Definite integral for exp(+/- i ax) Is the following correct?
$$\begin{align}
 \int_{-\infty}^{+\infty} dx\, e^{+iax} &=\int_{-\infty}^{+\infty} dx\,\cos{(ax)}+i\int_{-\infty}^{+\infty}dx\,\sin{(ax)}\\\\ 
&\overbrace{=}^{x=-y}\int_{+\infty}^{-\infty} d(-y)\,\cos{(a(-y))}+i\int_{+\infty}^{-\infty}d(-y)\,\sin{(a(-y))}  \\\\
&=\underbrace{\int_{-\infty}^{+\infty} d{y}\,\cos{(ay)} -i\int_{-\infty}^{+\infty} d{y}\,\sin{(ay)}}_{-\int_{+\infty}^{-\infty}=\int^{+\infty}_{-\infty}}\\\\
&=\int_{-\infty}^{+\infty} d{y}\, e^{-iay}\\\\
&=\int_{-\infty}^{+\infty} d{x}\, e^{-iax}.
\end{align}$$
 A: The object written $\int_{-\infty}^\infty e^{\pm iax}\,dx$ is either $(i)$ a divergent integral or $(ii)$ tempered distribution and not an integral.

It is easy to see from even-odd symmetry that for any $L\in \mathbb{R}$ $\int_{-L}^L e^{+ iax}\,dx=\int_{-L}^L e^{- iax}\,dx$, but the limit as $L\to \infty$ of both of these integrals fails to exist (unless the limit is a distributional limit).
However, if we interpret the object as a tempered distribution, then it is, in fact, the Fourier Transform of the constant $1$.  And in this case (i.e., in distribution) that the equality $\displaystyle \int_{-\infty}^\infty e^{+iax}\,dx=\int_{-\infty}^\infty e^{-iax}\,dx$ holds.

EVALUATION OF $\displaystyle \int_{-\infty}^\infty e^{\pm iax}\,dx$
If we interpret the object as a Fourier Transform, we can evaluate it as follows.  Let $\phi\in \mathbb{S}$ (i.e., a Schwarz Space Function).  Then, invoking the Fourier Inverse Theorem, we have
$$\begin{align}
\langle \mathscr{F}\{1\}, \phi\rangle&=\langle 1, \mathscr{F}\{\phi\}\rangle\\\\
&=\int_{-\infty}^\infty \int_{-\infty}^\infty \phi(x)e^{\pm iax}\,dx\,da\\\\
&=2\pi \phi(0)\tag1
\end{align}$$
Hence, in distribution we see from $(1)$ that $\mathscr{F}\{1\}=2\pi \delta$ where $\delta$ is the Dirac Delta distribution.
Note that we have tacitly shown that the equality $\displaystyle \int_{-\infty}^\infty e^{+iax}\,dx=\int_{-\infty}^\infty e^{-iax}\,dx$ holds in distribution.


ALTERNATIVE DEVELOPMENT:
Here, we represent the Fourier Transform of $1$ as the distributional limit $\displaystyle \lim_{L\to\infty}\int_{-L}^L e^{\pm iax}\,dx$.  Proceeding, we have for any $\phi\in \mathbb{S}$
$$\begin{align}
\langle \mathscr{F}\{1\}, \phi\rangle&=\lim_{L\to \infty}\int_{-\infty}^\infty \phi(x) \int_{-L}^L e^{\pm iax} \,da\,dx\tag2\\\\
&=\lim_{L\to \infty}\int_{-\infty}^\infty \phi(x) \frac{2\sin(Lx)}{x}\,dx\\\\
&\overbrace{=}^{\text{IBP}}-\lim_{L\to \infty}\int_{-\infty}^\infty \phi'(x) \int_{-\infty}^x \frac{2\sin(x'L)}{x'}\,dx'\,dx\\\\
&=-\lim_{L\to \infty}\int_{-\infty}^\infty \phi'(x) \int_{-\infty}^{Lx} \frac{2\sin(x')}{x'}\,dx'\,dx\\\\
&\overbrace{=}^{\text{DCT}}-\int_{-\infty}^\infty \phi'(x) \lim_{L\to \infty}\int_{-\infty}^{Lx} \frac{2\sin(x')}{x'}\,dx'\,dx\\\\
&=-\int_{-\infty}^\infty \phi'(x) 2\pi H(x)\,dx\\\\
&=-2\pi \int_0^\infty \phi'(x)\,dx\\\\
&=2\pi \phi(0)
\end{align}$$
from which we deduce that in distribution
$$\lim_{L\to \infty}\int_{-L}^L e^{\pm iax}\,dx=2\pi \delta(a)$$
where the limit is taken as in $(2)$.
