order of the pole for: $z^2 - z + 1$


$\displaystyle \frac{1}{z^3 + 1} = \frac{1}{(z+1)(z^2 - z + 1)}$, here $\displaystyle k = 1$


$\displaystyle \frac{1}{(z+1)(z^2 - z + 1)^2}$, here $\displaystyle k = 2$


We will use the formula to find the residue.


The residue at $\displaystyle z_0 = e^{\frac{i\pi}{3}}$ is $\displaystyle \lim_{z\rightarrow e^{\frac{i\pi}{3}}}\frac{1}{(1 - 1)!}\frac{d^{1 - 1}}{dz^{1 -1}}\left[(z - e^{\frac{i\pi}{3}})\frac{1}{z^3 + 1}\right]$

$\displaystyle = \lim_{z\rightarrow e^{\frac{i\pi}{3}}}\frac{z - e^{\frac{i\pi}{3}}}{z^3 + 1} = \lim_{z\rightarrow e^{\frac{i\pi}{3}}}\frac{1}{3z^2}$

And it is your job to solve the limit

 

[logistic_guy]

This is not an easy subject. It requires writing functions as power series
$$f(z)=\sum_{n=-\infty }^\infty a_n (z-z_0)^n$$and investigate the coefficients $a_n.$ I admit that the article I linked in post #15 isn't an easy read but it summarizes the concept. It is finally an integration in the complex number plane. Your integrand
$$\dfrac{1}{x^3+1}=\dfrac{1}{x+1}\cdot \dfrac{1}{x^2-x+1}=\dfrac{1}{x+1}\cdot \dfrac{1}{x-x_1} \cdot \dfrac{1}{x-x_2}$$with $x_{1,2}=\dfrac{1}{2}\pm \dfrac{i\sqrt{3}}{2}$ has three locations (one real number, $-1$, two complex numbers $x_1$ and $x_2$) where it is not defined and we have to deal with these so-called poles. We already ran into divergencies by real integration, now we have two more of them. The residue theorem and Cauchy's integral formula give us a method but it involves things like counting how often we circle around such a pole and to compute "residues" and finally a limit consideration since the integral includes an infinity $\int_0^\infty .$

thank

I will explain three methods to find the residues and you decide which one is easier for you to understand.

Let $\displaystyle z_1 = e^{\frac{i\pi}{3}} = \frac{1}{2} + \text{i}\frac{\sqrt{3}}{2} = \sqrt[3]{-1} \ \ \ \ \$ (Our main zero.)

First method.

It is exactly what professor fresh_42 wrote in post #46. We have a Laurent series $\displaystyle f(z) = \sum_{k=-\infty}^{\infty}a_k(z-z_0)^k$

And we want to use it to expand our function $\displaystyle \frac{1}{z^3 + 1} = \left(\frac{1}{z + 1}\right)\left(\frac{1}{z - z_1}\right)\left(\frac{1}{z - z_2}\right)$

where $z_2 = e^{\frac{i5\pi}{3}}$.

Since $\displaystyle \frac{1}{z + 1}$ is analytic at $\displaystyle z_1$, it has a Taylor expansion equal to:

$\displaystyle \frac{1}{1 + \sqrt[3]{-1}} - \frac{z - \sqrt[3]{-1}}{(1 + \sqrt[3]{-1})^2} + \frac{(z - \sqrt[3]{-1})^2}{(1 + \sqrt[3]{-1})^3} - . . . .$

Also since $\displaystyle \frac{1}{z - z_2}$ is analytic at $\displaystyle z_1$, it has a Taylor expansion equal to:

$\displaystyle -\frac{\text{i}}{\sqrt{3}} + \frac{1}{3}\left(z - \sqrt[3]{-1}\right) + \frac{\text{i}(z - \sqrt[3]{-1})^2}{3\sqrt{3}} - . . . .$

We are only interested in the first term from the expansions, so our Laurent expansion is:

$\displaystyle \frac{1}{z^3 + 1} =\left(\frac{1}{1 + \sqrt[3]{-1}}\right)\left(\frac{1}{z + z_1}\right)\left(-\frac{\text{i}}{\sqrt{3}}\right) + ....$


The residue of $\displaystyle \frac{1}{z^3 + 1}$ at $\displaystyle z_1$ is the coefficient of $\displaystyle \frac{1}{z + z_1}$ in the Laurent expansion which is $\displaystyle \left(\frac{1}{1 + \sqrt[3]{-1}}\right)\left(-\frac{\text{i}}{\sqrt{3}}\right) = \ \ \ ?$

It is your job to simplify it.

Second method.


Let $\displaystyle z_0 = e^{\frac{i\pi}{3}} = \frac{1}{2} + \text{i}\frac{\sqrt{3}}{2} = \sqrt[3]{-1}$

And

Let $\displaystyle \frac{1}{z^3 + 1} = \frac{p(z)}{q(z)}$

This means:

$\displaystyle p(z) = 1$
$\displaystyle q(z) = z^3 + 1$
$\displaystyle q'(z) = 3z^2$

If the condition below is satisfied:

$\displaystyle p(z_0) \neq 0$
$\displaystyle q(z_0) = 0$
$\displaystyle q'(z_0) \neq 0$

Then,

The residue of $\displaystyle \frac{1}{z^3 + 1}$ at $\displaystyle z_0$ is $\displaystyle \frac{p(z_0)}{q'(z_0)}$

Try it

Third method.

I think that this is the easiest one. Use this formula:

The residue of $\displaystyle \frac{1}{z^3 + 1}$ at $\displaystyle z_0 = \lim_{z\rightarrow z_0}\frac{1}{(k - 1)!}\frac{d^{k-1}}{dz^{k-1}}[(z - z_0)f(z)]$

where $\displaystyle k$ is the order of the pole
$\displaystyle z_0$ is our zero inside the contour
$\displaystyle \frac{d^{k-1}}{dz^{k-1}}$ is the kth derivative
$\displaystyle f(z)$ is our main function


Examples of the order of the pole for: $z^2 - z + 1$


$\displaystyle \frac{1}{z^3 + 1} = \frac{1}{(z+1)(z^2 - z + 1)}$, here $\displaystyle k = 1$


$\displaystyle \frac{1}{(z+1)(z^2 - z + 1)^2}$, here $\displaystyle k = 2$


We will use the formula to find the residue.


The residue at $\displaystyle z_0 = e^{\frac{i\pi}{3}}$ is $\displaystyle \lim_{z\rightarrow e^{\frac{i\pi}{3}}}\frac{1}{(1 - 1)!}\frac{d^{1 - 1}}{dz^{1 -1}}\left[(z - e^{\frac{i\pi}{3}})\frac{1}{z^3 + 1}\right]$

$\displaystyle = \lim_{z\rightarrow e^{\frac{i\pi}{3}}}\frac{z - e^{\frac{i\pi}{3}}}{z^3 + 1} = \lim_{z\rightarrow e^{\frac{i\pi}{3}}}\frac{1}{3z^2}$

And it is your job to solve the limit

i think i'm give up

thank mario99 very much
thank fresh_42 very much

if anyone of you will show the rest of the solution i'll be so glad