Peter Luschny schrieb:   
   >   
   > I'm much better in finding integrals than in solving them,   
   > therefore I ask here where the experts on this subject seem   
   > to sit in a row like migratory birds in autumn on the   
   > telephone wires.   
   >   
   > I have no problem to calculate the value, and indeed it is   
   > utterly trivial to solve this integral by numerical approximation.   
   >   
   > But I am interested in everything else which could be said about   
   > this integral and possible transformations of this integral.   
   >   
   > Of course integrability in elementary terms is one of these   
   > questions (I think it is not).   
   >   
   > The question remains why it is worth dealing with this integral.   
   > This much can be disclosed (so as not to spoil the fun): the   
   > value on the definite interval (-oo..+oo) is the quotient of   
   > two famous constants, which implies a bunch of some interesting   
   > new formulas.   
   >   
   > This is the integral in Maple parlance:   
   >   
   >     int( -log((z^2+1/4)^(1/4))*sech(Pi*z)^2, z = -infinity..infinity)   
   >   
      
   Rewriting the integrand functions, exploiting their symmetry about z=0,   
   integrating by parts, and then splitting the second factor:   
      
   INT(- LOG((z^2 + 1/4)^(1/4))*SECH(pi*z)^2, z, -inf, inf)   
      
   1/2*INT(- LOG(z^2 + 1/4)*(2*#e^(pi*z)/(#e^(2*pi*z) + 1))^2, z, 0, inf)   
      
   LN(2)/pi + 1/2*INT(- 8*z/(4*z^2 + 1)*(2/(pi*(#e^(2*pi*z) + 1))), z, 0,   
   inf)   
      
   LN(2)/pi + 1/2*INT(- 8*z/(4*z^2 + 1)*(2/(pi*(#e^(2*pi*z) - 1))), z, 0,   
   inf) - 1/2*INT(- 8*z/(4*z^2 + 1)*(4/(pi*(#e^(4*pi*z) - 1))), z, 0, inf)   
      
   results in a pair of integrals listed in Gradshteyn-Ryzhik under 3.415   
   with attribution to the "Nouvelles tables d'intégrales définies" by   
   Bierens de Haan (1867). So your "bunch of new formulas" appears to be   
   more than 150 years old.   
      
   Martin.   
      
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