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Michael E2
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PossibleZeroQ will numerically check for zero if standard transformations do not work. It, being numerical, is not foolproof, and so it is not a complete rigorous method. For that reason, @Artes' method is superior in this case.

Simplify[
 1/12 (\[Pi]^2π^2 + 3 ArcCosh[7/2] ArcCsch[2] - 
    12 ArcTanh[1/Sqrt[5]]^2),
 TransformationFunctions -> {Automatic, # /. _?PossibleZeroQ :> 0 &}]
`Simplify::ztest1`: Unable to decide whether numeric quantity `-3 (-Log[1+Times[<<2>>]]+Log[1+1/Sqrt[5]])^2+3 Log[1/2+Sqrt[5]/2] Log[7/2+(3 Sqrt[5])/2]` is equal to zero. Assuming it is.
(*  \[Pi]^2π^2/12  *)

Simplify caches results, so the message only appears on first execution.

PossibleZeroQ will numerically check for zero if standard transformations do not work. It, being numerical, is not foolproof, and so it is not a complete rigorous method. For that reason, @Artes' method is superior in this case.

Simplify[
 1/12 (\[Pi]^2 + 3 ArcCosh[7/2] ArcCsch[2] - 
    12 ArcTanh[1/Sqrt[5]]^2),
 TransformationFunctions -> {Automatic, # /. _?PossibleZeroQ :> 0 &}]
`Simplify::ztest1`: Unable to decide whether numeric quantity `-3 (-Log[1+Times[<<2>>]]+Log[1+1/Sqrt[5]])^2+3 Log[1/2+Sqrt[5]/2] Log[7/2+(3 Sqrt[5])/2]` is equal to zero. Assuming it is.
(*  \[Pi]^2/12  *)

Simplify caches results, so the message only appears on first execution.

PossibleZeroQ will numerically check for zero if standard transformations do not work. It, being numerical, is not foolproof, and so it is not a complete rigorous method. For that reason, @Artes' method is superior in this case.

Simplify[
 1/12 (π^2 + 3 ArcCosh[7/2] ArcCsch[2] - 
    12 ArcTanh[1/Sqrt[5]]^2),
 TransformationFunctions -> {Automatic, # /. _?PossibleZeroQ :> 0 &}]
`Simplify::ztest1`: Unable to decide whether numeric quantity `-3 (-Log[1+Times[<<2>>]]+Log[1+1/Sqrt[5]])^2+3 Log[1/2+Sqrt[5]/2] Log[7/2+(3 Sqrt[5])/2]` is equal to zero. Assuming it is.
(*  π^2/12  *)

Simplify caches results, so the message only appears on first execution.

Source Link
Michael E2
  • 245k
  • 18
  • 351
  • 775

PossibleZeroQ will numerically check for zero if standard transformations do not work. It, being numerical, is not foolproof, and so it is not a complete rigorous method. For that reason, @Artes' method is superior in this case.

Simplify[
 1/12 (\[Pi]^2 + 3 ArcCosh[7/2] ArcCsch[2] - 
    12 ArcTanh[1/Sqrt[5]]^2),
 TransformationFunctions -> {Automatic, # /. _?PossibleZeroQ :> 0 &}]
`Simplify::ztest1`: Unable to decide whether numeric quantity `-3 (-Log[1+Times[<<2>>]]+Log[1+1/Sqrt[5]])^2+3 Log[1/2+Sqrt[5]/2] Log[7/2+(3 Sqrt[5])/2]` is equal to zero. Assuming it is.
(*  \[Pi]^2/12  *)

Simplify caches results, so the message only appears on first execution.