Solving $\int Log(Sin(x))dx$ - Tips & Resources

[Luhter]

How can i solve this:

\int Log(Sin(x))dx

? And if someone knows a good material on this please share :). Thanks

 

[Mark44]

A good place to start would be integration by parts, with u = ln(sin x), dv = dx. I'm assuming you are working with the natural log, ln, rather than base-10 log.

 

[Luhter]

Pretty sure I end up with \int x*Cot(x) dx and then ... ?

 

[Mark44]

You could do integration by parts again, but neither possible partition seems to get you anywhere. One choice-- u = cot x, dv = xdx gets you a harder integral, and the other -- u = x, dv = cot x dx takes you right back to the original integral being equal to itself, which is no help. If this were a definite integral, you might be able to approximate the integrand with a few terms of a Taylor's series.

I'm stumped.

 

[Дьявол]

You end up with:

xln(sin(x)) - \int x*Cot(x) dx

Try using Cot(x)=\sqrt{\csc^2 (x) - 1}\

and substitute t=csc(x).

 

[Bohrok]

Wolframalpha doesn't give an elementary integral...

 

[Count Iblis]

You can compute the definite integral from zero to pi quite easily, but I don't think the indefinite integral can be expressed in terms of elementary functions.

In case of the definite integral, you substitute x = 2 t, then t goes from zero to pi/2, and you write Log[sin(2 t)] as Log(2) + Log[sin(t)] + Log[cos(t)]. By symmetry the two latter integrals are half the original integral. There is a factor 2 from dx = 2 dt, so you find that the integral I satisfies the equation:

I = pi Log(2) + 2 I ------->

I = - pi Log(2)

 

[Luhter]

Ty Count Iblis :),nice approach .
I need it undefined though. So I'm sure it can't be done with "ortodox" methods. I tried even with FT , but i don't have good hold of it. So i was wondering if you know something else?
Also I think the result should be complex since Sin(x) can take a negative value.

 

[queenofbabes]

FT won't help if you're looking for a indefinite integral...right?
I'm not sure if it can be expressed with just elementary functions.

 

[Luhter]

I guessed so. So I'll try [0,y] so it won't go complex

 

[RoyalCat]

I guessed so. So I'll try [0,y] so it won't go complex

For a non complex result, wouldn't you need sin(x) to remain positive?
That would mean you have to restrict yourself to [0,\pi], and not to [0,y], wouldn't it? In which case, the definite integral has been shown by Count Ilbis.

 

[Luhter]

Yes.True, forgot to mention. But for 0->\pi/3 it won't work for example.

The idea is I'm missing some course,so i xero-copied from a colleague.Yet the only thing i got for this part from his courses (between 2 mickey mouse figures) is this integral defined once from 0 to \pi/4 which gives : -(G/2 + \pi*Log[2]/4) . Which is wow , i mean what is G first.
The problem is their all on holiday , couldn't contact the ones that know. So I'm curious what is the method and what to learn
Yeah.. know sounds dumb, but it happens :|

 

[Count Iblis]

You can express the indefinite integral in terms of the Barnes G-function, http://arxiv.org/abs/math/0308086"