We show that the integral method used to prove the no-hair theorem for black holes (BH's) in spherically symmetric and static spacetimes within the framework of general relativity with matter composed by a complex-valued scalar-field does not lead to a straightforward conclusion about the absence of hair in the stationary and rotating (axisymmetric) scenario. We argue that such a failure can be used to justify in a simple and heuristic way the existence of nontrivial boson clouds or hair found numerically by Herdeiro and Radu [Phys. Rev. Lett. 112, 221101 (2014); Classical Quantum Gravity 32, 144001 (2015)] and analytically by Hod in the test field limit [Phys. Rev. D 86, 104026 (2012); Phys. Rev. D 86, 129902(E) (2012); Eur. Phys. J. C 73, 2378 (2013); J. High Energy Phys. 01 (2017) 030]. This is due to the presence of a contribution that is negative when rotation exists which allows for an integral to vanish even when a nontrivial boson hair is present. The presence of such a negative contribution that depends on the rotation properties of the BH is perfectly correlated with the eigenvalue problem associated with the boson-field equation. Conversely, when the rotation is absent the integral turns to be composed only by non negative (i.e., positive semidefinite) terms and thus the only way it can vanish is when the hair is completely absent. This analysis poses serious challenges and obstructions towards the elaboration of no-hair theorems for more general spacetimes endowed with a BH region even when including matter fields that obey the energy conditions. Thus rotating boson stars, if collapsed, may lead indeed to a new type of rotating BH, like the ones found in [Phys. Rev. Lett. 112, 221101 (2014); Classical Quantum Gravity 32, 144001 (2015)]. In order to achieve this analysis we solve numerically the eigenvalue problem for the boson field in the Kerr-BH background by imposing rigorous regularity conditions at the BH horizon for the non-extremal case ($0<a<M$) which include the near extremal one in the limit $M\ensuremath{\rightarrow}a$, as well as the small BH limit $M\ensuremath{\rightarrow}a\ensuremath{\rightarrow}0$.
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