1. ## tough integral?.

Here is an integral I have been wrestling with. Does anyone have a technique?.

I have attempted many different ways, but always get stuck.

I have tried all sorts of series, gamma/beta, etc.

I know there is a $\displaystyle {\zeta}(3)$ in the solution, but getting there..

Kriz?. PaulRS?.

Commutative on SOS posed this:

$\displaystyle \int_{0}^{\frac{\pi}{2}}xln(sin(x))dx$

Just plain ln(sinx) isn't a problem, but that x tacked on.....

I tried various series which are equaivalent:

$\displaystyle \frac{-1}{2}\int_{0}^{1}\frac{(sin^{-1}(x))^{2}}{x}dx$

$\displaystyle \frac{-1}{2}\int_{0}^{1}x^{2}cot(x)dx$

Making a u sub gives:

$\displaystyle \int_{0}^{1}\frac{sin^{-1}(u)ln(u)}{\sqrt{1-u^{2}}}du$

I found others as well, but I will stop here for now.

I get a good ways through it, but then hit a wall.

This has been bugging me. Anyone have some clever way.

Personally, I like the gamma/beta thing, but anyway will do.

Some workings may be found here:

http://www.mathhelpforum.com/math-he...tegrals-3.html

2. We have: $\displaystyle - \ln \left[ {\sin \left( x \right)} \right] - \ln \left( 2 \right) = \tfrac{{\cos \left( {2x} \right)}} {1} + \tfrac{{\cos \left( {4x} \right)}} {2} + \tfrac{{\cos \left( {6x} \right)}} {3} + ...$ for $\displaystyle 0 < x < \pi$

Thus: $\displaystyle - x \cdot \ln \left[ {\sin \left( x \right)} \right] - x \cdot \ln \left( 2 \right) = \tfrac{{x \cdot \cos\left( {2x} \right)}} {1} + \tfrac{{x \cdot \cos \left( {4x} \right)}} {2} + ...$

Integrate (assuming it's possible to exchange the series and the integral): $\displaystyle - \int_0^{\tfrac{\pi } {2}} {x \cdot \ln \left[ {\sin \left( x \right)} \right]dx} - \tfrac{{\pi ^2 }} {8} \cdot \ln \left( 2 \right) = \tfrac{{\int_0^{\tfrac{\pi } {2}} {x \cdot \cos\left( {2x} \right)dx} }} {1} + \tfrac{{\int_0^{\tfrac{\pi } {2}} {x \cdot \cos \left( {4x} \right)dx} }} {2} + ...$

Now $\displaystyle \int_0^{\tfrac{\pi } {2}} {x \cdot \cos \left( {2 \cdot k \cdot x} \right)dx} = - \left[ {\tfrac{{1 + \left( { - 1} \right)^{k + 1} }} {{\left( {2k} \right)^2 }}} \right]$ for all positive integers $\displaystyle k$

Thus: $\displaystyle \int_0^{\tfrac{\pi } {2}} {x \cdot \ln \left[ {\sin \left( x \right)} \right]dx} = \tfrac{1} {4} \cdot \sum\limits_{k = 1}^\infty {\tfrac{{1 + \left( { - 1} \right)^{k + 1} }} {{k^3 }}} - \tfrac{{\pi ^2 }} {8} \cdot \ln \left( 2 \right)$

And: $\displaystyle \int_0^{\tfrac{\pi } {2}} {x \cdot \ln \left[ {\sin \left( x \right)} \right]dx} = \tfrac{1} {2} \cdot \sum\limits_{k = 1}^\infty {\tfrac{1} {{\left( {2k + 1} \right)^3 }}} - \tfrac{{\pi ^2 }} {8} \cdot \ln \left( 2 \right)$

Since: $\displaystyle \sum\limits_{k = 1}^\infty {\tfrac{1} {{\left( {2k + 1} \right)^3 }}} = \tfrac{7} {8} \cdot \zeta \left( 3 \right)$ we get $\displaystyle \int_0^{\tfrac{\pi } {2}} {x \cdot \ln \left[ {\sin \left( x \right)} \right]dx} = \tfrac{7} {{16}} \cdot \zeta \left( 3 \right) - \tfrac{{\pi ^2 }} {8} \cdot \ln \left( 2 \right)$

3. Cool, Paul. I knew you would march through it with little effort.

There are some identities I didn't make the connection.

4. Originally Posted by PaulRS
We have: $\displaystyle - \ln \left[ {\sin \left( x \right)} \right] - \ln \left( 2 \right) = \tfrac{{\cos \left( {2x} \right)}} {1} + \tfrac{{\cos \left( {4x} \right)}} {2} + \tfrac{{\cos \left( {6x} \right)}} {3} + ...$ for $\displaystyle 0 < x < \pi$

Thus: $\displaystyle - x \cdot \ln \left[ {\sin \left( x \right)} \right] - x \cdot \ln \left( 2 \right) = \tfrac{{x \cdot \cos\left( {2x} \right)}} {1} + \tfrac{{x \cdot \cos \left( {4x} \right)}} {2} + ...$

Integrate (assuming it's possible to exchange the series and the integral): $\displaystyle - \int_0^{\tfrac{\pi } {2}} {x \cdot \ln \left[ {\sin \left( x \right)} \right]dx} - \tfrac{{\pi ^2 }} {8} \cdot \ln \left( 2 \right) = \tfrac{{\int_0^{\tfrac{\pi } {2}} {x \cdot \cos\left( {2x} \right)dx} }} {1} + \tfrac{{\int_0^{\tfrac{\pi } {2}} {x \cdot \cos \left( {4x} \right)dx} }} {2} + ...$

Now $\displaystyle \int_0^{\tfrac{\pi } {2}} {x \cdot \cos \left( {2 \cdot k \cdot x} \right)dx} = - \left[ {\tfrac{{1 + \left( { - 1} \right)^{k + 1} }} {{\left( {2k} \right)^2 }}} \right]$ for all positive integers $\displaystyle k$

Thus: $\displaystyle \int_0^{\tfrac{\pi } {2}} {x \cdot \ln \left[ {\sin \left( x \right)} \right]dx} = \tfrac{1} {4} \cdot \sum\limits_{k = 1}^\infty {\tfrac{{1 + \left( { - 1} \right)^{k + 1} }} {{k^3 }}} - \tfrac{{\pi ^2 }} {8} \cdot \ln \left( 2 \right)$

And: $\displaystyle \int_0^{\tfrac{\pi } {2}} {x \cdot \ln \left[ {\sin \left( x \right)} \right]dx} = \tfrac{1} {2} \cdot \sum\limits_{k = 1}^\infty {\tfrac{1} {{\left( {2k + 1} \right)^3 }}} - \tfrac{{\pi ^2 }} {8} \cdot \ln \left( 2 \right)$

Since: $\displaystyle \sum\limits_{k = 1}^\infty {\tfrac{1} {{\left( {2k + 1} \right)^3 }}} = \tfrac{7} {8} \cdot \zeta \left( 3 \right)$ we get $\displaystyle \int_0^{\tfrac{\pi } {2}} {x \cdot \ln \left[ {\sin \left( x \right)} \right]dx} = \tfrac{7} {{16}} \cdot \zeta \left( 3 \right) - \tfrac{{\pi ^2 }} {8} \cdot \ln \left( 2 \right)$
One question, please. Where did that $\displaystyle k^{3}$ come from after the $\displaystyle (2k)^{2}$ in the denominator?.

5. Originally Posted by galactus
One question, please. Where did that $\displaystyle k^{3}$ come from after the $\displaystyle (2k)^{2}$ in the denominator?.
Remember that the series on the RHS is: $\displaystyle \sum\limits_{k = 1}^\infty {\tfrac{{\int_0^{\tfrac{\pi } {2}} {x \cdot \cos \left( {2 \cdot k \cdot x} \right)} }} {k}}$

6. Paul, I gotta ask. Is it that you are just naturally brilliant and just 'see' these things, derive them, know them before hand from previous applications, etc?.

For instance, how did you know to use $\displaystyle -ln(sinx)-ln(2)=\sum_{k=1}^{\infty}\frac{cos(2kx)}{k}$?.

Where does that come from?. This is certainly one wacky identity I did not know.

It amazes me how you come up with these obscure series identities.

Using the series for ln(1-x) and subbing in $\displaystyle cos^{2}(x)$, I got

it down to $\displaystyle \frac{x}{2}ln(1-cos^{2}(x))$. Which leads to

$\displaystyle \frac{-1}{2}\int_{0}^{\frac{\pi}{2}}\sum_{k=1}^{\infty}\f rac{xcos^{2k}(x)}{k}dx$,

but got stuck there. I can't even get Maple to give me a solution of this. At least with your cos(2kx) identity it does give a closed form.

I will pick at it some more.

I am doing this to learn a little more about the way you, Kriz, PH and other gifted ones manipulate these series and identities so well.

Anyway, thanks for your input. It was great.

7. Originally Posted by galactus
There are some identities I didn't make the connection.
This is how it is derived.
(Hopefully no mistakes).

Remember that $\displaystyle \cos x = \cosh(ix)= \tfrac{1}{2}(e^{ix} + e^{-ix})$.

$\displaystyle \sum_{n=1}^{\infty} \frac{\cos 2nx}{n} = \frac{1}{2}\sum_{n=1}^{\infty}\frac{e^{2inx} + e^{-2inx}}{n} = \frac{1}{2}\sum_{n=1}^{\infty} \frac{(e^{2ix})^n}{n} + \frac{1}{2}\sum_{n=1}^{\infty} \frac{(e^{-2ix})^n}{n}$
This gives us,
$\displaystyle -\frac{1}{2}\left[ \log (1 - e^{2ix}) + \log (1 - e^{-2ix}) \right]$$\displaystyle = - \frac{1}{2} \left[ \log (-e^{ix}(e^{ix}-e^{-ix})) + \log (e^{-ix}(e^{ix} - e^{-ix})) \right]$.
Thus,
$\displaystyle -\frac{1}{2}\log (e^{ix} - e^{-ix})^2 = - \log (e^{ix} - e^{-ix}) = - \log (2 \sin x) = - \log 2 - \log (\sin x)$

Commutative on SOS posed this:
Did you know Commutative and NonCommutativeAlgebra is the same person?

8. Did you know Commutative and NonCommutativeAlgebra is the same person?
Yes, as a matter of fact, I did.

Thanks for the series derivation. You all are learning me some new things and I like that.

Opening my eyes to things I have seen but never really thought about. Stuck in a rut I suppose.

9. And this is my approach for that identity.

$\displaystyle \sum_{n=1}^{\infty} \frac{\cos 2nx}{n} = \text{Re} \left [ \sum_{n=1}^{\infty} \frac{e^{2inx}}{n} \right ] = \text{Re} \left [ -\ln (1-e^{2ix}) \right ]$ $\displaystyle ~ = \text{Re} \left [ -\ln|1-e^{2ix}|-i \text{arg}(1-e^{2ix}) \right ]= - \ln |1-e^{2ix}|$

Now $\displaystyle |1-e^{2ix}| = |\cos^2 x + \sin^2x - \cos 2x - i \sin 2x| = |2\sin^2x - i \sin 2x| = \sqrt{4\sin^4 x + \sin^2 2x}$

$\displaystyle = \sqrt{4\sin^4 x + 4\sin^2x \cos^2x} = 2\sin x \sqrt{\sin^2 x + \cos^2 x} = 2\sin x$

So $\displaystyle \sum_{n=1}^{\infty} \frac{\cos 2nx}{n} = -\ln (2\sin x) = -\ln 2 - \ln \sin x$

10. Originally Posted by wingless
Now $\displaystyle |1-e^{2ix}| = |\cos^2 x + \sin^2x - \cos 2x - i \sin 2x| = |2\sin^2x - i \sin 2x| = \sqrt{4\sin^4 x + \sin^2 2x}$
Nice approach wingless.

But you can do it easier. $\displaystyle |e^{ix}(e^{ix}-e^{-ix})| = |e^{ix}-e^{-ix}| = |2i\sin x| = 2\sin x$

11. Originally Posted by ThePerfectHacker
Nice approach wingless.

But you can do it easier. $\displaystyle |e^{ix}(e^{ix}-e^{-ix})| = |e^{ix}-e^{-ix}| = |2i\sin x| = 2\sin x$
Nice =)

One more way to find $\displaystyle |1-e^{2ix}|$. Let $\displaystyle A = 1$ and $\displaystyle B = e^{2ix}$ on the complex plane. Then $\displaystyle |1-e^{2ix}|$ is the distance between these points. As $\displaystyle |OA| = |OB|$, this is an isosceles triangle. $\displaystyle \angle AOB = 2x$ so the law of cosines gives $\displaystyle |1-e^{2ix}| = \sqrt{2-2\cos 2x} = 2 \sin x$.