I have encountered this problem while trying to evaluate $$ \int_0^\infty e^{ix^2}dx $$ by squaring it and using Fubini's theorem.
I just treat $i$ as a constant: $$ \int_0^\infty e^{i \xi} d \xi = \frac{1}{i} \left[ e^{\pm i\infty} -1 \right] $$ But what is '$i\infty$'?
This integral does not converge. We can associate a "meaning" to it by stepping into the world of generalised functions (distributions) via the Dirac Delta generalised function as
$$\delta (k) = \int_0^{+\infty} e^{\pm ikx}\ \text{d}x$$
but this is a completely different story.
Edit, for the sake of completeness
The Dirac delta is the distribution (generalised function) $\delta \in \mathcal{D}(\mathbb{R}) := C_c^\infty(\mathbb{R})^\prime$ defined by $$ \forall f \in C_c^\infty(\mathbb{R}), \quad \delta(f) := f(0). $$ For any $a \in \mathbb{R} \setminus \{0\}$, define a linear transformation $S_a : C_c^\infty(\mathbb{R}) \to C_c^\infty(\mathbb{R})$ by $$ \forall f \in C_c^\infty(\mathbb{R}), \; \forall x \in \mathbb{R}, \quad (S_a f)(x) := f(ax). $$ For any $a \neq 0$, one can use the change of coordinates $x^\prime = ax$ to show that for all $f,g \in C_c^\infty(\mathbb{R})$, $$ \begin{align} \int_{-\infty}^\infty (S_a f)(x)\, g(x)\,dx &= \int_{-\infty} f(ax)\,g(x)\,dx\\ &= \int_{-\infty}^\infty f(x^\prime) g(a^{-1}x^\prime)\frac{1}{\lvert a \rvert}dx^\prime\\ &= \int_{-\infty} f(x) \left(\frac{1}{\lvert a \rvert}S_{a^{-1}}g\right)(x) \,dx, \end{align} $$ i.e., that $S_a^\ast = \lvert a \rvert^{-1} S_{a^{-1}}$ with respect to the $L^2$ inner product on $\mathbb{R}$, so, by duality, we're obliged to define $S_a\delta$ (i.e., $\delta(ax)$) to be the distribution $$ S_a\delta := \delta \circ \frac{1}{\lvert a \rvert}S_{a^{-1}} \in \mathcal{D}(\mathbb{R}). $$ But now, for any $f \in C_c^\infty(\mathbb{R})$, $$ (S_a\delta)(f) = \delta\left(\frac{1}{\lvert a \rvert}S_{a^{-1}}f\right) = \left(\frac{1}{\lvert a \rvert}S_{a^{-1}}f\right)(0) = \frac{1}{\lvert a \rvert}f(a^{-1}0) = \frac{1}{\lvert a \rvert}f(0) = \left(\frac{1}{\lvert a \rvert}\delta\right)(f), $$ so that, indeed, $$ S_a\delta = \frac{1}{\lvert a \rvert}\delta, $$ which is the rigorous statement of the identity $\delta(ax) = \frac{1}{\lvert a \rvert}\delta(x)$.
For what concerns how to represent Dirac Delta, then we have the following:
Assume $h_n \in L^1_{\text {loc}} $ are such that
Then $h_n \to \delta $.
Proof
First notice that $$|\phi (x) - \phi (0)| = |\int_0^x \phi'(t) \, dt| \leq \int_0^x |\phi'(t)| \, dt \leq \int_0^x \|\phi'\| \, dt = \|\phi'\| x $$ where $\|\cdot\|$ is uniform/infinity norm.
Then notice that $$\langle \delta, \phi \rangle = \phi (0) = \phi (0) \int h_n (x) \, dx = \int h_n (x) \, \phi (0) \, dx $$
Therefore, $$|\langle h_n, \phi \rangle - \langle \delta, \phi \rangle| = | \int h_n (x) \, \phi (x) \, dx - \int h_n (x) \, \phi (0) \, dx| \leq \int h_n (x) |\phi (x) - \phi (0)| \, dx \\ \leq \int h_n (x) \, \|\phi'\| x \, dx = \|\phi'\| \int x h_n (x) \, dx \to 0$$ since $\|\phi'\| < \infty $.
Thus, $h_n \to \delta $ as $n \to \infty $.
Dirac Delta Representations*
For what concerns the integral the OP presented, we have when you compute it,
$$∫_{-L}^L e^{i\alpha x} \ \text{d}\alpha = \frac{e^{iLx} - e^{iLx}}{ix} = \frac{\sin(Lx)}{2x} = \frac{L}{2} \frac{\sin(Lx)}{Lx}$$ you see that there isn't a limit as $L\to +\infty$. The expression is more subtle than that.
In your case it's $\int_0^{+\infty}$ which has no problems in zero but when you compute $L\to +\infty$... eh!
You may want to see the following related question, Dirac Delta function and normal distribution Roughly, the Gaussian distribution with variance tending to zero, has the 'pointwise' behaviour one asks of the 'dirac delta function'.
Also, the following PDF may be helpful. http://www.cse.yorku.ca/~kosta/CompVis_Notes/fourier_transform_Gaussian.pdf
Advances
I find it to be most natural to define the Dirac delta $\delta_{x'}$ on a manifold $M$ so that for any smooth function $f:M\to\mathbb C$ the formal integral $\int_Mf(x)\delta_{x'}(x)dV(x)$ (or more precisely, the duality pairing $\langle\delta_{x'},f\rangle$) equals $f(x')$. Here $V$ denotes the volume measure of the manifold (which may be Riemannian or Lorentzian, for example). In fact, I find this the most natural way to go even over a Euclidean space. I would call this the definition of the Dirac delta.
I wrote $\delta_{x'}(x)$ instead of $\delta(x-x')$ on purpose. The difference $x-x'$ doesn't mean anything on a general manifold. However, $x-x'$ does mean something in local coordinates, but then both $x$ and $x'$ have to be within the same coordinate chart. In local coordinates $dV(x)=\sqrt{|g(x)|}dx$. Once you are in a local coordinate chart, you get $$ \int\sqrt{|g(x)|} \delta_{x'}(x)f(x)dx=f(x'), \tag{2} $$
(Again, this is just a formal integral.) However, I would not treat this as the definition, but as a local coordinate representation of the Dirac delta.
Once you are on a local coordinate chart, you can indeed formally write $$ \delta_{x'}(x) = |g(x')|^{-1/2}\int_{\mathbb R^d}\frac{1}{(2\pi)^d}e^{ik\cdot (x-x')}dk. $$ Here $x$ and $x'$ are vectors in $\mathbb R^d$ and are identified with two points on $M$ via a coordinate chart. (If you plug this representation to (2) with a smooth function $f$ supported in the coordinate chart, you obtain the correct result.) The integral is taken over the entire Euclidean space. I doubt this representation is of much use, since the Fourier transform tends to be less useful on manifolds. (It can be defined on coordinate charts, but it does not have the nice and simple global properties the usual Fourier transform does.)
Euclidean spaces have a lot of special structure. This structure allows us to speak of the difference $x-x'$ and also allows for (simple) Fourier analysis.
On general manifolds you lose both, so a Fourier representation and difference of two points are kind of lost at the same time. Locally a manifold can be identified with (a subset of) a Euclidean space, but only locally.
$$\int_0^p e^{+i x} dx=\sin (p)+i(1- \cos (p))$$ $$\int_0^p e^{-i x} dx=\sin (p)-i(1- \cos (p))$$
What happens when $p \to \infty$ ?
