TITLE: Find the value of $1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\frac{1}{23}+\frac{1}{25}....$
QUESTION [9 upvotes]: Find the value of  this :$$1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\frac{1}{23}+\frac{1}{25}....$$

Try: We can write the above series as
$${S} = \int^{1}_{0}\bigg[1-x^6+x^8-x^{14}+x^{16}-x^{22}+\cdots\bigg]dx$$
$$S = \int^{1}_{0}(1-x^6)\bigg[1+x^{8}+x^{16}+\cdots \cdots \bigg]dx$$
So $$S = \int^{1}_{0}\frac{1-x^6}{1-x^{8}}dx = \int^{1}_{0}\frac{x^4+x^2+1}{(x^2+1)(x^4+1)}dx$$
Now i am struck in that integration.
Did not understand how to solve it 
could some help me to solve it. Thanks in advance

REPLY [5 votes]: Another approach, which could be useful to check yours through integrals,
is same as per the similar post, using the Digamma function $\psi (z)$.
$$
\eqalign{
  & 1 - {1 \over {8 - 1}} + {1 \over {8 + 1}} - {1 \over {2 \cdot 8 - 1}} + {1 \over {2 \cdot 8 + 1}} +  \cdots  =   \cr 
  &  = 1 - \sum\limits_{1\, \le \,n} {{1 \over {8n - 1}}}  + \sum\limits_{1\, \le \,n} {{1 \over {8n + 1}}}  =   \cr 
  &  = 1 - {1 \over 8}\left( {\sum\limits_{1\, \le \,n} {{1 \over {n - 1/8}}}  - \sum\limits_{1\, \le \,n} {{1 \over {n + 1/8}}} } \right) =   \cr 
  &  = 1 - {1 \over 8}\left( {\sum\nolimits_{\;n = 7/8}^{\;\infty } {{1 \over n}}  - \sum\nolimits_{\;n = 9/8}^{\;\infty } {{1 \over n}} } \right) =   \cr 
  &  = 1 - {1 \over 8}\left( {\sum\nolimits_{\;n = 7/8}^{\;9/8} {{1 \over n}} } \right) =   \cr 
  &  = 1 - {1 \over 8}\left( {\psi (9/8) - \psi (7/8)} \right) =   \cr 
  &  = 1 - {1 \over 8}\left( {\psi \left( {1 + 1/8} \right) - \psi \left( {1 - 1/8} \right)} \right) =   \cr 
  &  = 1 - {1 \over 8}\left( {{1 \over {1/8}} + \psi \left( {1/8} \right) - \psi \left( {1 - 1/8} \right)} \right) =   \cr 
  &  = {1 \over 8}\left( {\psi \left( {1 - 1/8} \right) - \psi \left( {1/8} \right)} \right) =   \cr 
  &  = {1 \over 8}\left( {\pi \cot \left( {{\pi  \over 8}} \right)} \right) =   \cr 
  &  = {{\pi \left( {1 + \sqrt 2 } \right)} \over 8} \cr} 
$$
where:
$$
\Delta _{\,z} \psi \left( z \right) = \psi \left( {z + 1} \right) - \psi \left( z \right) = {1 \over z}
$$
is the functional equation for the Digamma;
which implies that Digamma is the Antidelta of $1/z$
$$
\eqalign{
  & \psi \left( z \right) = \Delta _{\,z} ^{ - 1} \left( {{1 \over z}} \right) = \sum\nolimits_z {{1 \over z}} \quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \sum\nolimits_{\;z = a}^{\;b} {{1 \over z}}  = \psi \left( b \right) - \psi \left( a \right) \cr} 
$$
and we used the Reflection formula for Digamma
$$
\psi \left( {1 - z} \right) = \psi \left( z \right) + \pi \cot \left( {\pi z} \right)
$$

Now, the above, suggests a way to solve the integral.

Let's replace $x^8$ with $y$
$$
\int_{x = 0}^{\,1} {{{1 - x^{\,6} } \over {1 - x^{\,8} }}dx} \quad \mathop
  = \limits^{x^{\,8}  = y} \quad {1 \over 8}\int_{y = 0}^{\,1} {{{1 - y^{\,3/4} } \over {\left( {1 - y} \right)y^{\,7/8} }}dy} 
$$
then consider that we have
$$
\eqalign{
  & {{1 - y^{\,3/4} } \over {\left( {1 - y} \right)y^{\,7/8} }} =   \cr 
  &  = \left( {\left( {1 - y} \right)^{ - 1} y^{\, - 7/8}  - \left( {1 - y} \right)^{ - 1} y^{\, - 1/8} } \right)  \cr 
  &  = \mathop {\lim }\limits_{\varepsilon \, \to \,0} \left( {\left( {1 - y} \right)^{\varepsilon  - 1} y^{\,1/8 - 1}
  - \left( {1 - y} \right)^{\varepsilon  - 1} y^{\,\,7/8 - 1} } \right) \cr} 
$$
so we are ready to use the integral and Gamma representation for the Beta function
$$
\eqalign{
  & 8 \;\int_{x = 0}^{\,1} {{{1 - x^{\,6} } \over {1 - x^{\,8} }}dx}  = \int_{y = 0}^{\,1} {{{1 - y^{\,3/4} } \over {\left( {1 - y} \right)y^{\,7/8} }}dy}  =   \cr 
  &  = \mathop {\lim }\limits_{\varepsilon \, \to \,0} \left( {\int_{y = 0}^{\,1} {\left( {1 - y} \right)^{\varepsilon  - 1} y^{\,1/8 - 1} dy}
  - \int_{y = 0}^{\,1} {\left( {1 - y} \right)^{\varepsilon  - 1} y^{\,\,7/8 - 1} dy} } \right) =   \cr 
  &  = \mathop {\lim }\limits_{\varepsilon \, \to \,0} \left( {{\rm B}\left( {1/8,\varepsilon } \right) - {\rm B}\left( {7/8,\varepsilon } \right)} \right) =   \cr 
  &  = \mathop {\lim }\limits_{\varepsilon \, \to \,0} \left( {{{\Gamma \left( {1/8} \right)\Gamma \left( \varepsilon  \right)}
  \over {\Gamma \left( {1/8 + \varepsilon } \right)}}
 - {{\Gamma \left( {7/8} \right)\Gamma \left( \varepsilon  \right)} \over {\Gamma \left( {7/8 + \varepsilon } \right)}}} \right) =   \cr 
  &  = \mathop {\lim }\limits_{\varepsilon \, \to \,0} \left( {{{\Gamma \left( {1/8} \right)\Gamma \left( \varepsilon  \right)}
 \over {\psi \left( {1/8} \right)\varepsilon \;\Gamma \left( {1/8} \right)
 + \Gamma \left( {1/8} \right)}} - {{\Gamma \left( {7/8} \right)\Gamma \left( \varepsilon  \right)}
 \over {\psi \left( {7/8} \right)\varepsilon \;\Gamma \left( {7/8} \right) + \Gamma \left( {7/8} \right)}}} \right) =   \cr 
  &  = \mathop {\lim }\limits_{\varepsilon \, \to \,0} \left( {{{\Gamma \left( \varepsilon  \right)} \over {\psi \left( {1/8} \right)\varepsilon \; + 1}}
 - {{\Gamma \left( \varepsilon  \right)} \over {\psi \left( {7/8} \right)\varepsilon \; + 1}}} \right) =   \cr 
  &  = \mathop {\lim }\limits_{\varepsilon \, \to \,0} \Gamma \left( \varepsilon  \right)\left( {\left( {1 - \psi \left( {1/8} \right)\varepsilon \;} \right)
 - \left( {1 - \psi \left( {7/8} \right)\varepsilon \;} \right)} \right) =   \cr 
  &  = \mathop {\lim }\limits_{\varepsilon \, \to \,0} \varepsilon \,\Gamma \left( \varepsilon  \right)\left( {\psi \left( {7/8} \right) - \psi \left( {1/8} \right)} \right) =   \cr 
  &  = \mathop {\lim }\limits_{\varepsilon \, \to \,0} \,\Gamma \left( {1 + \varepsilon } \right)\left( {\psi \left( {7/8} \right) - \psi \left( {1/8} \right)} \right) =   \cr 
  &  = \psi \left( {7/8} \right) - \psi \left( {1/8} \right) \cr} 
$$
which is the same result as above.

Let me add a more straight derivation of the above.  

We rewrite the integrand as
$$
\eqalign{
  & {{1 - y^{\,3/4} } \over {\left( {1 - y} \right)y^{\,7/8} }} = {{y^{\, - 7/8}  - y^{\, - 1/8} } \over {\left( {1 - y} \right)}} =   \cr 
  &  = {{1 - y^{\, - 1/8}  - \left( {1 - y^{\, - 7/8} } \right)} \over {\left( {1 - y} \right)}} \cr} 
$$
and compare with the integral representation of Digamma
$$
\psi (s + 1) =  - \gamma  + \int_0^1 {{{1 - x^{\,s} } \over {1 - x}}dx} 
$$