Solve Integral: ln x tan⁻¹⁴ x / 1 + x²

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In summary: It's not zero, but it's not very nice either. Maple gives the answer as a sum of polylogarithms, lograithams and riemann zeta functions.
  • #1
nasim
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can you solve:

[tex]\int_0^\infty\;\frac{\ln x\cdot\left( \tan^{-1} x\right)^3}{1+x^2}\;\mathrm dx[/tex]
 
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  • #2
I'm going to conjecture that it is zero.
 
  • #3
It's not zero, but it's not very nice either. Maple gives the answer as a sum of polylogarithms, lograithams and riemann zeta functions.
 
  • #4
cristo said:
It's not zero, but it's not very nice either. Maple gives the answer as a sum of polylogarithms, lograithams and riemann zeta functions.

really ? :frown: i was hoping the answer would have been a linear combination of [tex]\boxed{\pi^4 \ln 2}[/tex] and [tex]\boxed{\pi^2 \;\zeta{(3)}}[/tex] only
 
  • #5
I managed to get [itex]\frac{3}{64}[7\pi^2\zeta(3) - 31\zeta(5)][/itex] from Mathematica, though you might want to double check that.
 
  • #6
Parthalan said:
I managed to get [itex]\frac{3}{64}[7\pi^2\zeta(3) - 31\zeta(5)][/itex] from Mathematica, though you might want to double check that.


hey, thanks. :smile:

[ i don't have mathematica or maple, so i am doing it by hand... i am poor... :frown: ]


anyhow, i will also try and see if i can do

[tex]\int_0^\infty\;\;\frac{\ln {(1+x^2)}\cdot \left( \tan^{-1} x\right)^3}{1+x^2}\;\;\mathrm dx[/tex]
 
  • #7
Did you say
[tex]\frac{3}{64} \left(7 \pi ^2 \zeta (3)-31 \zeta (5)\right)[/tex]

That's what I got too... question is now, why is this logical? :)
 
  • #8
nasim said:
anyhow, i will also try and see if i can do

[tex]\int_0^\infty\;\;\frac{\ln {(1+x^2)}\cdot \left( \tan^{-1} x\right)^3}{1+x^2}\;\;\mathrm dx[/tex]
Probably that's easier. Incidentally, the integral of arctan x is 1/x2 which suggests partial integration; also there are a lot of 1 + x2 in there... with a little effort you might be able to do this analytically...

[edit] on second thought, the indefinite integral also contains a lot of PolyLogs and other creepy functions; the definite integral evaluates to
[tex]\frac{1}{64} \left(\pi^4 \log (4)+24 \pi^2 \zeta (3)-93 \zeta(5)\right)[/tex]
so I doubt my remarks above[/edit]
 
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  • #10
CompuChip said:
Probably that's easier. Incidentally, the integral of arctan x is 1/x2 which suggests partial integration; also there are a lot of 1 + x2 in there... with a little effort you might be able to do this analytically...

did you actually mean to say since [tex]\frac{\mathrm d}{\mathrm dx}\left( \tan^{-1} x\right) \;=\;\frac{1}{1+x^2}[/tex],
then i could do [tex]\textbf u\;=\;\ln {(1+x^2)}\;\;\;\;\;\textbf{and}\;\;\;\;\;\textbf v\;=\;\frac{\left( \tan^{-1} x\right)^3}{1+x^2}\;\;\;?[/tex]


but then you will get [tex]\lim_{R\to\infty}\;\frac{\pi^4}{8}\ln R[/tex] in the first part and you need to account for it somewhere along the rest of the integration process...


PS: [tex]\int \;\tan^{-1} x\;\;\mathrm dx\;=\;x\;\tan ^{-1} x -\frac{1}{2}\;\ln {(1+x^2)}[/tex]


CompuChip said:
[edit] on second thought, the indefinite integral also contains a lot of PolyLogs and other creepy functions; the definite integral evaluates to
[tex]\frac{1}{64} \left(\pi^4 \log (4)+24 \pi^2 \zeta (3)-93 \zeta(5)\right)[/tex]
so I doubt my remarks above[/edit]


oh darn ! there is that [tex]\zeta{(5)}[/tex] in the answer again ? :frown:

thanks anyway ! :biggrin:
 
  • #11
nasim said:
did you actually mean to say [...]
Errr yeah. My holiday officially started today, appearantly my brain immediately shut down :grumpy:
 
  • #12
If you have to do it with algebra, have you thought about using substitution with [itex]u = \tan^{-1}(x) \implies du = \frac{1}{1+x^2}\,dx[/itex] to get [itex]\int \ln(1+x^2) \cdot \left [ \tan^{-1}(x) \right]^2\cdot u\,du[/itex]?
 
  • #13
For: [tex]\int_0^{\infty} \frac{\ln x \cdot \left( \tan^{-1} x \right)^3}{1+x^2} \ dx [/tex]
The only idea I have so far to it use the substitution [tex]t=\tan^{-1} x[/tex] to get:
[tex]\int_0^{\pi/2} t^3 [ \ln (\sin t) - \ln (\cos t) ] \ dt[/tex]
 
  • #14
Ahh kummer how did you change that infinite bound to pi/2, is [itex]\lim_{x\to \infty} \tan^{-1} x = \frac{\pi}{2}[/itex]? Because I was working on a solution to that last night and I got stuck at changing the bounds.

And..
Parthalan said:
If you have to do it with algebra, have you thought about using substitution with [itex]u = \tan^{-1}(x) \implies du = \frac{1}{1+x^2}\,dx[/itex] to get [itex]\int \ln(1+x^2) \cdot \left [ \tan^{-1}(x) \right]^2\cdot u\,du[/itex]?

Did you mean [tex]\int \log (1+x^2) u^3 \frac{du}{dx} dx[/tex]? Either way that doesn't work, you still have the log as a function of x, whilst integrating with respect to u.
 
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  • #15
CompuChip said:
[edit] on second thought, the indefinite integral also contains a lot of PolyLogs and other creepy functions; the definite integral evaluates to
[tex]\frac{1}{64} \left(\pi^4 \log (4)+24 \pi^2 \zeta (3)-93 \zeta(5)\right)[/tex]
so I doubt my remarks above[/edit]

i was quite surprised to see that [tex]\zeta{(5)}[/tex] crept into the answer. :redface: i am still doing the manual calculation, things are getting quite messy, so i am taking a little break now... but i am still very interested to see how the 5th riemann zeta comes into the picture in both those integrals...
 
  • #16
Well the general solution we discovered involved polylogarithms, and apparently [itex]\mathrm{Li}_s(1) = \zeta(s)[/itex] where [itex]\Re(s) > 1[/itex], but that still doesn't really answer the question!

Gib Z said:
...
I can't even tell what I was thinking. Sorry.
 
  • #17
Gib Z said:
Ahh kummer how did you change that infinite bound to pi/2, is [itex]\lim_{x\to \infty} \tan^{-1} x = \frac{\pi}{2}[/itex]? Because I was working on a solution to that last night and I got stuck at changing the bounds.

Well, do you know the limit:
[tex]\lim_{x \rightarrow + \infty} \arctan (x) = \frac{\pi}{2}[/tex]?

Since arctan(x) only returns values on the interval [tex]\left] -\frac{\pi}{2} ; \ \frac{\pi}{2}\right[[/tex], and on that interval, it's a 1-to-1 function, we also know that: [tex]\lim_{x \rightarrow \frac{\pi}{2}} \tan (x) = + \infty[/tex], so, we have: [tex]\lim_{x \rightarrow + \infty} \arctan (x) = \frac{\pi}{2}[/tex].

You can prove that: [tex]\lim_{x \rightarrow - \infty} \arctan (x) = -\frac{\pi}{2}[/tex] in the same manner. :)
 
  • #18
Could you tell me how to derive that first limit in your post? I read something about that not being the actual limit, and that the limit does in fact not exist as the values oscillate or something like that, and the limits usually given to us, like the one in question, is the Principal Value of the limit. Although in this case I'm sure that is precisely what we want, The Principal Value of the Integral, as the integral clearly diverges otherwise.
 
  • #19
do you agree that

[tex]\lim_{x\to+\infty}\;\tan^{-1} x\;=\;\lim_{y\to 0^{+}}\;\tan^{-1} \frac{1}{y}\;=\;\lim_{y\to 0^{+}}\;\left[ \frac{\pi}{2}\;-\;\tan^{-1} y\right][/tex]

[tex]=\;\frac{\pi}{2}\;-\;\lim_{y\to 0^{+}}\;\sum_{j=0}^\infty\;\frac{(-1)^j\;y^{2j+1}}{2j+1}\;=\;\boxed{\frac{\pi}{2}}\;\;\;\quad (\;\;the\;\;series\;\;is\;\;valid\;\;for\;\;|y|<1\;\;)[/tex]

[ notice how i stated it, i said [tex]\lim_{y\to 0^{+}}[/tex], not [tex]\lim_{y\to 0}\;\;[/tex] ]

had i said [tex]\lim_{y\to 0}[/tex], then yes, the limit does not exist because if i approach 0 from the positive side, then i will get one limit, i.e. [tex]+\frac{\pi}{2}[/tex], which will be different than if i approach 0 from the negative side, where i will get a different limit, i.e. [tex]-\frac{\pi}{2}[/tex]

so, in conclusion,

[tex]\lim_{x\to+\infty}\;\tan^{-1} x\;=\;\lim_{y\to 0^{+}}\;\tan^{-1} \frac{1}{y}\;=\;\frac{\pi}{2}[/tex]

and

[tex]\lim_{x\to-\infty}\;\tan^{-1} x\;=\;\lim_{y\to 0^{-}}\;\tan^{-1} \frac{1}{y}\;=\;-\frac{\pi}{2}[/tex]
 
  • #20
I think the point Gib Z is making (correct me if I'm wrong) is that, since [itex]\tan^{-1}(x)[/itex] is multivalued, we're using the branch cut of [itex](-\frac{\pi}{2}, \frac{\pi}{2})[/itex]. This is our choice of principal values, and then the limit is indeed [itex]\lim_{x \rightarrow \infty} \tan^{-1}(x) = \frac{\pi}{2}[/itex].
 
  • #21
Gib Z said:
Could you tell me how to derive that first limit in your post? I read something about that not being the actual limit, and that the limit does in fact not exist as the values oscillate or something like that, and the limits usually given to us, like the one in question, is the Principal Value of the limit. Although in this case I'm sure that is precisely what we want, The Principal Value of the Integral, as the integral clearly diverges otherwise.

Well, I did some kind of proving that in the previous post. =.="

You should note that, tan(x) itself does not have an inverse function, since it is a many-to-one function. But, like sin(x), if we restrict the interval just to be [tex]\left] -\frac{\pi}{2} ; \ \frac{\pi}{2}\right[[/tex], then, on that interval, tan(x) meets all the requirements to have an inverse. Its inverse function is defined to be arctan(x), or some may even written that as tan-1(x) (i.e, the inversion of tan(x)).

The domain of arctan(x) is all the reals, and its image is on the interval: [tex]\left] -\frac{\pi}{2} ; \ \frac{\pi}{2}\right[[/tex]. We notice that:
[tex]\lim_{x \rightarrow \frac{\pi}{2} ^ -} \tan (x) = + \infty[/tex], so, that mean: [tex]\lim_{x \rightarrow + \infty} \arctan (x) = \frac{\pi}{2}[/tex] (since arctan is an inverse of tan).

Can you get this? :)
 
  • #22
It doesn't have an inverse in the typical sense of a function, however, it can be thought of as a multivalued function, and these have been studied extensively. Consider the example of [itex]x\mapstox^2[/itex]. Then its inverse is [itex]\pm \sqrt{x}[/itex]. If we choose by convention to take the positive roots, we have the inverse [itex]\sqrt{x}[/itex], and this is known as the principal square root. The same situation occurs with the inverse of [itex]\tan(x)[/itex]. The restriction of the range as we did earlier is called a branch cut, and the principal values are those which are chosen by convention as the result for a given argument.
 
  • #23
Parthalan said:
I think the point Gib Z is making (correct me if I'm wrong) is that, since [itex]\tan^{-1}(x)[/itex] is multivalued, we're using the branch cut of [itex](-\frac{\pi}{2}, \frac{\pi}{2})[/itex]. This is our choice of principal values, and then the limit is indeed [itex]\lim_{x \rightarrow \infty} \tan^{-1}(x) = \frac{\pi}{2}[/itex].

Well that was not my actual point, but you are not at fault because I was the one mistaken :) I confused it with the cases of tan (rather than arctan), sin and cos where [tex]\lim_{x\to\infty} \mbox{One of the Trig Functions}[/tex] has no actual value because of the functions periodic nature. So instead sometimes they are given principal values. But I see It was my confusion, thanks :)

O and VietDao29, sorry about the request for a proof when in was in my face, i feel so stupid! And thanks nasim for a longer version :P
 

1. What is an integral?

An integral is a mathematical concept that represents the area under a curve in a given interval. It is used to solve problems related to finding the total accumulation of a quantity over a given range.

2. What is ln x?

ln x, also known as the natural logarithm, is the inverse function of the exponential function. It represents the power to which the base e, a mathematical constant approximately equal to 2.718, must be raised to produce a given number x.

3. What is tan⁻¹⁴ x?

tan⁻¹⁴ x, also known as the inverse tangent function, is the angle whose tangent is x. It can be thought of as the opposite of the tangent function, which calculates the ratio of the opposite side to the adjacent side of a right triangle.

4. How do I solve this integral?

To solve this integral, you can use integration by parts or substitution. You can also use a calculator or computer software to find the numerical value of the integral.

5. What is the purpose of solving integrals?

Solving integrals has various applications in mathematics, physics, and engineering. It allows us to find the area under a curve, find the volume of a solid, and calculate work and displacement. It is also useful in solving differential equations, which are used to model various real-world phenomena.

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