Is there a fourier series for ln(x)?

[ChowPuppy]

If so, could it be used to integrate sin(x)/ln(x)

 

[LCKurtz]

You aren't going to find a simple antiderivative of that and if you are looking for a definite integral just use a numerical method directly on the integral.

 

[romsofia]

You aren't going to find a simple antiderivative of that and if you are looking for a definite integral just use a numerical method directly on the integral.

That was the second part of the question, I believe he was more looking for the first part... (And if you believe substituting a function with a Fourier series in order to integrate it is easy, then you must be a genius...)

Anyways, doesn't a function have to be oscillating to have a Fourier series?

 

[homeomorphic]

Yes, and yes, but both qualified yes's.

Any reasonable integrable function has a Fourier series (actually, I believe it's ANY Lebesgue integrable function, if you allow divergence on a set of measure 0 if I remember right). Dichirlet conditions. If you restrict to most intervals ln x will satisfy that.

http://en.wikipedia.org/wiki/Dirichlet_conditions

The only problem is around x = 0.

Actually, for the second part, you want a Fourier series for 1/ ln x. Then, all but one of the terms will be orthogonal to sin x on a well-chosen interval, so all you need is one of the Fourier coefficients. But for one thing, it's circular because the way you FIND the Fourier coefficient is to do that integral. And for another thing, you can only do it over a period of sin x because that is what orthogonality refers to. So, it doesn't really have any significance, as far as I'm aware.

Anyways, doesn't a function have to be oscillating to have a Fourier series?

Not exactly. It has to be periodic. But if you just want a Fourier series that works over an interval, just take that interval and look at that, and then take infinitely many copies of the function on that interval and it's periodic.

 

[CellCoree]

Yes, there is a Fourier series for ln(x). It is given by:

ln(x) = 2∑n=1 ∞ (-1)n+1 (x-1)n/n

This series can be used to approximate the natural logarithm function for values of x close to 1. However, it is not a valid representation for all values of x, as it diverges for x=0 and x=2.

As for integrating sin(x)/ln(x), we can use the Fourier series for ln(x) to rewrite the integral as:

∫sin(x)/ln(x) dx = ∫sin(x) * 2∑n=1 ∞ (-1)n+1 (x-1)n/n dx

= 2∑n=1 ∞ (-1)n+1 ∫sin(x) * (x-1)n/n dx

= 2∑n=1 ∞ (-1)n+1 ∫sin(x) * x^n/n dx - 2∑n=1 ∞ (-1)n+1 ∫sin(x) * x^(n-1)/n dx

= 2∑n=1 ∞ (-1)n+1 (-1)^n * n!/n - 2∑n=1 ∞ (-1)n+1 (-1)^n * (n-1)!/n

= 2∑n=1 ∞ (-1)n+1 (-1)^n * (n-1)!/n - 2∑n=1 ∞ (-1)n+1 (-1)^n * (n-1)!/n

= 2∑n=1 ∞ (-1)n+1 (-1)^n * (n-1)!/n - 2∑n=1 ∞ (-1)n+1 (-1)^n * (n-1)!/n

= 2∑n=1 ∞ (-1)n+1 (-1)^n * (n-1)!/n - 2∑n=1 ∞ (-1)n+1 (-1)^n * (n-1)!/n

= 2∑n=1 ∞ (-1)n+1 (-1)^n * (n-1)!/n - 2∑n=1 ∞ (-1)n+1 (-1)^n *