Complex integral: integral [0, infinity] (log x^3)/(8 + x^3)

[maxandri]

I have this integral (1)
\[\int_{0}^{\infty }\frac{logx^3}{8+x^3}\]
To start I would use
(2) \3[\int_{0}^{\infty }\frac{log^2z}{8+x^3}\]
So after some passages I have \[\int_{0}^{\infty }\frac{4\pi}{8+x^3} -\int_{0}^{\infty }\frac{logz}{8+x^3} =2\pi Res[f(2e^{i\pi/3}), f(2e^{i\pi}),(2e^{-i\pi/3}) ]\]
But now I have so strange results... for examples, I take the last Res.
\[\lim_{z->1-i\sqrt{3}}\frac{(ln2-i\sqrt{3})^2}{(1-i\sqrt{3}-(-2))(1-i\sqrt{3}-(1+i\sqrt{3}))}=?=(ln^2-\pi^2/9-21\pi/3ln2-i\sqrt{3}ln^2+\pi^2/9i\sqrt{3}-2\sqrt{3}\pi/3ln2)/24\]
Is this correct? and then the result of everything?
Thank you

 

[brianz24]

I suppose you meant the intergral with dx at the end.

Try

log(x³) = 3log(x)

x³ + 1 + 7

Factor x³ + 1 = (x+1)(x² - x + 1)

You will have

3log(x)/((x + 1)(x² - x + 1) + 7)

Use partial fraction.

 

[Steven G]

I suppose you meant the intergral with dx at the end.

Try

log(x³) = 3log(x)

x³ + 1 + 7

Factor x³ + 1 = (x+1)(x² - x + 1)

You will have

3log(x)/((x + 1)(x² - x + 1) + 7)

Use partial fraction.

Your denominator has two terms. You can only use partial fractions when the denominator has only one terms as then you can separate the factors. Alternative you can think of your denominator as one term but in this case you only have one factor--so nothing to separate.

 

[brianz24]

Your denominator has two terms. You can only use partial fractions when the denominator has only one terms as then you can separate the factors. Alternative you can think of your denominator as one term but in this case you only have one factor--so nothing to separate.

My bad, I wasn't thinking it through.

Well, you can use

Factor x³ + 2³ = (x + 2)(x² - 2x + 4)

You'll end up

3log(x)/(x+2)(x² - 2x + 4)

Then you can use partial fraction

 

[maxandri]

forgot...

sorry guys, I see now that my message was not written correctly.
I would use \[3\int_{0}^{\infty }\frac{log^2z}{8+x^3}\] to avoid to eliminate the ln. But if I want to use the 3 singularities I wrote, where is my mistake? Prof use always this method...

 

[brianz24]

sorry guys, I see now that my message was not written correctly.
I would use \[3\int_{0}^{\infty }\frac{log^2z}{8+x^3}\] to avoid to eliminate the ln. But if I want to use the 3 singularities I wrote, where is my mistake? Prof use always this method...

You are trying to find the integral with respect to what variable?

See, all integrals have a form with at least a dx or any ds within

the integral sign to denote what variable you are trying to

integrate.

 

[maxandri]

You are trying to find the integral with respect to what variable?

See, all integrals have a form with at least a dx or any ds within

the integral sign to denote what variable you are trying to

integrate.

Sorry dz, I mean the excercize is with x but then I change everything in z to solve it in complex numbers. So it is\[\frac{log^2z}{8+z^3}dz\]

 

[brianz24]

Sorry dz, I mean the excercize is with x but then I change everything in z to solve it in complex numbers. So it is\[\frac{log^2z}{8+z^3}dz\]

I can't help you if you can't give me the right integral.

Anyway, here's what might help you:

Integration http://www.wolframalpha.com/input/?i=integrate+((log(z))(log(z)))/(8+z^3)

Definite Integral http://www.wolframalpha.com/input/?i=integrate+0+to+infinity+((log(z))(log(z)))/(8+z^3

Better see for yourself.