= Solution
The same <integration by substitution> gives the <expected value>
$$
\mathbb E[X]=\frac12\int_{-\infty}^{\infty}(\mu+2st)\operatorname{sech}^2t\,dt
=\mu+s\int_{-\infty}^{\infty}t\operatorname{sech}^2t\,dt.
$$
The remaining integrand is an <odd function>. Its <integral> is absolutely convergent, since $\operatorname{sech}^2t\leq4e^{-2|t|}$ and $|t|e^{-2|t|}$ is integrable. Thus cancellation under <symmetry in integration> is legitimate, rather than merely a principal-value calculation. Consequently
$$
\boxed{\mathbb E[X]=\mu.}
$$
Equivalently, the <logistic distribution> is symmetric about $\mu$, and its exponentially decaying tails ensure that its <expected value> exists.
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