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## Proving identities of special functions

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### Proving identities of special functions

Thu Aug 29, 2013 11:39 am

Posts: 850
Location: Jaipur, India

Proving Identities of Special Functions

### Re: Proving identities of special functions

Thu Aug 29, 2013 11:42 am

Posts: 850
Location: Jaipur, India

Here we go!

Problem 1

Show that

$\displaystyle \text{Li}_2 \left(\frac{1}{3} \right)-\frac{1}{6}\text{Li}_2 \left(\frac{1}{9} \right)=\frac{\pi^2}{18}-\frac{\log^3 3}{6}$

### Re: Proving identities of special functions

Thu Aug 29, 2013 9:12 pm
Random Variable Integration Guru

Posts: 381
We will need two functional equations of the dilogarithm.

$\displaystyle \text{Li}_{2}(x) + \text{Li}_{2}(-x) = \frac{1}{2} \text{Li}_{2}(x^{2})$ (1)

$\displaystyle \text{Li}_{2}(1-x) + \text{Li}_{2} (1-x^{-1}) = - \frac{1}{2} \ln^{2}(x)$ (2)

$\displaystyle \text{Li}_2 \left(\frac{1}{3} \right)-\frac{1}{6}\text{Li}_2 \left(\frac{1}{9} \right) = \text{Li}_{2} \left(\frac{1}{3} \right) - \frac{1}{6} \Big[ 2 \text{Li}_2 \left(\frac{1}{3} \right) + 2 \text{Li}_2 \left(-\frac{1}{3} \right) \Big]$ (1)

$\displaystyle = \frac{2}{3} \text{Li}_2 \left(\frac{1}{3} \right) - \frac{1}{3} \text{Li}_2 \left(-\frac{1}{3} \right)$

$\displaystyle = \frac{2}{3} \text{Li}_2 \left(\frac{1}{3} \right) - \frac{1}{3} \Big[ -\text{Li}_2 \left(1-\frac{3}{4} \right) - \frac{1}{2} \ln^{2} \left(\frac{4}{3} \right) \Big]$ (2)

$\displaystyle = \frac{2}{3} \text{Li}_2 \left(\frac{1}{3} \right) + \frac{1}{6} \ln^{2} \left(\frac{4}{3} \right) + \frac{1}{3} \text{Li}_2 \left( \frac{1}{4} \right)$

$\displaystyle = \frac{2}{3} \text{Li}_2 \left(\frac{1}{3} \right) + \frac{1}{6} \ln^{2} \left(\frac{4}{3} \right) + \frac{1}{3} \Big[ 2\text{Li}_2 \left(\frac{1}{2} \right) + 2 \text{Li}_2 \left(-\frac{1}{2} \right) \Big]$ (1)

$\displaystyle = \frac{2}{3} \text{Li}_2 \left(\frac{1}{3} \right) + \frac{2}{3} \text{Li}_2 \left(\frac{1}{2 } \right) + \frac{1}{6} \ln^{2} \left(\frac{4}{3} \right) + \frac{2}{3} \text{Li}_2 \left(-\frac{1}{2} \right)$

$\displaystyle = \frac{2}{3} \text{Li}_2 \left(\frac{1}{3} \right) + \frac{2}{3} \text{Li}_2 \left(\frac{1}{2 } \right) + \frac{1}{6} \ln^{2} \left(\frac{4}{3} \right) + \frac{2}{3} \Big[ -\text{Li}_2 \left(1-\frac{2}{3} \right) - \frac{1}{2} \ln^{2} \left(\frac{3}{2} \right) \Big]$ (2)

$\displaystyle = \frac{2}{3} \text{Li}_2 \left(\frac{1}{2} \right) + \frac{1}{6} \ln^{2} \left(\frac{4}{3} \right) - \frac{1}{3} \ln^{2} \left(\frac{3}{2} \right)$

$\displaystyle = \frac{2}{3} \Big( \frac{\pi^{2}}{12} - \frac{1}{2} \ln^{2}(2) \Big) + \frac{1}{6} \ln^{2} \left(\frac{4}{3} \right) - \frac{1}{3} \ln^{2} \left(\frac{3}{2} \right)$

$\displaystyle = \frac{\pi^{2}}{18} - \frac{1}{3} \ln^{2}(2) + \frac{1}{6} \ln^{2} \left(\frac{4}{3} \right) - \frac{1}{3} \ln^{2} \left(\frac{3}{2} \right) = \frac{\pi^{2}}{18} - \frac{1}{3} \ln^{2}(2) + \frac{1}{6} (\ln 4 - \ln 3)^{2} - \frac{1}{3} (\ln 3 - \ln 2)^{2}$

$\displaystyle = \frac{\pi^{2}}{18} - \frac{1}{3} \ln^{2}(2) + \frac{1}{6} \Big( 4 \ln^{2}(2) - 4 \ln(2)\ln(3) - \ln^{2}(3) \Big) - \frac{1}{3} \Big( \ln^{2}(3) - 2 \ln(3) \ln(3) + \ln^{2}(2) \Big)$

$\displaystyle = \frac{\pi^{2}}{18} - \frac{1}{6} \ln^{2}(3)$

And I don't like the idea of posting comments and questions in an separate thread. It will make this thread feel lifeless.

### Re: Proving identities of special functions

Fri Aug 30, 2013 9:03 am

Posts: 850
Location: Jaipur, India

Random Variable wrote:

And I don't like the idea of posting comments and questions in an separate thread. It will make this thread feel lifeless.

I think you are right. It will become difficult to keep track.

### Re: Proving identities of special functions

Sat Aug 31, 2013 7:57 am
zaidalyafey
Global Moderator

Posts: 354
Shobhit wrote:
Random Variable wrote:

And I don't like the idea of posting comments and questions in an separate thread. It will make this thread feel lifeless.

I think you are right. It will become difficult to keep track.

When the thread is finished we can move the questions with solutions to another stand-alone thread.
$\displaystyle \sum_{k\geq 0}\frac{f^{(k)} (s)}{(s+1)_k} (-s)^{k}= \int^{1}_0x^{s} f(s x) \left( \frac{x}{s}\right) \, dx$

Wanna learn what we discuss , see tutorials

### Re: Proving identities of special functions

Sat Aug 31, 2013 8:00 am
zaidalyafey
Global Moderator

Posts: 354
Problem # 2

$\displaystyle \sum_{k\geq 0} \frac{(-1)^k}{zk+1}= \frac{1}{2z} \left( \psi_0\left(\frac{z+1}{2z} \right)- \psi_0\left(\frac{1}{2z} \right)\right)$
$\displaystyle \sum_{k\geq 0}\frac{f^{(k)} (s)}{(s+1)_k} (-s)^{k}= \int^{1}_0x^{s} f(s x) \left( \frac{x}{s}\right) \, dx$

Wanna learn what we discuss , see tutorials

### Re: Proving identities of special functions

Sat Aug 31, 2013 8:35 am

Posts: 850
Location: Jaipur, India

zaidalyafey wrote:

When the thread is finished we can move the questions with solutions to another stand-alone thread.

Yeah, that would be much better!

### Re: Proving identities of special functions

Sat Aug 31, 2013 8:58 am

Posts: 850
Location: Jaipur, India

Solution for Problem 2

My solution is based upon a known integral representation of the digamma function.

Note that

\displaystyle \begin{align*} \sum_{k=0}^\infty \frac{(-1)^k}{zk+1}&= \sum_{k=0}^\infty (-1)^k \int_0^1 x^{zk} \ dx \\ &= \int_0^1 \left( \sum_{k=0}^\infty (-1)^k x^{zk} \right) dx \\ &= \int_0^1 \left( \sum_{k=0}^\infty x^{2kz} -\sum_{k=0}^\infty x^{(2k+1)z} \right)dx \\ &= \int_0^1 \left( \frac{1}{1-x^{2z}}-\frac{x^z}{1-x^{2z}}\right)dx \\ &= \frac{1}{2z}\int_0^1 \frac{(1-\sqrt{x})x^{\frac{1}{2z}-1}}{1-x}dx \quad x \mapsto x^{\frac{1}{2z}}\\ &= \frac{1}{2z}\int_0^1 \frac{x^{\frac{1}{2z}-1} - x^{\frac{1}{2z}-\frac{1}{2} } }{1-x}dx \\ &= \frac{1}{2z}\int_0^1 \frac{\left(1-x^{\frac{1}{2z}-\frac{1}{2} } \right)- \left(1-x^{\frac{1}{2z}-1} \right)}{1-x}dx \\ &= \frac{1}{2z}\left\{ \psi_0 \left( \frac{z+1}{2z}\right)-\psi_0 \left( \frac{1}{2z}\right)\right\} \end{align*}

In the last step, I used that

$\displaystyle \psi_0(s+1)=-\gamma+\int_0^1 \frac{1-x^s}{1-x}dx$

### Re: Proving identities of special functions

Sat Aug 31, 2013 9:04 am

Posts: 850
Location: Jaipur, India

Here is a very interesting identity:

Problem 3

$\displaystyle \psi_0 \left(\frac{p}{q} \right)+\psi_0 \left(1-\frac{p}{q}\right)=-2\gamma-2\log q+\sum_{n=1}^{q-1} \cos \left(\frac{2\pi n p}{q} \right)\log \left(2-2 \cos \left(\frac{2\pi n }{q} \right)\right)$

### Re: Proving identities of special functions

Sat Aug 31, 2013 11:46 am
Random Variable Integration Guru

Posts: 381
Another solution for problem 2

$\displaystyle \psi(x) = - \gamma + \sum_{k=0}^{\infty} \Big( \frac{1}{k+1} - \frac{1}{z+k} \Big)$

$\displaystyle \psi \left(\frac{z+1}{2z} \right) = - \gamma + \sum_{k=0}^{\infty} \Big( \frac{1}{k+1} - \frac{2z}{z(2k+1) +1} \Big)$

$\displaystyle \psi \left(\frac{1}{2z} \right) = - \gamma + \sum_{k=0}^{\infty} \Big( \frac{1}{k+1} - \frac{2z}{2zk +1} \Big)$

So $\displaystyle \frac{1}{2z} \Big[ \psi \Big(\frac{z+1}{2z} \Big)- \psi \Big(\frac{1}{2z} \Big)\Big] = \sum_{k=0}^{\infty} \Big( \frac{1}{2kz+1} - \frac{1}{(2k+1)z + 1} \Big) = \sum_{k=0}^{\infty} \frac{(-1)^{k}}{zk+1}$

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