Theorem: Variance of Logistic Distribution

Let $X$ be a continuous random variable which satisfies the logistic distribution:
:$X \sim \map {\operatorname {Logistic} } {\mu, s}$
The variance of $X$ is given by:
:$\var X = \dfrac {s^2 \pi^2} 3$

Proof:
From the definition of the logistic distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac {\map \exp {-\dfrac {\paren {x - \mu} } s} } {s \paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^2}$
From Variance as Expectation of Square minus Square of Expectation: 
:$\ds \var X = \int_{-\infty}^\infty x^2 \, \map {f_X} x \rd x - \paren {\expect X}^2$
So:
{{begin-eqn}}
{{eqn | l = \var X
      | r = \frac 1 s \int_{-\infty}^\infty \dfrac {x^2 \map \exp {-\dfrac {\paren {x - \mu} } s} } {\paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^2} \rd x - \mu^2
      | c = 
}}
{{eqn | r = \frac 1 s \int_{-\infty}^\mu \dfrac {x^2 \map \exp {-\dfrac {\paren {x - \mu} } s} } {\paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^2} \rd x + \frac 1 s \int_\mu^\infty \dfrac {x^2 \map \exp {-\dfrac {\paren {x - \mu} } s} } {\paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^2} \rd x\ - \mu^2
      | c = Sum of Integrals on Adjacent Intervals for Integrable Functions
}}
{{eqn | r = \frac 1 s \int_{-\infty}^0 \dfrac {\paren {sz + \mu}^2 \map \exp {-z} } {\paren {1 + \map \exp {-z} }^2} \paren {s \rd z} + \frac 1 s \int_0^\infty \dfrac {\paren {sz + \mu}^2 \map \exp {-z} } {\paren {1 + \map \exp {-z} }^2} \paren {s\rd z } - \mu^2
      | c = $z = \dfrac {\paren {x - \mu} } s$; $s\rd z = \rd x$
}}
{{eqn | r = \int_{-\infty}^0 \dfrac {\paren {s^2z^2 + 2sz\mu} \map \exp {-z} } {\paren {1 + \map \exp {-z} }^2} \rd z + \int_0^\infty \dfrac {\paren {s^2z^2 + 2sz\mu} \map \exp {-z} } {\paren {1 + \map \exp {-z} }^2} \rd z
      | c = cleaning up; s and $\mu^2$ cancel
}}
{{eqn | r = \int_0^\infty \dfrac {\paren {s^2z^2 - 2sz\mu} \map \exp {z} } {\paren {1 + \map \exp {z} }^2} \rd z + \int_0^\infty \dfrac {\paren {s^2z^2 + 2sz\mu} \map \exp {-z} } {\paren {1 + \map \exp {-z} }^2} \rd z
      | c = rewriting the first integral
}}
{{eqn | r = \int_0^\infty \dfrac {\paren {s^2z^2 - 2sz\mu} \map \exp {z} } {\map \exp {2 z} \paren {1 + \map \exp {-z} }^2} \rd z + \int_0^\infty \dfrac {\paren {s^2z^2 + 2sz\mu} \map \exp {-z} } {\paren {1 + \map \exp {-z} }^2} \rd z
      | c = extracting $\map \exp {2 z}$ from the denominator of the first integral
}}
{{eqn | r = 2s^2 \int_0^\infty \dfrac {z^2 \map \exp {-z} } {\paren {1 + \map \exp {-z} }^2} \rd z
      | c = simplifying
}}
{{end-eqn}}
From Sum of Infinite Geometric Sequence, we have:
:$\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \map \exp {-z}^n = \dfrac 1 {1 + \map \exp {-z} }$
Taking the derivative of both sides, we have:
{{begin-eqn}}
{{eqn	| l = \sum_{n \mathop = 0}^\infty \paren {-n} \paren {-1}^n \map \exp {-z}^n
        | r = \dfrac { \map \exp {-z} } {\paren {1 + \map \exp {-z} }^2}
	| c = 
}}
{{eqn	| l = \sum_{n \mathop = 1}^\infty n \paren {-1}^{n + 1} \map \exp {-z}^n
        | r = \dfrac { \map \exp {-z} } {\paren {1 + \map \exp {-z} }^2}
	| c = simplifying
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn	| l = \var X
	    | r = 2 s^2 \paren {\int_0^\infty z^2 \sum_{n \mathop = 1}^\infty n \paren {-1}^{n + 1} \map \exp {-z}^n \rd z}
        | c = substitution from above
}}
{{eqn	| r = 2 s^2 \paren {\sum_{n \mathop = 1}^\infty n \paren {-1}^{n + 1} \int_0^\infty z^2 \map \exp {-n z} \rd z}
        | c = Fubini's Theorem
}}
{{eqn	| r = 2 s^2 \paren {\sum_{n \mathop = 1}^\infty n \paren {-1}^{n + 1} \bigintlimits {-\dfrac 1 n \map \exp {-nz} \paren {z^2 + \frac {2 z} n + \frac 2 {n^2} } } 0 \infty }
        | c = Primitive of x squared by Exponential of a x
}}
{{eqn	| r = 2 s^2 \paren {\sum_{n \mathop = 1}^\infty n \paren {-1}^{n + 1} \paren {0 + \dfrac 1 n \paren {0^2 + \frac 0 n + \frac 2 {n^2} } } }
        | c = 
}}
{{eqn	| r = 4 s^2 \paren {\sum_{n \mathop = 1}^\infty \paren {-1}^{n + 1} \frac 1 {n^2}  }
        | c = 
}}
{{eqn	| r = 4 s^2 \paren {\dfrac {\pi^2} {12} }
        | c = Sum of Reciprocals of Squares Alternating in Sign
}}
{{eqn	| r = \dfrac {s^2 \pi^2} 3
        | c = 
}}
{{end-eqn}}
{{qed}}