It's important to distinguish different uses of "wave functions" in this context. It took me a while to understand it, when I learned quantum theory, because it's sometimes not made too clear in textbooks.
In wave mechanics you start with the idea that (for a single non-relativistic particle) there's a wave function describing the (pure) state of the particle, ##\psi(t,\vec{x})##. The physical meaning is that for a particle in this state the probability distribution for its position is given by
$$P(t,\vec{x})=|\psi(t,\vec{x})|^2$$
provided you have properly normalized the wave function,
$$\int_{\mathbb{R}^3} \mathrm{d}^3 x P(t,\vec{x})=1.$$
Further the observables are represented by self-adjoint operators on the corresponding Hilbert space of square-integrable functions, e.g., ##\hat{\vec{x}}=\vec{x}## (i.e., the position operator, applied to the wave function is just the multiplication of the wave function with ##\vec{x})## or ##\hat{\vec{p}}=-\mathrm{i} \hbar \vec{\nabla}## (i.e., the momentum operator is given by the gradient of the wave function times ##-\mathrm{i} \hbar##).
The self-adjoint operators have eigenfunctions, sometimes generalized ones if the eigenvalues form a continuous set. Let's take the momentum eigenstates, which must fulfill the equation
$$-\mathrm{i} \hbar \vec{\nabla} u_{\vec{p}}(\vec{x})=\vec{p} u_{\vec{p}}(\vec{x}).$$
The solution is
$$u_{\vec{p}}(\vec{x})=N \exp(\mathrm{i} \vec{p} \cdot \vec{x}/\hbar), \quad \vec{p} \in \mathbb{R}^3.$$
These momentum eigenfunctions are not square integrable but "normalizable to a ##\delta## distribution", i.e., the usual choice of the phase factor is such that
$$\int_{\mathbb{R}^3} \mathrm{d}^3 x u_{\vec{p}}^*(\vec{x}) u_{\vec{p}'}(\vec{x})=\delta^{(3)}(\vec{p}-\vec{p}').$$
From the theory of Fourier integrals one gets ##N=(2 \pi \hbar)^{-3/2}##, i.e., the conveniently normalized momentum eigenfunctions are
$$u_{\vec{p}}(\vec{x})=\frac{1}{\sqrt{(2 \pi \hbar)^3}} \exp(\mathrm{i} \vec{p} \cdot \vec{x}/\hbar).$$
The importance of such eigenfunctions of self-adjoint operators that represent observables is that you get the probabilities (or probability distribution for continuous eigenvalues) by the corresponding projections. E.g., the probability distribution for momentum for a particle prepared in the state represented by the square-integrable wave function ##\psi## is given by
$$\tilde{P}(t,\vec{p})=|\tilde{\psi}(t,\vec{p})|^2$$
where the momentum-space wave function is given by the projection to the momentum eigenfunctions,
$$\tilde{\psi}(t,\vec{p})=\langle u_{\vec{p}}|\psi(t) \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 x u_{\vec{p}}^*(\vec{x}) \psi(t,\vec{x}) = \int_{\mathbb{R}^3} \mathrm{d}^3 x \frac{1}{\sqrt{(2 \pi \hbar)}^3} \exp(-\mathrm{i} \vec{p} \cdot \vec{x}) \psi(t,\vec{x}).$$
It is also important to know that usually the eigenfunctions of self-adjoint operators are "complete" in the sense that you can write any wave function (to be more careful any wave function in the domain, where the self-adjoint operator is well defined, which build a dense subspace of the Hilbert space of square-integrable functions) as "superposition" of the eigenstates. In the case of operators with continuous eigenvalues the "superposition" is rather an integral than a sum. For the momentum space eigenfunktions you indeed get
$$\int_{\mathbb{R}^3} \mathrm{d}^3 p |u_{\vec{p}} \rangle \langle u_{\vec{p}}|=\hat{1}.$$
This is the "completeness relation" for the momentum eigenstates.
For the wave functions this means that
$$\psi(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 p \langle u_{\vec{p}}|\psi(t) \rangle u_{\vec{p}}(\vec{x}) = \int_{\mathbb{R}^3} \mathrm{d}^3 p \frac{1}{\sqrt{(2 \pi \hbar)^3}} \tilde{\psi}(t,\vec{p}) \exp(\mathrm{i} \vec{p} \cdot \vec{x}/\hbar).$$