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

[Najmoddin]

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

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 realize 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: http://www.goiit.com/posts/list/integral-calculus-how-can-i-integrate-ln-cos-x-with-respect-1012410.htm

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

 

[micromass]

The integral \int \log(\cos(x))dx 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]

The integral \int \log(\cos(x))dx 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.

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

And more importantly, when do we use it?

 

[Byron Chen]

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

I haven't thought of the method yet, but the answer looks complicated:
http://www.wolframalpha.com/input/?i=integrate+log(cos(x))

Courtesy Wolfram Alpha.

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

 

[micromass]

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

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

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

for |z|<1. It is defined for |z|\geq 1 by "analytic continuation".

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

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

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]

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

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

for |z|<1. It is defined for |z|\geq 1 by "analytic continuation".

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

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

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)

Thanks a lot.

Can you also explain when we usually use this function?

 

[micromass]

Thanks a lot.

Can you also explain when we usually use this function?

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

Li_s(1)=\zeta (s)

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]

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

Li_s(1)=\zeta (s)

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.

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]

I haven't thought of the method yet, but the answer looks complicated

<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]

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?

Well, for example, if you encounter an integral like

\int \frac{\log(x)}{x}dx

then you can solve it using Li_2(x). 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]

...

For a function like log(g(x)) you could be tempted try for g(x)=e^h(x)
and realize that log(x)=ln(x)/ln(10).

...

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

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

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

for |z|<1. It is defined for |z|\geq 1 by "analytic continuation".

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

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

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)

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

\mbox{Li}_2(x) = - \int_0^x dt~\frac{\ln(1-t)}{t},
(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 \int_{0}^{\frac{\pi}{2}} {\ln(\cos(x))dx} = *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]

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

For example \int_{0}^{\frac{\pi}{2}} {\ln(\cos(x))dx} = *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.

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