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Proving identities of special functions

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Post Thu Aug 29, 2013 11:39 am
Shobhit Site Admin
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Proving Identities of Special Functions

Shobhit Site Admin
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Posts: 850
Location: Jaipur, India

Here we go!

Problem 1

Show that

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

Random Variable Integration Guru
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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.

Shobhit Site Admin
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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.

zaidalyafey Global Moderator
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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.
Wanna learn what we discuss , see Book

zaidalyafey Global Moderator
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Problem # 2

\(\displaystyle \sum_{k\geq 0} \frac{(-1)^k}{zk+1}= \frac{1}{2z} \left( \psi_0\left(\frac{z+1}{2z} \right)- \psi_0\left(\frac{1}{2z} \right)\right)\)
Wanna learn what we discuss , see Book

Shobhit Site Admin
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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!

Shobhit Site Admin
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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

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

Shobhit Site Admin
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Posts: 850
Location: Jaipur, India

Here is a very interesting identity:

Problem 3

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

Random Variable Integration Guru
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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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