Moment of inertia when velocity is zero The most basic form of the moment of inertia comes from fix axis rotation, that is
$$L=\int rv\, dm=\int(r\omega)(r\,dm)=\omega \int r^2\, dm\tag{1}$$ Here $r$ is the perpendicular distance from the axis.
Now the point I want to address here is that this derivation explicitly requires the object to be rotating with some velocity $v$, thus the existence of $r\omega$. But this begs the question about the nature of moment of inertia that is, it exists whether or not the object is rotating, one has to simply calculate the integral $\int r^2\, dm$, but this integral itself comes from assuming some non zero $v$. Thus if we look at the math, the moment of inertia is independent of the velocity of the object, but its derivation is not, which seems strange to me. This also fundamentally touches on the question of the actual physical meaning behind the moment of inertia.
Just as most of us are taught about the center of mass as a definition without motivation or by very weak motivations such as saying there is an analogy between force equation for a single mass point and force equation for the whole system with a net mass of the system, but the real physical motivation comes from choosing a frame where net momentum is zero, similarly, I believe here too is the same case. The above derivation just seems to be a weak motivation as described above, instead of the real physical content.
Edit: Since this question is being identified as a duplicate I want to make clear that it is not a duplicate as my main concern over here is on the derivation and the existence of moment of inertia when velocity is zero, either of which is not satisfied with any of the questions mentioned or suggested by the users.
 A: 
The most basic form of the moment of
inertia comes from
fix axis rotation, that is  $$L=\int rv
dm=\int(r\omega)(r\,dm)=\omega \int r^2\, dm\tag{1}$$ Here $r$ is the
perpendicular distance from the axis.

It seems like you are saying $L$ in the above equation is moment of inertia. Just to be clear, it is angular momentum. The moment of inertia is
$$I=\int r^{2}dm$$
Angular momentum is related to moment of inertia by
$$L=I\omega$$

But this begs the question about the nature of moment of inertia that
is, it exists whether or not the object is rotating,

Moment of inertia for rotational motion and angular momentum is analogous to mass for linear motion and linear momentum. Moment of inertia is a measure of resistance to change in rotational motion whereas mass is a measure of resistance to change in linear motion.
In the same sense that mass exists whether or not it is in linear motion the moment of inertia of an object exists whether or not the object rotating.

This also fundamentally touches on the question of the actual physical
meaning behind the moment of inertia.

The physical meaning of moment of inertia is it is a measure of resistance to change in rotational motion analogous to mass being a measure of resistance to change in linear motion.
Hope this helps.
A: In classical physics, and object of mass $m$ and moment of inertia $I$ has translational kinetic energy $\frac12mu^2$ when its centre of mass has translational velocity $u$, and rotational kinetic energy $\frac12I\omega^2$ when its rotational velocity about a relevant axis is $\frac12I\omega^2$. (If it's rotated some other way, $I$ would change.) Your objection is that we can determine $I$ even when $\omega=0$; that's no more to the point we can determine $m$ even when $u=0$.
Either way, all we doing is computing the entire kinetic energy by summing and/or integrating, and transaltional and rotational kinetic energy result from individual pieces, be they particles or of infinitesimal volume, each undergoing whatever motion they do. Since you asked about the centre of mass, say $\bar{x}$, its definition is analogous: we can show $m\dot{\bar{x}}$ is the total momentum.
A: 
But this begs the question about the nature of moment of inertia that
is, it exists whether or not the object is rotating, one has to simply
calculate the integral ∫r2dm, but this integral itself comes from
assuming some non zero v.

We also could say that mass has its meaning from $p = mv$.  Mass is a property of an object that allows us to calculate the linear momentum, knowing also the velocity. But we know that its utility comes from the fact that once we know the relation $\frac{p}{v}$ from a given velocity, it is valid for any velocity. The object doesn't change its mass when changing velocity.
The same way, once the moment of inertia is known for a given angular velocity, the relation between different angular velocities and angular momentum can be known.
A: The general state of a rigid body is combined rotation and translation, and the case of pure translation where MMOI is not important is only a special case.
In fact, rotation is the defining state, as when the center of mass is translating and the body is rotating about the center of mass, there always exists an axis in space where the body is rotating about (the instantaneous axis of rotation) and translates along at the same time (the so called screw motion).
To derive MMOI we ignore the translation of the center of mass, since it is accounted for with linear momentum and Newton's 2nd law.
The assumption that each particle i on the body moves with rotation $$ \vec{v}_i = \vec{\omega} \times \vec{r}_i$$ is actually a requirement for being a rigid body and maintaining fixed distances between all particles.
Now considering the angular momentum of each particle i as being defined as $$ \vec{L}_i = \vec{r}_i \times \vec{p}_i$$ where $\vec{p}_i= m_i \vec{v}_i$ is the momentum of the particle, then when all the contributions of all the particles are summed up we can derive mass moment of inertia from factoring the rotation $\vec{\omega}$ out.
$$ \vec{L} = \sum_i m \vec{r}_i \times ( \vec{\omega} \times \vec{r}_i) = \left( \sum_i m_i ( \vec{r}_i \cdot \vec{r}_i - \vec{r}_i \odot \vec{r}_i) \right) \vec{\omega} = \mathrm{I} \;\vec{\omega}$$
If you are questioning why ignore the motion of the center of mass, then you can easily show that any translation of the center of mass $\vec{c}_{\rm COM}$ would cancel out of the above expression. Remember due to $\vec{r}_i$ being defined from the COM you have $\sum_i m_i \vec{r}_i =0$ by definition.
