integration involving fraction

[fresh_42]

Where did you get [imath] x+1 [/imath] in the denominator from? You correctly split the quotient
[math] \dfrac{1}{x(x^2+1)} =\dfrac{1}{x}-\dfrac{x}{x^2+1}[/math]This can be solved by logarithms, the first one is directly a logarithm, and the second one needs a substitution [imath] u=x^2+1. [/imath]

Did you switch to your original problem at this stage without mention?

The original problem has a nasty anti-derivative with logs and arcus tangent.

[math] \int \dfrac{x-2}{x^2-x+1}\, dx = \dfrac{1}{2} (\log(x^2 - x + 1) - 2 \sqrt{3} \tan^{-1}((2 x - 1)\sqrt{3})) + constant [/math]
You could probably do it by looking up the anti-derivatives of [imath] \dfrac{1}{x^2-x+1} [/imath] and [imath] \dfrac{x}{x^2-x+1} [/imath] eventually after another substitution. However, it is a definite integral. It is easier to solve by the residue method than by partial fraction decomposition. Did you read the article I linked? It contains an example of how it is done, although it only mentions that [imath] \int (x^3+1)^{-1} dx [/imath] can be solved accordingly.

Allow me a remark: stop using [imath] \infty [/imath] as if it were a number, it is not! [imath] \int_0^\infty \dfrac{1}{x+1}\,dx [/imath] does not converge! This is another problem with the partial fraction decomposition: it leaves you with two divergent integrals! Since [imath] \int_0^\infty (x^3+1)^{-1} dx [/imath] is finite, the two "divergencies" cancel each other what cannot be seen by the partial fractions.

 

[logistic_guy]

Where did you get [imath] x+1 [/imath] in the denominator from? You correctly split the quotient
[math] \dfrac{1}{x(x^2+1)} =\dfrac{1}{x}-\dfrac{x}{x^2+1}[/math]This can be solved by logarithms, the first one is directly a logarithm, and the second one needs a substitution [imath] u=x^2+1. [/imath]

Did you switch to your original problem at this stage without mention?

The original problem has a nasty anti-derivative with logs and arcus tangent.

[math] \int \dfrac{x-2}{x^2-x+1}\, dx = \dfrac{1}{2} (\log(x^2 - x + 1) - 2 \sqrt{3} \tan^{-1}((2 x - 1)\sqrt{3})) + constant [/math]
You could probably do it by looking up the anti-derivatives of [imath] \dfrac{1}{x^2-x+1} [/imath] and [imath] \dfrac{x}{x^2-x+1} [/imath] eventually after another substitution. However, it is a definite integral. It is easier to solve by the residue method than by partial fraction decomposition. Did you read the article I linked? It contains an example of how it is done, although it only mentions that [imath] \int (x^3+1)^{-1} dx [/imath] can be solved accordingly.

Allow me a remark: stop using [imath] \infty [/imath] as if it were a number, it is not! [imath] \int_0^\infty \dfrac{1}{x+1}\,dx [/imath] does not converge! This is another problem with the partial fraction decomposition: it leaves you with two divergent integrals! Since [imath] \int_0^\infty (x^3+1)^{-1} dx [/imath] is finite, the two "divergencies" cancel each other what cannot be seen by the partial fractions.

sorry i switch to the original question wihtout mentioning it

why say nasty anti-derivative? means i can't solve it?

i can't understand how you solve this
\(\displaystyle \int \dfrac{x-2}{x^2-x+1}\, dx = \dfrac{1}{2} (\log(x^2 - x + 1) - 2 \sqrt{3} \tan^{-1}((2 x - 1)\sqrt{3})) + constant \)

from where the log and tangent come? i think log only come with structure \(\displaystyle \int \frac{1}{f(x)} \ dx\)

can you please divide your steps into detailed steps?

i'll read your article and anything about residues after i solve the integration with partial fraction decomposition

 

[fresh_42]

sorry i switch to the original question wihtout mentioning it

why say nasty anti-derivative? means i can't solve it?

I find combinations with logarithms and arcus tangents nasty.

i can't understand how you solve this
\(\displaystyle \int \dfrac{x-2}{x^2-x+1}\, dx = \dfrac{1}{2} (\log(x^2 - x + 1) - 2 \sqrt{3} \tan^{-1}((2 x - 1)\sqrt{3})) + constant \)

I cheated.

integral ((x-2)/(x^2-x+1)) dx = - Wolfram|Alpha

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.

www.wolframalpha.com

from where the log and tangent come? i think log only come with structure \(\displaystyle \int \frac{1}{f(x)} \ dx\)

can you please divide your steps into detailed steps?

I made no steps. I looked up the result.

i'll read your article and anything about residues after i solve the integration with partial fraction decomposition

You can't. Partial fraction decomposition gives you a problem you can't solve:

[math] \dfrac{2\pi}{3\sqrt{3}}=\int_0^\infty \dfrac{1}{x^3+1}\,dx =\dfrac{1}{3}\underbrace{\int_0^\infty \dfrac{1}{x+1}\,dx}_{divergent}+\dfrac{2}{3}\underbrace{\int_0^\infty \dfrac{1}{x^2-x+1}\,dx}_{=\dfrac{4\pi}{3\sqrt{3}}}-\dfrac{1}{3}\underbrace{\int_0^\infty \dfrac{x}{x^2-x+1}\,dx}_{divergent} [/math]How will you deal with the divergent integrals? You cannot subtract one divergency from another and get zero. And even if, it won't produce the correct result. All you can do is solve the integral in the middle which gives you an arcus tangent, but this isn't the result of the entire integral.

Here are a couple of useful links where you can look up integrals:

https://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Integralrechnung
https://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Integrale
https://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Bestimmte_Integrale
https://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Integraltransformationen

It's German and there is unfortunately no English version, but it is so full of formulas that you won't need a translation, and the technical terms are more or less identical. The only word you possibly need is "Beweis" which means "Proof".

 

[mario99]

Because I am a fan of integration, I will continue solving the integral in details from where everyone got stuck. I will start from here:

[imath]\displaystyle \frac{1}{3}\int_{0}^{\infty}\frac{1}{x + 1} \ dx - \frac{1}{3}\int_{0}^{\infty}\frac{x - 2}{x^2 - x + 1} \ dx[/imath]

We know that the derivative of [imath]x^2 - x + 1 \ \text{is} \ \rightarrow 2x - 1[/imath].

So, do this trick:

[imath]\displaystyle x - 2 = \frac{1}{2}(2x - 1 - 3)[/imath]

Our integral will be:

[imath]\displaystyle \frac{1}{3}\int_{0}^{\infty}\frac{1}{x + 1} \ dx - \frac{1}{3}\frac{1}{2}\int_{0}^{\infty}\frac{2x - 1 - 3}{x^2 - x + 1} \ dx [/imath]


[imath]\displaystyle = \frac{1}{3}\int_{0}^{\infty}\frac{1}{x + 1} \ dx - \frac{1}{6}\int_{0}^{\infty}\frac{2x - 1}{x^2 - x + 1} \ dx - \frac{1}{6}\int_{0}^{\infty}\frac{-3}{x^2 - x + 1} \ dx[/imath]


[imath]\displaystyle = \frac{1}{3}\int_{0}^{\infty}\frac{1}{x + 1} \ dx - \frac{1}{6}\int_{0}^{\infty}\frac{2x - 1}{x^2 - x + 1} \ dx + \frac{1}{2}\int_{0}^{\infty}\frac{1}{x^2 - x + 1} \ dx[/imath]

We cannot factor [imath]x^2 - x + 1 \rightarrow[/imath] This gives us an idea to complete the square!

[imath]\displaystyle x^2 - x + 1 = \left(x - \frac{1}{2}\right)^2 + 1 - \frac{1}{4} = \left(x - \frac{1}{2}\right)^2 + \frac{3}{4}[/imath]

The main idea of completing the square is to have a form that looks like this: [imath]\displaystyle \left( .... \right)^2 + 1[/imath]
So, we need to do some manipulation.

[imath]\displaystyle \left(x - \frac{1}{2}\right)^2 + \frac{3}{4} = \left(\frac{2x}{2} - \frac{1}{2}\right)^2 + \frac{3}{4} = \frac{1}{4}\left(2x - 1\right)^2 + \frac{3}{4} = \frac{3}{4}\frac{\left(2x - 1\right)^2}{3} + \frac{3}{4} = \frac{3}{4}\left[\left(\frac{2x - 1}{\sqrt{3}}\right)^2 + 1\right][/imath]

We have achieved what we wanted.

\(\displaystyle \displaystyle x^2 - x + 1 = \frac{3}{4}\left[\left(\frac{2x - 1}{\sqrt{3}}\right)^2 + 1\right]\)

Our integral will be:

[imath]\displaystyle \frac{1}{3}\int_{0}^{\infty}\frac{1}{x + 1} \ dx - \frac{1}{6}\int_{0}^{\infty}\frac{2x - 1}{x^2 - x + 1} \ dx + \frac{1}{2}\frac{4}{3}\int_{0}^{\infty}\frac{1}{\left(\frac{2x - 1}{\sqrt{3}}\right)^2 + 1} \ dx[/imath]


[imath]\displaystyle = \frac{1}{3}\int_{0}^{\infty}\frac{1}{x + 1} \ dx - \frac{1}{6}\int_{0}^{\infty}\frac{2x - 1}{x^2 - x + 1} \ dx + \frac{2}{3}\int_{0}^{\infty}\frac{1}{\left(\frac{2x - 1}{\sqrt{3}}\right)^2 + 1} \ dx[/imath]

Our integral is ready for cooking

At this stage beginner students will need to do [imath]u[/imath] substitution on paper. The substitution will give them this known form:

[imath]A\displaystyle \int \frac{1}{u} \ du + B\int \frac{1}{u} \ du + C\int \frac{1}{u^2 + 1} \ du[/imath]

Where [imath]A, B, \text{and} \ C[/imath] are some constants.

Advanced students and I will compute the integral directly.

[imath]\displaystyle \frac{1}{3}\int_{0}^{\infty}\frac{1}{x + 1} \ dx - \frac{1}{6}\int_{0}^{\infty}\frac{2x - 1}{x^2 - x + 1} \ dx + \frac{2}{3}\int_{0}^{\infty}\frac{1}{\left(\frac{2x - 1}{\sqrt{3}}\right)^2 + 1} \ dx[/imath]


[imath]\displaystyle = \frac{1}{3}\ln(x + 1)\bigg|_{0}^{\infty} - \frac{1}{6}\ln(x^2 - x + 1)\bigg|_{0}^{\infty} + \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x - 1}{\sqrt{3}}\right)\bigg|_{0}^{\infty}[/imath]

We can combine the natural logarithmic functions together.

[imath]\displaystyle = \frac{1}{6}\ln\left(\frac{(x + 1)^2}{x^2 - x + 1}\right)\bigg|_{0}^{\infty} + \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x - 1}{\sqrt{3}}\right)\bigg|_{0}^{\infty}[/imath]

We will switch to limit notation.

[imath]\displaystyle = \lim_{R\rightarrow \infty}\frac{1}{6}\ln\left(\frac{(x + 1)^2}{x^2 - x + 1}\right)\bigg|_{0}^{R} + \lim_{R\rightarrow \infty} \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x - 1}{\sqrt{3}}\right)\bigg|_{0}^{R}[/imath]


[imath]\displaystyle = \lim_{R\rightarrow \infty}\frac{1}{6}\ln\left(\frac{R^2 + 2R + 1}{R^2 - R + 1}\right) - \frac{1}{6}\ln 1 + \lim_{R\rightarrow \infty} \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2R - 1}{\sqrt{3}}\right) - \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{-1}{\sqrt{3}}\right)[/imath]


[imath]\displaystyle = \frac{1}{6}\ln 1 - \frac{1}{6}\ln 1 + \frac{1}{\sqrt{3}}\frac{\pi}{2} + \frac{1}{\sqrt{3}}\frac{\pi}{6}[/imath]


[imath]\displaystyle = \frac{1}{\sqrt{3}}\frac{\pi}{2} + \frac{1}{\sqrt{3}}\frac{\pi}{6}[/imath]


[imath]\displaystyle = \frac{3}{\sqrt{3}}\frac{\pi}{6} + \frac{1}{\sqrt{3}}\frac{\pi}{6}[/imath]


[imath]\displaystyle = \frac{4}{\sqrt{3}}\frac{\pi}{6}[/imath]


[imath]\displaystyle = \frac{2\pi}{3\sqrt{3}}[/imath]

 

[fresh_42]

Wow! And some nice tricks! I liked the step where you got rid of the two divergencies by using the properties of the logarithm!

My math teacher used to say: Anyone can differentiate, but integration is for artists!

 

[mario99]

Wow! And some nice tricks! I liked the step where you got rid of the two divergencies by using the properties of the logarithm!

Thanks fresh_42.

I am sure that you would have figured out all these tricks and manipulations if you took the integral seriously. But it seems that you and I are interested more in the complex analysis part because it is very rare to see a student discusses contours.

My math teacher used to say: Anyone can differentiate, but integration is for artists!

Believe it or not, a few weeks ago Mr. logistic_guy told us that he is an Artist. Therefore, I think that your teacher meant that integrals can only be solved by an Artist, but not all Artists can solve integrals.

The most fascinating thing here is that an Artist who studies engineering. A very rare thing to happen.

 

[khansaheb]

...... an Artist who studies engineering. A very rare thing to happen.

Sounds like you have not met Architectural Engineers.....

 

[logistic_guy]

i'm back

thank mario99

few remarks i don't understand

the completeing the square
how you jump from \(\displaystyle x^2 - x + 1\) to \(\displaystyle (x - \frac{1}{2})^2 + 1 - \frac{1}{4}\)
it's not clear at all

how you decide this integration \(\displaystyle \int_{0}^{\infty}\frac{1}{(\frac{x - 1}{\sqrt{3}})^2 + 1} \ dx = \frac{1}{\sqrt{3}}\tan^{-1}(\frac{x-1}{\sqrt{3}})\)

i know tangent inverse is the angle inside a right triangle. i don't have triangle in the question

how this limit \(\displaystyle \lim_{R\to\infty}\frac{1}{6}\ln(\frac{R^2 + 2R + 1}{R^2 - R + 1}) = \frac{1}{6}\ln 1\)

how this limit \(\displaystyle \lim_{R\to\infty}\tan^{-1}(\frac{2R - 1}{\sqrt{3}}) = \frac{\pi}{2}\)

how \(\displaystyle \tan^{-1}(\frac{-1}{\sqrt{3}}) = \frac{\pi}{6}\)

you don't explain the steps of how you get them, so i got lost. i think i'm understand other calculations

 

[mario99]

the completeing the square
how you jump from \(\displaystyle x^2 - x + 1\) to \(\displaystyle (x - \frac{1}{2})^2 + 1 - \frac{1}{4}\)
it's not clear at all

I have just added a quarter and subtracted a quarter.

[imath]\displaystyle x^2 - x + 1 = x^2 - x + 1 + \frac{1}{4} - \frac{1}{4} \ \ \ \ \ [/imath] (I did not change anything.)

And from algebra class you should know: [imath]\displaystyle x^2 - x + \frac{1}{4} = \left(x - \frac{1}{2}\right)\left(x - \frac{1}{2}\right) = \left(x - \frac{1}{2}\right)^2[/imath].


Then, [imath]\displaystyle x^2 - x + 1 = x^2 - x + 1 + \frac{1}{4} - \frac{1}{4} = \left(x - \frac{1}{2}\right)^2 + 1 - \frac{1}{4}[/imath]

how you decide this integration \(\displaystyle \int_{0}^{\infty}\frac{1}{(\frac{x - 1}{\sqrt{3}})^2 + 1} \ dx = \frac{1}{\sqrt{3}}\tan^{-1}(\frac{x-1}{\sqrt{3}})\)

i know tangent inverse is the angle inside a right triangle. i don't have triangle in the question

You are right about the triangle, but we can also get the inverse tangent from other places. In the first day of integral calculus class, you should have been given the solution of some elementary integrals such as [imath]\displaystyle \int \frac{1}{u^2 + 1} \ du = \tan^{-1} u + c.[/imath]
(Don't worry about the [imath]c[/imath], it is just some constant.)

If don't remember, you can follow this concept:

If the derivative of [imath]x[/imath] is [imath]\rightarrow 1[/imath], then the integral of [imath]1[/imath] is [imath]\rightarrow x + c. [/imath]

This means:

If the derivative of [imath]\tan^{-1}x[/imath] is [imath]\displaystyle \rightarrow \frac{1}{x^2 + 1}[/imath], then the integral of [imath]\displaystyle \frac{1}{x^2 + 1}[/imath] is [imath]\rightarrow \tan^{-1}x + c.[/imath]

Have you ever differentiated the inverse tangent function? Let me guess. No.

Then, it is time for differentiation

Let [imath]y = \tan^{-1}x[/imath]

[imath]\tan y = x[/imath]

[imath]\displaystyle \frac{d}{dx}(\tan y = x)[/imath]

[imath]\displaystyle \sec^2 y \ \frac{dy}{dx} = 1[/imath]

[imath]\displaystyle \frac{dy}{dx} = \frac{1}{\sec^2 y}[/imath]

And we know from trigonometry class, [imath]\tan^2 y + 1 = \sec^2 y[/imath]

Then,

[imath]\displaystyle \frac{dy}{dx} = \frac{1}{\sec^2 y} = \frac{1}{\tan^2 y + 1} = \frac{1}{x^2 + 1} \ \ \ \ \ \ \ \ [/imath] (And the proof is completed.)

how this limit \(\displaystyle \lim_{R\to\infty}\frac{1}{6}\ln(\frac{R^2 + 2R + 1}{R^2 - R + 1}) = \frac{1}{6}\ln 1\)

Limit involving infinity needs a lot of practice. But here we just have a fraction of polynomials. If the leading terms have the same power, the limit is just the coefficient of the leading terms division. This just means [imath]\displaystyle \frac{R^2}{R^2} = \frac{1}{1} = 1[/imath].

how this limit \(\displaystyle \lim_{R\to\infty}\tan^{-1}(\frac{2R - 1}{\sqrt{3}}) = \frac{\pi}{2}\)

Here we have [imath]\tan^{-1}\infty[/imath] and we want to know what happens to the inverse tangent function as [imath]x[/imath] goes to [imath]\infty[/imath]. One way to see that [imath]y[/imath] goes to [imath]\displaystyle \frac{\pi}{2}[/imath] by graphing the function.

Another way is to write [imath]\tan^{-1}\infty[/imath] as:

[imath]\displaystyle \tan y = \frac{\sin y}{\cos y} = \infty[/imath]

What is the first angle that will make the denominator [imath]= 0[/imath]. Well, it is obvious [imath]90 \ \text{degrees}[/imath].

Then, [imath]\displaystyle \tan \frac{\pi}{2} = \frac{\sin \frac{\pi}{2}}{\cos \frac{\pi}{2}} = \frac{1}{0} = \infty[/imath].

This means [imath]\displaystyle \frac{\pi}{2} = \tan^{-1} \infty[/imath].

Some people (such as fresh_42) will hate you if you write infinity [imath]\infty[/imath] as if it is a number. Therefore, it is a good practice to always use a limit:

[imath]\displaystyle \lim_{x\rightarrow \infty}\tan^{-1}x = \frac{\pi}{2}[/imath]

how \(\displaystyle \tan^{-1}(\frac{-1}{\sqrt{3}}) = \frac{\pi}{6}\)

You mean [imath]\displaystyle \tan^{-1} \frac{-1}{\sqrt{3}} = -\frac{\pi}{6}.[/imath]
This last one will depend on how quickly you can calculate some angles without calculator such as, [imath]0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}, \ \text{and} \ 90^{\circ}[/imath].

Since the kindergarten, I have known:

[imath]\displaystyle \sin 0^{\circ} = 0[/imath]
[imath]\displaystyle \sin 30^{\circ} = \frac{1}{2}[/imath]
[imath]\displaystyle \sin 45^{\circ} = \frac{\sqrt{2}}{2}[/imath]
[imath]\displaystyle \sin 60^{\circ} = \frac{\sqrt{3}}{2}[/imath]
[imath]\displaystyle \sin 90^{\circ} = 1[/imath]

And

[imath]\displaystyle \cos 0^{\circ} = 1[/imath]
[imath]\displaystyle \cos 30^{\circ} = \frac{\sqrt{3}}{2}[/imath]
[imath]\displaystyle \cos 45^{\circ} = \frac{\sqrt{2}}{2}[/imath]
[imath]\displaystyle \cos 60^{\circ} = \frac{1}{2}[/imath]
[imath]\displaystyle \cos 90^{\circ} = 0[/imath]

I need a fraction from above that has [imath]\displaystyle \sqrt{3}[/imath], so I have [imath]\displaystyle \sin 60^{\circ} = \frac{\sqrt{3}}{2}[/imath] and [imath]\displaystyle \cos 30^{\circ} = \frac{\sqrt{3}}{2}[/imath].

Let us try both:

[imath]\displaystyle \tan 60^{\circ} = \frac{\sin 60^{\circ}}{\cos 60^{\circ}} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}[/imath]

This gives us: [imath]\displaystyle 60^{\circ} = \tan^{-1} \sqrt{3} \ \ \ \ \ [/imath] (Not what I wanted.)

Let us try the second one:

[imath]\displaystyle \tan 30^{\circ} = \frac{\sin 30^{\circ}}{\cos 30^{\circ}} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}[/imath]

This gives us: [imath]\displaystyle 30^{\circ} = \tan^{-1} \frac{1}{\sqrt{3}} \ \ \ \ \ [/imath] (Exactly what I wanted.)

This means: [imath]\displaystyle \tan^{-1} \frac{-1}{\sqrt{3}} = -\tan^{-1} \frac{1}{\sqrt{3}} = -30^{\circ} = -\frac{\pi}{6}\ \ \ \ \ [/imath]

 

[logistic_guy]

I have just added a quarter and subtracted a quarter.

[imath]\displaystyle x^2 - x + 1 = x^2 - x + 1 + \frac{1}{4} - \frac{1}{4} \ \ \ \ \ [/imath] (I did not change anything.)

And from algebra class you should know: [imath]\displaystyle x^2 - x + \frac{1}{4} = \left(x - \frac{1}{2}\right)\left(x - \frac{1}{2}\right) = \left(x - \frac{1}{2}\right)^2[/imath].


Then, [imath]\displaystyle x^2 - x + 1 = x^2 - x + 1 + \frac{1}{4} - \frac{1}{4} = \left(x - \frac{1}{2}\right)^2 + 1 - \frac{1}{4}[/imath]


You are right about the triangle, but we can also get the inverse tangent from other places. In the first day of integral calculus class, you should have been given the solution of some elementary integrals such as [imath]\displaystyle \int \frac{1}{u^2 + 1} \ du = \tan^{-1} u + c.[/imath]
(Don't worry about the [imath]c[/imath], it is just some constant.)

If don't remember, you can follow this concept:

If the derivative of [imath]x[/imath] is [imath]\rightarrow 1[/imath], then the integral of [imath]1[/imath] is [imath]\rightarrow x + c. [/imath]

This means:

If the derivative of [imath]\tan^{-1}x[/imath] is [imath]\displaystyle \rightarrow \frac{1}{x^2 + 1}[/imath], then the integral of [imath]\displaystyle \frac{1}{x^2 + 1}[/imath] is [imath]\rightarrow \tan^{-1}x + c.[/imath]

Have you ever differentiated the inverse tangent function? Let me guess. No.

Then, it is time for differentiation

Let [imath]y = \tan^{-1}x[/imath]

[imath]\tan y = x[/imath]

[imath]\displaystyle \frac{d}{dx}(\tan y = x)[/imath]

[imath]\displaystyle \sec^2 y \ \frac{dy}{dx} = 1[/imath]

[imath]\displaystyle \frac{dy}{dx} = \frac{1}{\sec^2 y}[/imath]

And we know from trigonometry class, [imath]\tan^2 y + 1 = \sec^2 y[/imath]

Then,

[imath]\displaystyle \frac{dy}{dx} = \frac{1}{\sec^2 y} = \frac{1}{\tan^2 y + 1} = \frac{1}{x^2 + 1} \ \ \ \ \ \ \ \ [/imath] (And the proof is completed.)


Limit involving infinity needs a lot of practice. But here we just have a fraction of polynomials. If the leading terms have the same power, the limit is just the coefficient of the leading terms division. This just means [imath]\displaystyle \frac{R^2}{R^2} = \frac{1}{1} = 1[/imath].


Here we have [imath]\tan^{-1}\infty[/imath] and we want to know what happens to the inverse tangent function as [imath]x[/imath] goes to [imath]\infty[/imath]. One way to see that [imath]y[/imath] goes to [imath]\displaystyle \frac{\pi}{2}[/imath] by graphing the function.

Another way is to write [imath]\tan^{-1}\infty[/imath] as:

[imath]\displaystyle \tan y = \frac{\sin y}{\cos y} = \infty[/imath]

What is the first angle that will make the denominator [imath]= 0[/imath]. Well, it is obvious [imath]90 \ \text{degrees}[/imath].

Then, [imath]\displaystyle \tan \frac{\pi}{2} = \frac{\sin \frac{\pi}{2}}{\cos \frac{\pi}{2}} = \frac{1}{0} = \infty[/imath].

This means [imath]\displaystyle \frac{\pi}{2} = \tan^{-1} \infty[/imath].

Some people (such as fresh_42) will hate you if you write infinity [imath]\infty[/imath] as if it is a number. Therefore, it is a good practice to always use a limit:

[imath]\displaystyle \lim_{x\rightarrow \infty}\tan^{-1}x = \frac{\pi}{2}[/imath]


You mean [imath]\displaystyle \tan^{-1} \frac{-1}{\sqrt{3}} = -\frac{\pi}{6}.[/imath]
This last one will depend on how quickly you can calculate some angles without calculator such as, [imath]0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}, \ \text{and} \ 90^{\circ}[/imath].

Since the kindergarten, I have known:

[imath]\displaystyle \sin 0^{\circ} = 0[/imath]
[imath]\displaystyle \sin 30^{\circ} = \frac{1}{2}[/imath]
[imath]\displaystyle \sin 45^{\circ} = \frac{\sqrt{2}}{2}[/imath]
[imath]\displaystyle \sin 60^{\circ} = \frac{\sqrt{3}}{2}[/imath]
[imath]\displaystyle \sin 90^{\circ} = 1[/imath]

And

[imath]\displaystyle \cos 0^{\circ} = 1[/imath]
[imath]\displaystyle \cos 30^{\circ} = \frac{\sqrt{3}}{2}[/imath]
[imath]\displaystyle \cos 45^{\circ} = \frac{\sqrt{2}}{2}[/imath]
[imath]\displaystyle \cos 60^{\circ} = \frac{1}{2}[/imath]
[imath]\displaystyle \cos 90^{\circ} = 0[/imath]

I need a fraction from above that has [imath]\displaystyle \sqrt{3}[/imath], so I have [imath]\displaystyle \sin 60^{\circ} = \frac{\sqrt{3}}{2}[/imath] and [imath]\displaystyle \cos 30^{\circ} = \frac{\sqrt{3}}{2}[/imath].

Let us try both:

[imath]\displaystyle \tan 60^{\circ} = \frac{\sin 60^{\circ}}{\cos 60^{\circ}} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}[/imath]

This gives us: [imath]\displaystyle 60^{\circ} = \tan^{-1} \sqrt{3} \ \ \ \ \ [/imath] (Not what I wanted.)

Let us try the second one:

[imath]\displaystyle \tan 30^{\circ} = \frac{\sin 30^{\circ}}{\cos 30^{\circ}} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}[/imath]

This gives us: [imath]\displaystyle 30^{\circ} = \tan^{-1} \frac{1}{\sqrt{3}} \ \ \ \ \ [/imath] (Exactly what I wanted.)

This means: [imath]\displaystyle \tan^{-1} \frac{-1}{\sqrt{3}} = -\tan^{-1} \frac{1}{\sqrt{3}} = -30^{\circ} = -\frac{\pi}{6}\ \ \ \ \ [/imath]

thank mario99

i think i'm understanding

My point was if you write the fraction as [imath]\displaystyle \frac{1}{6}[/imath] when it is meant to be [imath]\displaystyle \frac{5}{6}[/imath], you are missing some terms in the numerator. These terms are not difficult to find when you are calculating simple fractions as [imath]\displaystyle \frac{1}{2} + \frac{1}{3}[/imath]. But when the fractions are polynomials, you will need to do some work to find the missing terms as was explained to you in post #18 by fresh_42.

As professor Dave has told you, it is good to know how to solve by partial fraction, but in this problem you have to put this method aside as you were told not to use it. If I were you, I would directly solve the problem by complex analysis.


Thank you for uploading this picture. It helps a lot to understand what to calculate. In the contour, you have three paths, so you will need to solve three integrals:

[imath]\displaystyle \int_{C_1} f(z) \ dz + \int_{C_2} f(z) \ dz + \int_{C_3} f(z) \ dz = 2\pi iB[/imath]

If you have solved contours before, you would already know that the curve path would get cancelled as it would be equal to zero. Therefore, two paths would remain:

[imath]\displaystyle \int_{C_1} f(z) \ dz + \int_{C_3} f(z) \ dz = 2\pi iB[/imath]

Your first task is to find [imath]\displaystyle B[/imath] which is the residues at the zeros that are located inside the contour. To find [imath]\displaystyle B[/imath], you need first to find the zeros which is the easiest part in the problem.

[imath]\displaystyle x^3 + 1 = 0[/imath] (Since this is a cubic function, you should get three zeros. Some are imaginary.)

I remember that there is a post written by fresh_42 where he showed the method of finding all the roots (real and complex) of a function. You can use it as a quick guidance. If you got stuck in finding the complex roots, use W|A.

Note: It is not difficult to show that the integral of the curve path will be zero as [imath]\displaystyle R \rightarrow \infty.[/imath]

\(\displaystyle x^3 + 1 = 0\)
\(\displaystyle x^3 = -1\)
\(\displaystyle x = \sqrt[3]{-1}\)
\(\displaystyle \sqrt[3]{-1} = \sqrt[3]{1}e^{\frac{\pi i + 2k\pi i}{3}}\)
\(\displaystyle x_1 = e^{\frac{\pi i}{3}} = \cos \frac{\pi}{3} + i\sin \frac{\pi}{3} = \frac{1}{2} + i\frac{\sqrt{3}}{2}\)
\(\displaystyle x_2 = e^{\pi i} = \cos \pi + i\sin \pi = -1\)
\(\displaystyle x_3 = e^{\frac{5\pi i}{3}} = \cos \frac{5\pi}{3} + i\sin \frac{5\pi}{3} = \frac{1}{2} - i\frac{\sqrt{3}}{2}\)

i find the roots

Integration is a complicated subject. There are many tips and tricks but there is no golden way to solve them all.
Here is an overview of the residue method:

An Overview of Complex Differentiation and Integration

I want to shed some light on complex analysis without getting all the technical details in the way which are necessary for the precise treatments that can be found in many excellent standard textbooks.

www.physicsforums.com

thank fresh_42

i read the site. it's very complicated

 

[mario99]

\(\displaystyle x_1 = \frac{1}{2} + i\frac{\sqrt{3}}{2}\)
\(\displaystyle x_2 = -1\)
\(\displaystyle x_3 = \frac{1}{2} - i\frac{\sqrt{3}}{2}\)

i find the roots

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

 

[fresh_42]

thank fresh_42

i read the site. it's very complicated

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:

Because I am a fan of integration, I will continue solving the integral in details from where everyone got stuck.

 

[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 [imath]\displaystyle \frac{1}{2} + \text{i}\frac{\sqrt{3}}{2}[/imath] lies inside the bounded region of the contour. But before we decide that we have to know more about [imath]R[/imath] which you did not say any information about it. If [imath]R < 1[/imath], 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 [imath]R > 1[/imath]. Why did I choose [imath]1[/imath] and not other numbers? Because I have calculated the magnitude of the point [imath]\displaystyle \frac{1}{2} + \text{i}\frac{\sqrt{3}}{2}[/imath] which is [imath]\displaystyle \sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} = 1[/imath].

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 [imath]\displaystyle \frac{1}{2} + \text{i}\frac{\sqrt{3}}{2}[/imath] lies inside the bounded region of the contour. But before we decide that we have to know more about [imath]R[/imath] which you did not say any information about it. If [imath]R < 1[/imath], 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 [imath]R > 1[/imath]. Why did I choose [imath]1[/imath] and not other numbers? Because I have calculated the magnitude of the point [imath]\displaystyle \frac{1}{2} + \text{i}\frac{\sqrt{3}}{2}[/imath] which is [imath]\displaystyle \sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} = 1[/imath].

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
[math] f(z)=\sum_{n=-\infty }^\infty a_n (z-z_0)^n [/math]and investigate the coefficients [imath] a_n. [/imath] 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
[math] \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} [/math]with [imath] x_{1,2}=\dfrac{1}{2}\pm \dfrac{i\sqrt{3}}{2} [/imath] has three locations (one real number, [imath] -1 [/imath], two complex numbers [imath] x_1 [/imath] and [imath] x_2 [/imath]) 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 [imath] \int_0^\infty . [/imath]

 

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

First method.

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

And we want to use it to expand our function [imath]\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)[/imath]

where [imath]z_2 = e^{\frac{i5\pi}{3}}[/imath].

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

[imath]\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} - . . . . [/imath]

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

[imath]\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}} - . . . .[/imath]

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

[imath]\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) + ....[/imath]


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

It is your job to simplify it.

Second method.


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

And

Let [imath]\displaystyle \frac{1}{z^3 + 1} = \frac{p(z)}{q(z)}[/imath]

This means:

[imath]\displaystyle p(z) = 1[/imath]
[imath]\displaystyle q(z) = z^3 + 1[/imath]
[imath]\displaystyle q'(z) = 3z^2[/imath]

If the condition below is satisfied:

[imath]\displaystyle p(z_0) \neq 0[/imath]
[imath]\displaystyle q(z_0) = 0[/imath]
[imath]\displaystyle q'(z_0) \neq 0[/imath]

Then,

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

Try it

Third method.

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

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

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


Examples of the order of the pole for: [imath]z^2 - z + 1[/imath]


[imath]\displaystyle \frac{1}{z^3 + 1} = \frac{1}{(z+1)(z^2 - z + 1)}[/imath], here [imath]\displaystyle k = 1[/imath]


[imath]\displaystyle \frac{1}{(z+1)(z^2 - z + 1)^2}[/imath], here [imath]\displaystyle k = 2[/imath]


We will use the formula to find the residue.


The residue at [imath]\displaystyle z_0 = e^{\frac{i\pi}{3}}[/imath] is [imath]\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][/imath]

[imath]\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}[/imath]

And it is your job to solve the limit

 

[logistic_guy]

This is not an easy subject. It requires writing functions as power series
[math] f(z)=\sum_{n=-\infty }^\infty a_n (z-z_0)^n [/math]and investigate the coefficients [imath] a_n. [/imath] 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
[math] \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} [/math]with [imath] x_{1,2}=\dfrac{1}{2}\pm \dfrac{i\sqrt{3}}{2} [/imath] has three locations (one real number, [imath] -1 [/imath], two complex numbers [imath] x_1 [/imath] and [imath] x_2 [/imath]) 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 [imath] \int_0^\infty . [/imath]

thank

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

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

First method.

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

And we want to use it to expand our function [imath]\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)[/imath]

where [imath]z_2 = e^{\frac{i5\pi}{3}}[/imath].

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

[imath]\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} - . . . . [/imath]

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

[imath]\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}} - . . . .[/imath]

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

[imath]\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) + ....[/imath]


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

It is your job to simplify it.

Second method.


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

And

Let [imath]\displaystyle \frac{1}{z^3 + 1} = \frac{p(z)}{q(z)}[/imath]

This means:

[imath]\displaystyle p(z) = 1[/imath]
[imath]\displaystyle q(z) = z^3 + 1[/imath]
[imath]\displaystyle q'(z) = 3z^2[/imath]

If the condition below is satisfied:

[imath]\displaystyle p(z_0) \neq 0[/imath]
[imath]\displaystyle q(z_0) = 0[/imath]
[imath]\displaystyle q'(z_0) \neq 0[/imath]

Then,

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

Try it

Third method.

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

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

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


Examples of the order of the pole for: [imath]z^2 - z + 1[/imath]


[imath]\displaystyle \frac{1}{z^3 + 1} = \frac{1}{(z+1)(z^2 - z + 1)}[/imath], here [imath]\displaystyle k = 1[/imath]


[imath]\displaystyle \frac{1}{(z+1)(z^2 - z + 1)^2}[/imath], here [imath]\displaystyle k = 2[/imath]


We will use the formula to find the residue.


The residue at [imath]\displaystyle z_0 = e^{\frac{i\pi}{3}}[/imath] is [imath]\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][/imath]

[imath]\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}[/imath]

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