Indefinite Integration of function's like log(cos(x))???

Najmoddin

Can anyone please help me to calculate Indefinite Integration of function's like log(cos(x))???.

Simon Bridge

To integrate something like f(g(x)) you need to use a trick - and it is not always easy.
Usually you hope that you can make a substitution like u=g(x) so that dx=du/g'(x) and g'(x) can be expressed in terms of u.
If h(u)=g'(x) then f(g(x))dx becomes f(u)du/h(u) which may or may not be an improvement.

For a function like log(g(x)) you could be tempted try for g(x)=e^h(x)
and realise that log(x)=ln(x)/ln(10).
then log(g(x))dx becomes h(x)dx/ln(10)
... but just try doing that non-trivially ... since you are just saying that h(x)=ln(g(x)) which is proportional to what you started with.

For a sin or a cos inside the log, express the function as a sum of exponentials.
Then substitute (in the case of cosine x) u=e^ix+e^-ix

ref: AIEEE forum

[edit: why is this a poll? It's not a matter of opinion...]

micromass

The integral [itex]\int \log(\cos(x))dx[/itex] can not be solved by elementary functions. To solve the integral, we need the polylogarithmic function.

So the integral does exist, but we just can't find it.

Byron Chen

Sorry if this question sounds stupid, but what is polylogarithmic function?

And more importantly, when do we use it?

Byron Chen

I haven't thought of the method yet, but the answer looks complicated:
http://www.wolframalpha.com/input/?i...8cos%28x%29%29

Courtesy Wolfram Alpha.

Even the graph of the integral is not simple and continuous and there is an imaginary component.

micromass

It is not stupid at all. It's an advanced function. It is defined as

[tex]Li_s(z)=\sum_{n=1}^{+\infty}{\frac{z^k}{k^s}}[/tex]

for |z|<1. It is defined for [itex]|z|\geq 1[/itex] by "analytic continuation".

An example: if s=1, we get [itex]Li_1(z)=\log(z)[/itex] the ordinary logarithm.
For s=2, we get
[tex]Li_2(z)=\int_0^z \frac{\log(z)}{z}dz[/tex]
For s=3, we have
[tex]Li_3(z)=\int_0^z \frac{Li_2(z)}{z}dz[/tex]

In general, we always have
[tex]Li_s(z)=\int_0^z \frac{Li_{s-1}(z)}{z}dz[/tex]

This might look confusing. The only thing I want to make clear is that the polylogarithm is usually not always a elementary function (except for special values such as s=1)

Byron Chen

Thanks a lot.

Can you also explain when we usually use this function?

micromass

You can use it for a lot of things. One example is of course to calculate integrals. Many integrals can be calculated using polylogarithmic functions.

Another example is the connection between the Riemann-zeta function. We have that

[tex]Li_s(1)=\zeta (s)[/tex]

and the zeta function has of course many applications in number theory. As such, the polylogarithmic functions become important in number theory.

But to be honest, you'll never see these functions unless reading quite advanced books in complex analysis or (analytic) number theory.

Byron Chen

Just a little curious about the integration part. In which cases do you usually use the polylogarithmic functions when you do integration? Like is there certain types of functions where taking the integral in such a manner is useful?

Simon Bridge

<mutter> I tells people the method, I gives them a link to a walkthrough, does anybody follow the link? Hah! When I was young we followed advice oh yes ... we didn't have links: we had to go to these big buildings with papery things in them and thumb through drawers and drawers of small typed cards to find our information and we were better for it <grumble> time for bed...

micromass

Well, for example, if you encounter an integral like

[tex]\int \frac{\log(x)}{x}dx[/tex]

then you can solve it using [itex]Li_2(x)[/itex]. Likewise, if you can reduce an integral to that, then you can also solve it.

I'm sure there's more to it, but I'm not an expert on this (by far).

Mute

I would assume the OP was using the mathematician's convention of [itex]\log x = \ln x[/itex] rather than the engineering convention that [itex]\log(x) = \log_{10}(x)[/itex].

Actually, just to correct a slight mistake, the sum for s =1 gives [itex]\mbox{Li}_1(z) = -\log(1-z)[/itex], and so

[tex]\mbox{Li}_2(x) = - \int_0^x dt~\frac{\ln(1-t)}{t},[/tex]
(reverting to ln notation because I'm using real variables for the integral representation).

Boorglar

On a related note, the definite integral can sometimes be calculated for such functions.

For example [tex] \int_{0}^{\frac{\pi}{2}} {\ln(\cos(x))dx} = [/tex] *something involving pi and the natural log of 2 (but I don't remember right now). It can be proven using a substitution trick, and trig identities.

Mandlebra

I think this guy gets the u= x-pi/2 treatment. symmetry :)