# definite integral #4

• Feb 14th 2011, 05:59 PM
Random Variable
definite integral #4
$\displaystyle \displaystyle \int^{\pi/2}_{0} x^{2} \cot x \ dx$

I have a solution using contour integration. You could either try to replicate that solution, or preferably find a simpler approach.

Spoiler:
Integrate $\displaystyle \displaystyle f(z) = \frac{z^{2}}{e^{z}-1}$ around the proper contour in the complex plane.
• Feb 14th 2011, 09:24 PM
Krizalid
Integrate by parts and the integral equals $\displaystyle -2\displaystyle\int_0^{\frac\pi2}x\ln(\sin x)\,dx,$ now use $\displaystyle \displaystyle\ln (\sin x)=-\ln 2-\sum\limits_{j=1}^{\infty }{\frac{\cos (2jx)}{j}},$ and that gives the answer $\displaystyle \dfrac18(2\pi^2\ln2-7\zeta(3)).$
• Feb 15th 2011, 09:53 AM
Random Variable
Quote:

Originally Posted by Krizalid
Integrate by parts and the integral equals $\displaystyle -2\displaystyle\int_0^{\frac\pi2}x\ln(\sin x)\,dx,$ now use $\displaystyle \displaystyle\ln (\sin x)=-\ln 2-\sum\limits_{j=1}^{\infty }{\frac{\cos (2jx)}{j}},$ and that gives the answer $\displaystyle \dfrac18(2\pi^2\ln2-7\zeta(3)).$

From where is that series coming?
• Feb 15th 2011, 10:27 AM
Krizalid
\displaystyle \begin{aligned} \ln (\cos x)&=\text{Re}\ln \left( \frac{1+{{e}^{2ix}}}{2{{e}^{2ix}}} \right) \\ & =\text{Re}\left( \ln \left( 1+{{e}^{2ix}} \right)-2ix-\ln 2 \right) \\ & =\text{Re}\left( -\sum\limits_{j=1}^{\infty }{\frac{{{(-1)}^{j}}{{e}^{2ijx}}}{j}} \right)-\ln 2 \\ & =-\sum\limits_{j=1}^{\infty }{\frac{{{(-1)}^{j}}\cos (2jx)}{j}}-\ln 2, \\ \end{aligned}

now put $\displaystyle x\mapsto\dfrac\pi2-x$ and get

$\displaystyle \displaystyle\ln (\sin x)=-\sum\limits_{j=1}^{\infty }{\frac{\cos (2jx)}{j}}-\ln 2.$
• Feb 15th 2011, 02:46 PM
topsquark
Quote:

Originally Posted by Krizalid
Integrate by parts and the integral equals $\displaystyle -2\displaystyle\int_0^{\frac\pi2}x\ln(\sin x)\,dx,$ now use $\displaystyle \displaystyle\ln (\sin x)=-\ln 2-\sum\limits_{j=1}^{\infty }{\frac{\cos (2jx)}{j}},$ and that gives the answer $\displaystyle \dfrac18(2\pi^2\ln2-7\zeta(3)).$

I'm not as quite accomplished as you Kriz, but couldn't we have saved all this trouble and used, for example, a power series solution right from the start? Wouldn't this then give us two inequivalent series with the same sum?

-Dan
• Feb 15th 2011, 02:52 PM
Krizalid
You mean power series for contangent? I don't know it.

Perhaps it's better you to share your solution. :D
• Feb 15th 2011, 03:39 PM
topsquark
Quote:

Originally Posted by Krizalid
You mean power series for contangent? I don't know it.

Perhaps it's better you to share your solution. :D

(Doh) Okay, okay! (chuckles) I guess I didn't think that one through very far, did I?

-Dan
• Feb 15th 2011, 09:03 PM
CaptainBlack
Quote:

Originally Posted by Krizalid
You mean power series for contangent? I don't know it.

Perhaps it's better you to share your solution. :D

It lurks about halfway down >>this<< page.

CB
• Feb 18th 2011, 06:18 PM
chiph588@
Quote:

Originally Posted by Krizalid
You mean power series for contangent? I don't know it.

Perhaps it's better you to share your solution. :D

$\displaystyle \displaystyle \frac{\sin\pi x}{\pi x} = \prod_{n=1}^\infty \left(1-\frac{x^2}{n^2}\right) \implies \log(\pi x) - \log(\sin\pi x) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{2n}$

$\displaystyle \displaystyle \implies \frac1x-\pi\cot(\pi x) = 2\sum_{n=1}^\infty \zeta(2n)x^{2n-1} \quad\quad \text{(for } |x|<1\text{)}$