integration involving fraction

[logistic_guy]

here is the question.

Show that the integral \(\displaystyle \int_{0}^{\infty}\frac{1}{x^3 + 1} \ dx = \frac{2\pi}{3\sqrt{3}}\) without using partial fraction decomposition.

Hint: Use the contour given in sec. 2.7, example 13 and use the method of residues.

my attemb
i don't understand what is the meaning of partial fraction decomposition so i make some search and i found if i can factor \(\displaystyle x^3 + 1\) i can solve the integration. my problem is i can't factor that

 

[Dr.Peterson]

here is the question.

Show that the integral \(\displaystyle \int_{0}^{\infty}\frac{1}{x^3 + 1} \ dx = \frac{2\pi}{3\sqrt{3}}\) without using partial fraction decomposition.

Hint: Use the contour given in sec. 2.7, example 13 and use the method of residues.

my attempt
i don't understand what is the meaning of partial fraction decomposition so i make some search and i found if i can factor \(\displaystyle x^3 + 1\) i can solve the integration. my problem is i can't factor that

You were told NOT to use that method, so why does it matter?

(It's worth learning for other problems; clearly they expect you to know it. In addition, it is possible to factor the denominator, as a sum of cubes -- look it up, and learn that, too! But set it aside for now.)

What you need to do is to read "sec. 2.7" and learn about the method of residues, as they say. Presumably you've learned all that, unless you dove into the middle of a book without reading it, which would be a very foolish thing to do.

 

[logistic_guy]

You were told NOT to use that method, so why does it matter?

(It's worth learning for other problems; clearly they expect you to know it. In addition, it is possible to factor the denominator, as a sum of cubes -- look it up, and learn that, too! But set it aside for now.)

What you need to do is to read "sec. 2.7" and learn about the method of residues, as they say. Presumably you've learned all that, unless you dove into the middle of a book without reading it, which would be a very foolish thing to do.

thank

i don't trust the auther and i want to learn the method of partial fraction decomposition before using the method of residues

my experience in algebra give me this idea of factoring \(\displaystyle (x + A)(x^2 + B) = x^3 + Bx + Ax^2 + AB\)

this give me
\(\displaystyle A = 0\)
\(\displaystyle B = 0\)
\(\displaystyle AB = 1\)

i'm stuck what two things are zero when i multiply them give me one

 

[blamocur]

my experience in algebra give me this idea of factoring (x+A)(x2+B)=x3+Bx+Ax2+AB\displaystyle (x + A)(x^2 + B) = x^3 + Bx + Ax^2 + AB

It gave you a wrong idea. Why [imath]x^2+B[/imath] and not [imath]x^2+Cx+b[/imath] ? I.e., why do you assume that C = 0?

Also, have you looked up sum of cubes suggested in post #2?

 

[logistic_guy]

It gave you a wrong idea. Why [imath]x^2+B[/imath] and not [imath]x^2+Cx+b[/imath] ? I.e., why do you assume that C = 0?

Also, have you looked up sum of cubes suggested in post #2?

thank

i don't assume \(\displaystyle C = 0\). when we factor in algebra we've only two things inside the brackets.

so it's ok to use three things, \(\displaystyle (x + A)(x^2 + Bx + C) = x^3 + Bx^2 + Cx + Ax^2 + ABx + AC = x^3 + (B + A)x^2 + (C + AB)x + AC\)

this give me
\(\displaystyle B + A = 0\)
\(\displaystyle C + AB = 0\)
\(\displaystyle AC = 1\)
\(\displaystyle A = \frac{1}{C}\)
\(\displaystyle A = -B\)
\(\displaystyle -B = \frac{1}{C}\)
\(\displaystyle C + \frac{1}{C}\frac{-1}{C} = 0\)
\(\displaystyle C^3 = 1\)
\(\displaystyle C = 1\)
\(\displaystyle A = 1\)
\(\displaystyle B = -1\)

\(\displaystyle (x + 1)(x^2 - x + 1) = x^3 - x^2 + x + x^2 - x + 1 = x^3 + 1\)

wikipedia say: Every sum of cubes may be factored according to the identity \(\displaystyle (a^3 + b^3) = (a + b)(a^2 - ab + b^2)\)

unfortunately this identity won't work in my factor because i don't have the \(\displaystyle b^3\)

my integration
\(\displaystyle \int_{0}^{\infty}\frac{1}{(x + 1)(x^2 - x + 1)} \ dx\)

i think i can continue from here

\(\displaystyle \int_{0}^{\infty}\frac{1}{(x + 1)(x^2 - x + 1)} \ dx = \int_{0}^{\infty}\frac{1}{x + 1} \ dx + \int_{0}^{\infty}\frac{1}{x^2 - x + 1} \ dx = \ln(x + 1) + \ln(x^2 - x + 1) = \ln(\infty + 1) + \ln(\infty^2 - \infty + 1)\)

how to know what is the answer of \(\displaystyle \ln(\infty + 1)\)?

 

[Steven G]

You CAN factor sum of cubes (although you were told NOT to use that method)

 

[lookagain]

i think i can continue from here

\(\displaystyle \int_{0}^{\infty}\frac{1}{(x + 1)(x^2 - x + 1)} \ dx = \int_{0}^{\infty}\frac{1}{x + 1} \ dx + \int_{0}^{\infty}\frac{1}{x^2 - x + 1} \ dx = \ln(x + 1) + \ln(x^2 - x + 1) = \ln(\infty + 1) + \ln(\infty^2 - \infty + 1)\)

The fraction does not decompose into those fractions, and the steps after
those are not correct. You need to look at a tutorial for partial fraction
decomposition when a factor is linear and the other factor is quadratic.

 

[Steven G]

how to know what is the answer of \(\displaystyle \ln(\infty + 1)\)?

You use the definition of logs and decide what happens when the argument gets larger and larger.

 

[mario99]

i think i can continue from here

\(\displaystyle \int_{0}^{\infty}\frac{1}{(x + 1)(x^2 - x + 1)} \ dx = \int_{0}^{\infty}\frac{1}{x + 1} \ dx + \int_{0}^{\infty}\frac{1}{x^2 - x + 1} \ dx = \ln(x + 1) + \ln(x^2 - x + 1) = \ln(\infty + 1) + \ln(\infty^2 - \infty + 1)\)

If

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

It means

[imath]\displaystyle \frac{1}{6} = \frac{1}{(2)(3)} = \frac{1}{2} + \frac{1}{3}[/imath]

But

[imath]\displaystyle \frac{1}{2} + \frac{1}{3} = \frac{5}{6}[/imath]

Hint: Use the contour given in sec. 2.7, example 13 and use the method of residues.

Can you please show us a picture of the contour?

 

[blamocur]

wikipedia say: Every sum of cubes may be factored according to the identity (a3+b3)=(a+b)(a2−ab+b2)\displaystyle (a^3 + b^3) = (a + b)(a^2 - ab + b^2)(a3+b3)=(a+b)(a2−ab+b2)

unfortunately this identity won't work in my factor because i don't have the \(b^3\).

Of course you do have [imath]b^3[/imath] in your case.

 

[fresh_42]

[math] \dfrac{1}{(x+1)(x^2-x+1)}=\dfrac{1}{3}\cdot\left(\dfrac{1}{x+1}+\dfrac{-x+2}{x^2-x+1}\right) \neq \dfrac{1}{x+1}+\dfrac{1}{x^2-x+1}[/math]

 

[logistic_guy]

You CAN factor sum of cubes (although you were told NOT to use that method)

i did. \(\displaystyle (x^3 + 1) = (x + 1)(x^2 - x + 1)\)

The fraction does not decompose into those fractions, and the steps after
those are not correct. You need to look at a tutorial for partial fraction
decomposition when a factor is linear and the other factor is quadratic.

i'm trying to learn this new method. i'll look for some videos. if you can show me anything, i'll be glad

You use the definition of logs and decide what happens when the argument gets larger and larger.

i don't have logs but i don't mind to know the definition. what's the definition?

If

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

It means

[imath]\displaystyle \frac{1}{6} = \frac{1}{(2)(3)} = \frac{1}{2} + \frac{1}{3}[/imath]

But

[imath]\displaystyle \frac{1}{2} + \frac{1}{3} = \frac{5}{6}[/imath]

fraction with numbers is different than fraction with letters. that's what i learn in algebra. look at \(\displaystyle x\), it's variable not number

Can you please show us a picture of the contour?

Of course you do have [imath]b^3[/imath] in your case.

\(\displaystyle (x^3 + 1) \to 1\)

[math] \dfrac{1}{(x+1)(x^2-x+1)}=\dfrac{1}{3}\cdot\left(\dfrac{1}{x+1}+\dfrac{-x+2}{x^2-x+1}\right) \neq \dfrac{1}{x+1}+\dfrac{1}{x^2-x+1}[/math]

i don't understand how do you get \(\displaystyle \frac{1}{3}\) and \(\displaystyle -x+2\). teach me how to do this if you don't mind

 

[fresh_42]

i don't understand how do you get \(\displaystyle \frac{1}{3}\) and \(\displaystyle -x+2\). teach me how to do this if you don't mind

I first observed that [math] \dfrac{1}{(x+1)(x^2-x+1)}\neq \dfrac{1}{x+1}\cdot \dfrac{1}{x^2-x+1} [/math]and then looked for the correct partition. The idea is to write
[math] \dfrac{1}{(x+1)(x^2-x+1)}= \dfrac{f(x)}{x+1}\cdot \dfrac{g(x)}{x^2-x+1}. [/math]
The only restriction is, that [imath] \deg f(x)<\deg(x+1)=1 [/imath] and [imath] \deg g(x)<\deg(x^2-x+1)=2. [/imath] We have therefore the ansatz
[math]\begin{array}{lll} \dfrac{1}{(x+1)(x^2-x+1)}&= \dfrac{A}{x+1}\cdot \dfrac{Bx+C}{x^2-x+1}\\[16pt] &=\dfrac{A(x^2-x+1)+(Bx+C)(x+1)}{(x+1)(x^2-x+1)}\\[16pt] &=\dfrac{x^2(A+B)+x(-A+B+C)+(A+C)}{(x+1)(x^2-x+1)}\\[16pt] \end{array}[/math]and by comparison [imath] A+B=0\, , \,-A+B+C=0\, , \,A+C=1. [/imath] Hence, [imath] B=-A\, , \,C=1-A [/imath] and [imath] 0=-A+B+C=-A+(-A)+(1-A)=1-3A [/imath] or [imath]A=1/3.[/imath] Then [imath] B=-1/3 [/imath] and [imath] C=2/3. [/imath] Thus
[math]\begin{array}{lll} \dfrac{1}{(x+1)(x^2-x+1)}&= \dfrac{A}{x+1}\cdot \dfrac{Bx+C}{x^2-x+1}\\[16pt] &=\dfrac{(1/3)}{x+1}\cdot \dfrac{(-1/3)x+(2/3)}{x^2-x+1}\\[16pt] &=\dfrac{1}{3}\left(\dfrac{1}{x+1}+\dfrac{-x+2}{x^2-x+1}\right) \end{array}[/math]

 

[logistic_guy]

thank

so this partial fraction decomposition, that's why i don't know it. very complicated method. not what i expect. i'll study it deeply today and i hope to understand some of it. i insist to solve the integration with this method before the residues

 

[fresh_42]

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

 

[lookagain]

I first observed that [math] \dfrac{1}{(x+1)(x^2-x+1)}\neq \dfrac{1}{x+1}\cdot \dfrac{1}{x^2-x+1} [/math]and then looked for the correct partition.

There is a typo. That multiplication dot should be a plus sign instead, because
you are pointing out that the left-hand side is not equal to it.

 

[fresh_42]

There is a typo. That multiplication dot should be a plus sign instead, because
you are pointing out that the left-hand side is not equal to it.

Too late to edit, but thanks for mentioning it. Unfortunately, the dot was a victim of copy and paste and occurred more than once.

 

[fresh_42]

Corrected version:

I first observed that [math] \dfrac{1}{(x+1)(x^2-x+1)}\neq \dfrac{1}{x+1}+ \dfrac{1}{x^2-x+1} [/math]and then looked for the correct partition. The idea is to write
[math] \dfrac{1}{(x+1)(x^2-x+1)}= \dfrac{f(x)}{x+1}+ \dfrac{g(x)}{x^2-x+1}. [/math]
The only restriction is, that [imath] \deg f(x)<\deg(x+1)=1 [/imath] and [imath] \deg g(x)<\deg(x^2-x+1)=2. [/imath] We have therefore the ansatz
[math]\begin{array}{lll} \dfrac{1}{(x+1)(x^2-x+1)}&= \dfrac{A}{x+1}+ \dfrac{Bx+C}{x^2-x+1}\\[16pt] &=\dfrac{A(x^2-x+1)+(Bx+C)(x+1)}{(x+1)(x^2-x+1)}\\[16pt] &=\dfrac{x^2(A+B)+x(-A+B+C)+(A+C)}{(x+1)(x^2-x+1)}\\[16pt] \end{array}[/math]and by comparison [imath] A+B=0\, , \,-A+B+C=0\, , \,A+C=1. [/imath] Hence, [imath] B=-A\, , \,C=1-A [/imath] and [imath] 0=-A+B+C=-A+(-A)+(1-A)=1-3A [/imath] or [imath]A=1/3.[/imath] Then [imath] B=-1/3 [/imath] and [imath] C=2/3. [/imath] Thus
[math]\begin{array}{lll} \dfrac{1}{(x+1)(x^2-x+1)}&= \dfrac{A}{x+1}+ \dfrac{Bx+C}{x^2-x+1}\\[16pt] &=\dfrac{(1/3)}{x+1}+ \dfrac{(-1/3)x+(2/3)}{x^2-x+1}\\[16pt] &=\dfrac{1}{3}\left(\dfrac{1}{x+1}+\dfrac{-x+2}{x^2-x+1}\right) \end{array}[/math]

 

[mario99]

fraction with numbers is different than fraction with letters. that's what i learn in algebra. look at \(\displaystyle x\), it's variable not number

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.

View attachment 38504

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]

 

[logistic_guy]

Corrected version:

I first observed that [math] \dfrac{1}{(x+1)(x^2-x+1)}\neq \dfrac{1}{x+1}+ \dfrac{1}{x^2-x+1} [/math]and then looked for the correct partition. The idea is to write
[math] \dfrac{1}{(x+1)(x^2-x+1)}= \dfrac{f(x)}{x+1}+ \dfrac{g(x)}{x^2-x+1}. [/math]
The only restriction is, that [imath] \deg f(x)<\deg(x+1)=1 [/imath] and [imath] \deg g(x)<\deg(x^2-x+1)=2. [/imath] We have therefore the ansatz
[math]\begin{array}{lll} \dfrac{1}{(x+1)(x^2-x+1)}&= \dfrac{A}{x+1}+ \dfrac{Bx+C}{x^2-x+1}\\[16pt] &=\dfrac{A(x^2-x+1)+(Bx+C)(x+1)}{(x+1)(x^2-x+1)}\\[16pt] &=\dfrac{x^2(A+B)+x(-A+B+C)+(A+C)}{(x+1)(x^2-x+1)}\\[16pt] \end{array}[/math]and by comparison [imath] A+B=0\, , \,-A+B+C=0\, , \,A+C=1. [/imath] Hence, [imath] B=-A\, , \,C=1-A [/imath] and [imath] 0=-A+B+C=-A+(-A)+(1-A)=1-3A [/imath] or [imath]A=1/3.[/imath] Then [imath] B=-1/3 [/imath] and [imath] C=2/3. [/imath] Thus
[math]\begin{array}{lll} \dfrac{1}{(x+1)(x^2-x+1)}&= \dfrac{A}{x+1}+ \dfrac{Bx+C}{x^2-x+1}\\[16pt] &=\dfrac{(1/3)}{x+1}+ \dfrac{(-1/3)x+(2/3)}{x^2-x+1}\\[16pt] &=\dfrac{1}{3}\left(\dfrac{1}{x+1}+\dfrac{-x+2}{x^2-x+1}\right) \end{array}[/math]

thank

i think i'm understanding. i practice to solve \(\displaystyle \frac{1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}\)

\(\displaystyle 1 = A(x^2 + 1) + (Bx + C)x = Ax^2 + A + Bx^2 + Cx = (A + B)x^2 + Cx + A\)

\(\displaystyle A + B = 0\)
\(\displaystyle C = 0\)
\(\displaystyle A = 1\)
\(\displaystyle B = -1\)

\(\displaystyle \frac{1}{x(x^2+1)} = \frac{1}{x} + \frac{-x + 0}{x^2 + 1}\)

the integration is \(\displaystyle \int_0^{\infty} \dfrac{1}{3}\left(\dfrac{1}{x+1}+\dfrac{-x+2}{x^2-x+1}\right) \ dx\)

\(\displaystyle = \frac{1}{3}\int_{0}^{\infty}\frac{1}{x + 1} \ dx - \frac{1}{3}\int_{0}^{\infty}\frac{x - 2}{x^2 - x + 1} \ dx\)

the first integration is easy \(\displaystyle \frac{1}{3}\ln(\infty + 1)\) Steven G say this can be solved by the log definition

\(\displaystyle -\frac{1}{3}\int_{0}^{\infty}\frac{x - 2}{x^2 - x + 1} \ dx\)

this i don't know how to solve

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]

thank

hold on mario99, i'll get back to you once i solve the integration