Proving that the delta function is symmetric to prove that the delta function is symmetric, I need to show that $\delta(x) = \delta(-x)$ by employing a change in variables.
$$\delta(x) = {1\over 2\pi}\int_{-\infty}^\infty\exp(ikx)dk\tag{1}$$
$$\delta(-x) = {1\over 2\pi}\int_{-\infty}^\infty\exp(-ikx)dk$$
Let $k = -l$,
$$\delta(-x) = {-1\over 2\pi}\int_\infty^{-\infty}\exp(ilx)dl 
={1\over 2\pi}\int_{-\infty}^\infty\exp(ilx)dl\tag2 $$
$$=\delta(x)$$
Am I right to say that it doesn't matter whether it is k or l, so long as they produce the same expression, the 2 integrals (1) & (2) are equal? But why is it the case, when k is not equal to l?
 A: Beware that $\delta$ is not a function but a distribution and you need to be careful with what you mean by expressions such as "$\delta(-x)$".
To do this more rigorously, lets start by defining an operation on smooth functions: if $\phi$ is smooth, let $\check\phi$ be the function defined by $\check\phi(x) = \phi(-x)$. If $f$ is a distribution that comes from a smooth (or even $L^1_{loc}$ function), then for any test function $\phi$
$$ 
\langle \check f, \phi \rangle = \int_{-\infty}^\infty f(-x)\phi(x)\,dx = \int_{-\infty}^\infty f(x)\phi(-x)\,dx = \langle f, \check \phi \rangle
$$
by a change of variables. Hence, it's reasonable to define $\check U$ for any distribution by $\langle \check U, \phi \rangle = \langle U, \check \phi \rangle$. For $\delta$, we get
$$ 
\langle \check\delta, \phi \rangle = \langle \delta, \check\phi \rangle = 
\check \phi(0) = \phi(0) = \langle \delta, \phi \rangle.
$$
Hence $\check \delta = \delta$, which is the distribution theory version of even.
A: Instead of using the Fourier transform to prove symmetry, I used the definition of the delta function to prove symmetry
$$f(a) = \int_{-\infty}^{\infty}f(x)\delta (x-a) \; dx$$
and
$$f(a) = \int_{-\infty}^{\infty}f(x)\delta (a-x) \; dx$$
Therefore
$$\int_{-\infty}^{\infty}f(x)\delta (x-a) \; dx = \int_{-\infty}^{\infty}f(x)\delta (a-x) \; dx$$
Finally
$$\delta (x-a) = \delta (a-x)$$
or 
$$\delta (x) = \delta (-x)$$
