Gravitational field within hollow cylinder Gauss says that there should be no field within a hollow shell, spherical or cylindrical. But when I integrate the contributions to gravitational potential from a cylindrical shell to an off-axis point within the cylinder, I get an integral that looks like this:$$
\phi ~~= ~~ \int^{\pi}_{0} {\dfrac{\mathrm{d}x}{\sqrt{\alpha ^{2}+\alpha }~-~2\alpha \cos {\left( x \right)}}}
\,,$$
where $\alpha$ is the ratio between the radius of the off-axis point to the cylinder radius.
The fact that this integral produces different potentials for different values of $\alpha$ means that there is a potential gradient within the cylinder, and that a particle will feel a gravitational force.
Where did Gauss (or maybe me) go wrong?
 A: This mixes up several distinct ideas, but makes a good conceptual or educational question. Firstly, "Newton's shell theorem" says the gravitational field inside a spherically symmetric distribution of mass is zero. This will not be true for general shapes; in particular intuitively I wouldn't expect it for a cylinder, apart from at the very centre. So where is the misconception?
Gauss's law for gravity (integral version) says$$
\oint_{{\partial}V} \vec{g} \cdot \mathrm{d} \vec{A} ~~=~~-4\pi GM
\,.$$
In words, the total flux of the gravitational field across a closed surface equals (up to a constant) the enclosed mass. So if we take a surface inside the cylinder, we know the enclosed mass is $M=0$, so the law tells us the total flux of the gravitational field across this surface is zero, but not that the field itself is zero.
A: The integral in your question looks like the potential within a ring
of matter,
$$ \phi(\alpha) = -\frac{G M}{\pi R} \int_0^\pi \mathrm{d}x \frac{1}{\sqrt{1 + \alpha^2 + 2 \alpha \cos(x)}}. $$
($\alpha$: distance from center relative to radius; $R$: radius; $M$:
total mass of ring)
Inside a ring the gravitational field is indeed not zero (except for
the center).  Gauss’s law still applies, but as Colin MacLaurin
pointed out there will be parts of the integration surface where the
field points outwards and parts where it points inwards, which cancel
each other.  Here is a qualitative picture of the fieldlines in a
plane perpendicular to the ring:

To obtain the potential within a cylinder, you must still integrate
over the direction along its axis (taken as $z$ coordinate here):
$$ \phi(\alpha, z) = -2 G R \sigma \cdot \int_{z_0}^{z_1} \mathrm{d}z' \int_0^\pi \mathrm{d}x \frac{1}{\sqrt{R^2 \left[ 1 + \alpha^2 + 2 \alpha \cos(x) \right] + (z-z')^2}}. $$
($z$: coordinate along cylinder axis; $z_0$, $z_1$: bottom and top of
cylinder; $\sigma$: mass per surface area)
If the cylinder is short, the pattern of fieldlines looks similar to
the ring case.  However, the center region where the field is weak
becomes larger.  In a cylinder of infinite length the field inside
actually vanishes and the potential is constant.  (Though the above
integral won't give you any valid potential at all, because it
diverges for $z_0 \to -\infty$, $z_1 \to +\infty$.  You'd have to
modify it a bit.)
In the case of infinite length you can in fact use Gauss's law to show
that the field vanishes.  For the integration surface, choose a
smaller concentric cylinder.  Then, for symmetry reasons, the field
must have the same strength everywhere on it and either point radially
inwards or outwards everywhere.  No cancellations can happen.  And
since there is no mass inside the cylinder, you can conclude that the
total flux as well as the field itself must be zero.
