■ある無限級数(その168)

右辺=(π/(8a^3)){tanh(aπ/2)-aπ/2・sech^2(aπ/2)}

 sinh(ix)=isinx

 cosh(ix)=cosx

 tanh(ix)=itanx

より,a→aiとおくと

=(-iπ/(8a^3)){itan(aπ/2)-aiπ/2・sec^2(aπ/2)}

=(π/(8a^3)){tan(aπ/2)-aπ/2・sec^2(aπ/2)}

 a=1/2とおくと,=π{π/2-1}

左辺=(p^2/2q)^2{{1/(p−q)−1/(p+q)}^2+{1/(3p−q)−1/(3p+q)}^2+{1/(5p−q)−1/(5p+q)}^2+・・・}

=4{{1/1−1/3}^2+{1/5−1/7}^2+{1/9−1/11}^2+・・・}

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