Does the exponential operator $e^{tA}$ defined via functional calculus and via semi group coincide?

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Given $X$ is a Hilbert space and consider the operator $A$ on $X$.

Functional calculus: given an unbounded operator $A$ (densely defined, closed and self-adjoint), we can define $e^{tA}$ by the spectral measure.

Semi-group: given an unbounded operator $A$ (densely defined, closed) which generates a continuous semi-group, we can define $e^{tA}$ by the Laplace transform of the resolvent operator.

Question: Do these two approaches give the same definition in the case where $A$ is densely defined, closed, self-adjoint and generates a continuous semi-group?

Trivial case: if $A$ is bounded or $A$ has compact resolvent, I think this question is rather trivial in view of the behavior of the point spectrum.