= Solution
For a <force> $\mathbf F\delta(\mathbf x)$ acting on an unbounded incompressible fluid, the <Stokeslet> and its <pressure> are
$$
\boxed{u_i(\mathbf x)=G_{ij}(\mathbf x)F_j,\quad G_{ij}=\frac1{8\pi\mu}\left(\frac{\delta_{ij}}r+\frac{x_ix_j}{r^3}\right),\quad p=\frac{\mathbf F\cdot\mathbf x}{4\pi r^3}.}
$$
They solve $-\nabla p+\mu\nabla^2\mathbf u+\mathbf F\delta=0$ and $\nabla\cdot\mathbf u=0$. Approximate a slender body by a line distribution of these point <forces> on its centreline. At a point labelled by arclength $s$,
$$
\dot{\mathbf X}(s)\simeq\int G(\mathbf X(s)-\mathbf X(s'))\mathbf f(s')\,ds'.
$$
To extract the large logarithm, write $s'=s+\ell$ and approximate $\mathbf X(s+\ell)-\mathbf X(s)\simeq\ell\mathbf t$, $\mathbf f(s+\ell)\simeq\mathbf f(s)$, where $\mathbf t=\mathbf X'$ is the unit <tangent vector>. The near-line contribution has kernel $(I+\mathbf t\mathbf t)/(8\pi\mu|\ell|)$. Cut it off at the wire radius $R$ and a macroscopic length $L$. The two sides of the line give
$$
\dot{\mathbf X}\sim\frac{\ln(L/R)}{4\pi\mu}(I+\mathbf t\mathbf t)\mathbf f.
$$
The far contribution and changes in the cutoff add terms without the leading logarithm. Since $(I+\mathbf t\mathbf t)^{-1}=I-\mathbf t\mathbf t/2$, inversion proves the leading <slender-body force density>
$$
\boxed{\mathbf f\sim\frac{2\pi\mu}{\ln(L/R)}(2I-\mathbf t\mathbf t)\dot{\mathbf X}.}
$$
This is <force> on the fluid; the fluid's <force> on the wire has the opposite sign. It gives twice the resistance to transverse motion as to tangential motion at this order.
The stated arclength describes the circular <helix> $\mathbf X=(a\cos\theta,a\sin\theta,b\theta)$. The PDF omits the second $a$ in the coordinates, so the literal printed curve and its arclength are inconsistent unless $a=1$. First use the circular interpretation consistent with the given arclength. Put $q=\sqrt{a^2+b^2}$ and $c=2\pi\mu/\ln(L/R)$. Then
$$
\mathbf t=\frac{(-a\sin\theta,a\cos\theta,b)}q,\qquad ds=q\,d\theta.
$$
Rotation about $z$ has <velocity> $\mathbf v=\Omega(-a\sin\theta,a\cos\theta,0)$, so $\mathbf t\cdot\mathbf v=\Omega a^2/q$. The axial <force> density is $f_z=-c\Omega a^2b/q^2$. Integrating over one turn gives
$$
\boxed{F_z^{\rm rotation}=-\frac{4\pi^2\mu a^2b}{\ln(L/R)\sqrt{a^2+b^2}}\,\Omega.}
$$
For translation, $\mathbf v=W\widehat{\mathbf z}$ and $\mathbf f=cW(2\widehat{\mathbf z}-(b/q)\mathbf t)$. Only the tangential term contributes to axial couple. Since $(\mathbf X\times\mathbf t)_z=a^2/q$, its density is $-cWa^2b/q^2$, giving
$$
\boxed{G_z^{\rm translation}=-\frac{4\pi^2\mu a^2b}{\ln(L/R)\sqrt{a^2+b^2}}\,W.}
$$
These are the off-diagonal coefficients of the <axial resistance matrix of a slender helix>. Their equality does not depend on this particular geometry: with symmetric local resistance $D=c(2I-\mathbf t\mathbf t)$, the rotation-induced axial <force> is $\int\widehat{\mathbf z}\cdot D(\widehat{\mathbf z}\times\mathbf X)ds$, whereas the translation-induced axial <torque> is $\int(\widehat{\mathbf z}\times\mathbf X)\cdot D\widehat{\mathbf z}\,ds$. They agree pointwise by <symmetry> of $D$. The full Stokes-flow statement is the <Lorentz reciprocal theorem>.
For completeness, retaining the literal elliptic <helix> instead gives $s'=\sqrt{a^2\sin^2\theta+\cos^2\theta+b^2}$ and the <axial coupling of an elliptic helix>
$$
\boxed{\frac{F_z}{\Omega}=\frac{G_z}{W}=-\frac{2\pi\mu ab}{\ln(L/R)}\int_{-\pi}^{\pi}\frac{d\theta}{\sqrt{a^2\sin^2\theta+\cos^2\theta+b^2}}.}
$$
Thus reciprocity holds for the literal curve as well; only the geometric coefficient and arclength change. In all these formulas the equal and opposite <force> or couple on the wire reverses the displayed sign.
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