- #1
nasim
- 9
- 0
can you solve:
[tex]\int_0^\infty\;\frac{\ln x\cdot\left( \tan^{-1} x\right)^3}{1+x^2}\;\mathrm dx[/tex]
[tex]\int_0^\infty\;\frac{\ln x\cdot\left( \tan^{-1} x\right)^3}{1+x^2}\;\mathrm dx[/tex]
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.
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.
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...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]
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...
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]
Errr yeah. My holiday officially started today, appearantly my brain immediately shut down :grumpy:nasim said:did you actually mean to say [...]
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]?
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 can't even tell what I was thinking. Sorry.Gib Z said:...
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.
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.
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].
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.
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.
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.
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.
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.