# Hard integral, help. (1 Viewer)

#### 5uMath

##### Member

Can someone please help with the question above, but replace ln( x ) with ln^2 (x).

I am asked to prove it equals pi cubed on 8, but when I calculate residue, this is what I get up (maybe with negative) until i apply contour integration, to which the integral converges to 0.

Also, you must use cauchy residue theorem, cant be real analysis.

I dont know why I keep getting 0, I am following methods correctly and using formulae correctly...

#### blyatman

##### Well-Known Member
My guess is that you're getting zero because you assumed that
$\bg_white \int_0^\infty\frac{\ln^2x}{1+x^2}dx=\frac{1}{2}\int_{-\infty}^\infty\frac{\ln^2x}{1+x^2}dx$
If you then apply the residue theorem using the 2 poles at $\bg_white z=\pm i$, they end up cancelling out.

If that's the step you took, then I believe the mistake is in the very first line. You cannot simply halve the integral from $\bg_white -\infty$ to $\bg_white \infty$ because $\bg_white \ln z\neq\ln(-z)$. The integral needs to be split into 2 domains and evaluated separately. The video below shows how to integrate the expression in the image, but can be extended to the integral involving the $\bg_white \ln^2x$ by the same process.

It's been a while since I've touched this stuff, so I could be wrong.

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#### 5uMath

##### Member
My guess is that you're getting zero because you assumed that
$\bg_white \int_0^\infty\frac{\ln^2x}{1+x^2}dx=1/2\int_{-\infty}^\infty\frac{\ln^2x}{1+x^2}dx$
If you then apply the residue theorem using the 2 poles at $\bg_white z=\pm i$, they end up cancelling out.

If that's the step you took, then I believe the mistake is in the very first line. You cannot simply halve the integral from $\bg_white -\infty$ to $\bg_white \infty)$ because $\bg_white \ln z\neq\ln(-z)$. The integral needs to be split into 2 domains and evaluated separately. The video below shows how to integrate the expression in the image, but can be extended to the integral involving the $\bg_white \ln^2x$ by the same process.

It's been a while since I've touched this stuff, so I could be wrong.
I cannot believe I forgot to say this. Contour defined on epsilon <= |z| <= R, epsilon goes to 0 and R goes to infinity.

If you integrate over the full contour you would get 0 according to jordans lemma

#### blyatman

##### Well-Known Member
I cannot believe I forgot to say this. Contour defined on epsilon <= |z| <= R, epsilon goes to 0 and R goes to infinity.

If you integrate over the full contour you would get 0 according to jordans lemma
The integrand is non-negative, so you can't get zero lol. Throw it into wolframalpha and you get 3.7578.

#### 5uMath

##### Member
My guess is that you're getting zero because you assumed that
$\bg_white \int_0^\infty\frac{\ln^2x}{1+x^2}dx=1/2\int_{-\infty}^\infty\frac{\ln^2x}{1+x^2}dx$
If you then apply the residue theorem using the 2 poles at $\bg_white z=\pm i$, they end up cancelling out.

If that's the step you took, then I believe the mistake is in the very first line. You cannot simply halve the integral from $\bg_white -\infty$ to $\bg_white \infty)$ because $\bg_white \ln z\neq\ln(-z)$. The integral needs to be split into 2 domains and evaluated separately. The video below shows how to integrate the expression in the image, but can be extended to the integral involving the $\bg_white \ln^2x$ by the same process.

It's been a while since I've touched this stuff, so I could be wrong.
I didnt halve. I integrated over the full contour defined, got the integral over the two semi circles =0, applied let z=-z then combined the integrals to get limits epsilon to R. In the process you get these other integrals, some which go to zero, and some which go to, collectively, pi cubed on 3 and then the 0 would appear

#### 5uMath

##### Member
The integrand is non-negative, so you can't get zero lol. Throw it into wolframalpha and you get 3.7578.
Thats exactly right. I know im wrong, but i dont want to result to calculators

#### 5uMath

##### Member
I cant see your latex text

#### blyatman

##### Well-Known Member
Oh right sorry, I misunderstood what you meant. I thought you realized that the integral must equal zero, rather than stating what you were doing. Either way, if you use the approach in the video, it should lead to the correct result.

#### 5uMath

##### Member
Oh right sorry, I misunderstood what you meant. I thought you realized that the integral must equal zero, rather than stating what you were doing. Either way, if you use the approach in the video, it should lead to the correct result.

The question is meant to be loge x squared. ive seen that and applied the same thing but get 0... ill send my answers tomorrow

#### blyatman

##### Well-Known Member
I cant see your latex text
Yeh it took a while for it to load on my end, tex commands seem to be really slow on bos these days. I can see it now after I waited a bit and hit refresh a few times.

#### 5uMath

##### Member
Yeh it took a while for it to load on my end, tex commands seem to be really slow on bos these days. I can see it now after I waited a bit and hit refresh a few times.
Yes I see it now.

I did split ln (-z) to ipi + lnz, and used it wherever it needed to be. That was the reason i kept getting 0

#### blyatman

##### Well-Known Member
The question is meant to be loge x squared. ive seen that and applied the same thing but get 0... ill send my answers tomorrow
Yeh might help if you detail your working. Either way it's been a long time since I've done this stuff so I don't know how much of it I can remember. Had to google the formula for the residue theorem to refresh my memory lol. Maybe someone else can jump in and offer assistance.

#### 5uMath

##### Member
Yeh might help if you detail your working. Either way it's been almost a long time since I've done this stuff so I don't know how much of it I can remember. Had to google the formula for the residue theorem was to refresh my memory.
Oh thats fine. I hope one day to take analysis after school. Not real analysis, never really into it, but i like complex and functional

#### blyatman

##### Well-Known Member
Oh thats fine. I hope one day to take analysis after school. Not real analysis, never really into it, but i like complex and functional
lol I don't miss it at all, too rigorous for me. I prefer the more applied subjects. The only aspect of complex analysis I enjoyed was its application to real integrals, as well as a some applications in studying certain fluid flows, but that's about it.

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#### 5uMath

##### Member
I have attached a copy of my solution.

Turns out i didnt get 0 this time, but got pi^3/4... do you see mistakes?

Thanks

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#### blyatman

##### Well-Known Member
Your working out seems fine, not sure where the missing factor of a half is meant to be lol.

#### 5uMath

##### Member
Your working out seems fine, not sure where the missing factor of a half is meant to be lol.
I honestly dont, but pi^3/8 is the value you got from wolfram yesterday, so definitely a mistake. But where......

#### 5uMath

##### Member
I found the mistake!!! Where i evaluated the integral containing (ipi + lnx)^2, i forgot to reconsider the second integral, I2, after proving I1 =-pi^3/2, which happened to be I. Therefore you are left with ipi^3/2 + 2I = -pi^3/4, and I = pi^3/8.

Noice one haha