= Solution
Take zero mean flow locally and incompressible fluctuations. Subtracting the averaged induction equation gives
$$
(\partial_t-\eta\nabla^2)\mathbf b=\nabla\times(\mathbf u\times\overline{\mathbf B})+\nabla\times(\mathbf u\times\mathbf b-\langle\mathbf u\times\mathbf b\rangle).
$$
The <first-order smoothing approximation> omits the second term on the right when solving for $\mathbf b$, but retains $\langle\mathbf u\times\mathbf b\rangle$ in the mean equation. For a locally uniform, slowly changing mean <magnetic field> the forcing becomes $(\overline{\mathbf B}\cdot\nabla)\mathbf u$.
There are two controlled ways to make the omitted fluctuation term small. At small eddy <magnetic Reynolds number> $R_m=u_{\mathrm{rms}}\ell/\eta$, diffusion gives $b/B\sim R_m$, so the omitted $ub/\ell$ is smaller than the retained $uB/\ell$ by $R_m$. Alternatively, if $u_{\mathrm{rms}}\tau/\ell\ll1$ for the fluctuation response time $\tau$, then $b/B\sim u_{\mathrm{rms}}\tau/\ell$ and the short-memory response is perturbative. Mean-field scale separation is additionally required. Merely calling the <turbulence> weak or assuming that the mean <magnetic field> is small does not justify the omission.
With homogeneous initial transients removed, the causal linear solution is
$$
b_i(t)=\overline B_j\int_0^\infty e^{\eta\tau\nabla^2}\partial_j u_i(t-\tau)\,d\tau.
$$
Insert this into the electromotive correlation. <Isotropy> gives $\alpha_{ij}=\alpha\delta_{ij}$, so taking the <trace> yields the <helicity formula for isotropic first-order smoothing>:
$$
3\alpha=\int_0^\infty\epsilon_{imn}\left\langle u_m(t)\partial_i e^{\eta\tau\nabla^2}u_n(t-\tau)\right\rangle d\tau.
$$
Using the alternating-tensor sign gives
$$
\boxed{\alpha=-\frac13\int_0^\infty\left\langle\mathbf u(t)\cdot\nabla\times e^{\eta\tau\nabla^2}\mathbf u(t-\tau)\right\rangle d\tau.}
$$
The minus sign follows from $\epsilon_{imn}=-\epsilon_{min}$. This derivation specifies the diffusion and correlation weighting, rather than asserting that alpha is proportional to instantaneous helicity for arbitrary <turbulence>.
If $\eta\tau/\ell^2\ll1$, diffusion over a <correlation time> is negligible. Define the helicity <correlation time> by
$$
\tau_H\langle\mathbf u\cdot\boldsymbol\varpi\rangle=\int_0^\infty\langle\mathbf u(t)\cdot\boldsymbol\varpi(t-\tau)\rangle d\tau,\qquad\boldsymbol\varpi=\nabla\times\mathbf u.
$$
For an ordinary decaying correlation with nonzero mean <kinetic helicity>, this gives the familiar estimate
$$
\boxed{\alpha=-\frac{\tau_H}{3}\langle\mathbf u\cdot\boldsymbol\varpi\rangle.}
$$
It is the <kinetic helicity>, not <magnetic helicity>. Opposite flow handedness reverses the <alpha effect>. The time-integral formula remains meaningful if the equal-time helicity vanishes or its correlation changes sign, when replacing it by a single positive <correlation time> may fail. In the same weak-diffusion short-memory approximation the gradient forcing $-(\mathbf u\cdot\nabla)\overline{\mathbf B}$ gives $\beta=\frac13\int_0^\infty\langle\mathbf u(t)\cdot\mathbf u(t-\tau)\rangle d\tau\simeq\tau_Uu_{\mathrm{rms}}^2/3$; the two <correlation times> need not be equal.
In the small-$R_m$, slowly varying quasistatic regime, diffusion is instead essential. The response gives
$$
\alpha=-\frac1{3\eta}\left\langle\mathbf u\cdot(-\nabla^2)^{-1}\boldsymbol\varpi\right\rangle.
$$
For velocity concentrated at wavenumber $k_0$, this becomes $\alpha=-\langle\mathbf u\cdot\boldsymbol\varpi\rangle/(3\eta k_0^2)$. Its effective memory is the <magnetic diffusion> time on the eddy scale. The controlled low-$R_m$ setting is developed in https://www.damtp.cam.ac.uk/user/hkm2/PDFs/Moffatt_1970_JFM_41_435.pdf[Moffatt's analysis of turbulent dynamo action]; the causal-response calculation above also covers short-memory <turbulence> without assuming that diffusion is always negligible.
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