= Solution
For a deterministic $s>0$, the <positive part> $(s-K)_+$ vanishes for $K\ge s$, so direct integration gives
$$
\int_0^\infty K^{\varepsilon-1}(s-K)_+\,dK
=s\frac{s^\varepsilon}{\varepsilon}-\frac{s^{1+\varepsilon}}{1+\varepsilon}
=\frac{s^{1+\varepsilon}}{\varepsilon(1+\varepsilon)}.
$$
At $s=0$ both sides vanish. Applying this pointwise to the nonnegative random variable proves the <power payoff static call representation>
$$
\boxed{S^{1+\varepsilon}=\varepsilon(1+\varepsilon)\int_0^\infty K^{\varepsilon-1}(S-K)_+\,dK.}
$$
By <Tonelli theorem>, the corresponding moment identity is
$$
M(1+\varepsilon)=\varepsilon(1+\varepsilon)\int_0^\infty K^{\varepsilon-1}C(K)\,dK,
$$
including the possibility that both sides are infinite. No higher-moment assumption is needed to interchange these nonnegative integrals.
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