Integral of (ln(cos x))^3

Finishing our epic saga, we present the integral of (ln(cos x))^3 over the interval 0 to pi/2.

[TheRandomFool]

I found a generalization for the integral log(cos(x))^n from 0 to pi/2 where n is a positive real number, but in terms of an infinite series. Non-integer values of n will produce a complex number, though. It is exp(pi*i*n)*gamma(n+1)*sum from k=0 to inf of (2k-1)!! / ( 2^(k)*k!*(2k+1)^(n+1) ). It looks like it should converge relatively quick.

[carlosgiovanardi]

awesome result!
please, more super integrals like this!

[gregsarnecki]

If you study multiple zeta values the double sum at 26:30 is just zeta(2, 1) which Euler showed many years ago is z(3).

[technoguyx]

That was monstruous, nice job. I wonder if there's a general formula for \int ln(cos(x))^n dx.

[sanjaysuresh]

Wow am I early :)) You should make some more IMO videos!

[gregsarnecki]

Also, I think it helps see trends better if you look for instances of (pi^2)/6 and replace it with another symbol. You could use zeta(2) of course, but I prefer to refer to (pi^2)/6 as B^ (in honour of The Basel Problem); A^ is zeta (3) after Apery. So, (pi^3)/24 can be written as (pi/4)*B^.

[mathunt]

I would make two successive integrals, u=cos(x), and then v=log(u), this is left with the following integral:



The next thing to do is to expand the square root as a power series and the integrals are then of the form v^3*exp(k*v) which are relatively simple to do. It leaves a sum as the answer though which isn't all that satisfactory an answer.

[seanmacfoy]

13:59 Thinking about it, your way might be slightly less labour intensive when you consider all the substitution, but it looks like if you substituted in the series, you'd end up with the negative of the original integral + some integrals you've evaluated in previous videos. Could one not just collect the original integral on the left hand side and divide both sides by 2, yielding the value of the original integral?

[kyintegralson]

Nice video & content. However, even though you present the correct answer at the end, there's an error in your calculation, which is not noticed b/c you don't go through it. The error 1st shows up in the last equation you display on the left-hand column. The integral of the product of the cosines should be (pi/4) x (del(m, n)+del(m, -n)), where "del" means the Kronecker delta. If you only had a double infinite sum (over m & n), then del(m, -n) wouldn't contribute & you'd be correct to drop it. But b/c you're gonna do triple inf't sum, you can't drop that term.
So @ 21:48 in the bottom equation in the right column (blue *), the 1st cos product gives not only del(n, 2m), but also del(n, 0), which doesn't make a contribution to the triple sum. But from the 2nd cos product in that expression, not only you get del(n, 2m-2l), but also del(n, 2l), which does make a contribution in your triple sum. Each of the 3 Kron. deltas contribute equally, hence, you end up with factor of 3 multiplying the Riemann Zeta. But, since @ 27:30 you skip the 2nd Kron. delta calculation on the board, the error is not noticed.

[nareshmehndiratta]

now i have begin to see your every lecture

[manthing]

Will you elaborate on "plothole" at 27:08 ? Also, the triple sum furmula?

[jonathanbekaert]

Is there maybe a general formula for (ln(cos(x)))^n?

[jotajaviergonzalezgarcia]

how can I put this on wolfrang alpha? integrate (ln(cos x))^3 dx from x=0 to pi/2 gives me integral_0^(π/2) log^3(cos(x)) dx = -6.04188

[polychromaa]

How the hell did he transform a sum cubed into a triple sum??

[CoderboyPB]

Nice, that trying by myself, I'd have burned my chalk board, at latest at 5:42 and would have been starting drinking :D
(Sorry, for this bad grammar, I am not a native english speaker, as you can see ... ;-))

[popalofiti]

Might as well just do a general case next!

[ywjk]

how you change sum^3 to triple sum .What form?

[ecoidea]

When m=1, the harmonic number H_m-1 does not exist, how is that treated?

[holyshit]

Int(ln(cos(x))^3, x=0..Pi/2)
t=-ln(cos(x))
-t=ln(cos(x))
exp(-t)=cos(x)
-exp(-t)dt=-sin(x)dx
exp(-t)dt=sin(x)dx
exp(-t)dt=sqrt(1-cos(x)^2)dx
exp(-t)dt = sqrt(1-exp(-2t))dx
dx=exp(-t)/sqrt(1-exp(-2t))dt
Int(-t^3exp(-t)/sqrt(1-exp(-2t)), t=0..infinity)
Let F(s) = L(1/sqrt(1-exp(-2t)))
then calculate d^3/ds^3 F(s) and evaluate it at s = 1
F(s) can be expressed with Beta function

[ZIN]

Sorry your solution this time is not clear.