= Solution
Let $\langle\cdot\rangle$ denote the spatial average. Assume periodicity, or an equivalent homogeneous averaging limit in which total derivatives average to zero and <integration by parts> is valid. Fix the induced field to have zero mean. Expand the solution of the <resistive induction equation> in small <magnetic Reynolds number>:
$$
\mathbf b=R_m\mathbf b_1+R_m^2\mathbf b_2+\cdots,\qquad
\Delta\mathbf b_1=-(\mathbf B\cdot\nabla)\mathbf u,\qquad
\Delta\mathbf b_2=-\nabla\times(\mathbf u\times\mathbf b_1).
$$
The uniform test <magnetic field> commutes with differentiation. Since $\Delta\mathbf u=-\mathbf u$, the first equation has solution $\mathbf b_1=(\mathbf B\cdot\nabla)\mathbf u$.
The quadratic second-order forcing need not be monochromatic, so one must not simply replace its inverse <Laplacian> by multiplication by $-1$. Instead use the <self-adjointness> of the <Laplacian> under the average:
$$
\langle\mathbf u\times\mathbf b_2\rangle
=-\langle\Delta\mathbf u\times\mathbf b_2\rangle
=-\langle\mathbf u\times\Delta\mathbf b_2\rangle
=\langle\mathbf u\times\nabla\times(\mathbf u\times\mathbf b_1)\rangle.
$$
This proves the <monochromatic small-Reynolds-number mean electromotive force> expansion
$$
\boxed{\boldsymbol{\mathcal E}
=R_m\langle\mathbf u\times(\mathbf B\cdot\nabla)\mathbf u\rangle
+R_m^2\langle\mathbf u\times\nabla\times[\mathbf u\times(\mathbf B\cdot\nabla)\mathbf u]\rangle
+O(R_m^3).}
$$
For the <symmetry of the first-order monochromatic alpha tensor>, write
$$
\alpha^{(1)}_{ij}=\epsilon_{ikl}\langle u_k\partial_j u_l\rangle.
$$
Its antisymmetric part is determined by contraction with the <Levi-Civita symbol>. The contraction identity gives
$$
\epsilon_{pij}\alpha^{(1)}_{ij}
=\langle u_j\partial_j u_p-u_p\partial_j u_j\rangle.
$$
The second term vanishes by <incompressibility>; the first is $\langle\partial_j(u_ju_p)\rangle=0$. Since every three-dimensional <antisymmetric second-rank tensor> is equivalent to its contracted axial vector, \b[$\alpha^{(1)}$ is symmetric]. No <Fourier series> or <Fourier transform> is required.
To derive the next identity, set $\mathbf C=\mathbf u\times(\mathbf B\cdot\nabla)\mathbf u$. Expanding the <curl> and <cross product> in components,
$$
(\mathbf u\times\nabla\times\mathbf C)_i
=u_j\partial_iC_j-u_j\partial_jC_i.
$$
The average of the second term is zero: <integration by parts> makes it $-\langle C_i\partial_j u_j\rangle$. Integrating the first term by parts gives
$$
\boxed{\mathcal E_i^{(2)}=-\langle(\partial_i u_j)\,[\mathbf u\times(\mathbf B\cdot\nabla)\mathbf u]_j\rangle.}
$$
Extracting the coefficient of $B_j$ now yields
$$
\alpha^{(2)}_{ij}
=-\langle\partial_i\mathbf u\cdot(\mathbf u\times\partial_j\mathbf u)\rangle
=\langle\mathbf u\cdot(\partial_i\mathbf u\times\partial_j\mathbf u)\rangle.
$$
Interchanging $i$ and $j$ reverses the <cross product>, proving the <antisymmetry of the second-order monochromatic alpha tensor>:
$$
\boxed{\alpha^{(1)}_{ij}=\alpha^{(1)}_{ji},\qquad
\alpha^{(2)}_{ij}=-\alpha^{(2)}_{ji}.}
$$
The second coefficient of the <alpha tensor> can therefore contribute <turbulent magnetic pumping>, while its symmetric part vanishes at this order.
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