Is the property $\int_{-a}^{a}f(x)dx=\int_{-a}^{a}f(-x)dx$ still applicable when $a \to \infty$ Is the property $\int_{-a}^{a}f(x)dx=\int_{-a}^{a}f(-x)dx$ still applicable when $a \to \infty$
that is can we say:
$$\int_{-\infty}^{\infty}f(x)dx=\int_{-\infty}^{\infty}f(-x)dx$$
because $\infty -\infty \ne 0$
 A: I'm going to answer your question in the comments first, because I think it's more fundamental.

What is the difference between $\lim_{a \to \infty} \int_{-a}^a f(x) \; \mathrm{d}x$ and $\int_{-\infty}^\infty f(x) \; \mathrm{d}x$.

Some might say "nothing", but it's not quite right. If the latter expression exists, then it is equal to the former. However, it is possible for the former to exist, but not the latter.
The definition of an improper integral $\int_{a_0}^\infty f(x) \; \mathrm{d}x$ is a limit $\lim_{a \to \infty} \int_{a_0}^a f(x) \; \mathrm{d}x$. If this limit exists, we say the improper integral exists, and define its value to be the value of the limit. Similarly, we define $\int_{-\infty}^{a_0} f(x) \; \mathrm{d}x$ to be $\lim_{a \to -\infty} \int_a^{a_0} f(x) \; \mathrm{d}x$.
You might guess that the definition of $\int_{-\infty}^\infty f(x) \; \mathrm{d}x$ is $\lim_{a \to \infty} \int_{-a}^a f(x) \; \mathrm{d}x$, but that's not how we define it. We actually say that it's equal to
$$\int_{-\infty}^{a_0} f(x) \; \mathrm{d}x + \int_{a_0}^\infty f(x) \; \mathrm{d}x,$$
where $a_0$ is any point in $\Bbb{R}$ (it doesn't matter which). So, it's equal to the sum of two limits, both of which have to exist and be finite independently of each other.
Here's an example: $f(x) = x^3$. Then
$$\int_{a_0}^a x^3 \; \mathrm{d}x = \frac{a^4 - a_0^4}{4} \to \infty$$
as $a \to \infty$, but
$$\int_{-a}^a x^3 \; \mathrm{d}x = 0 \to 0$$
as $a \to \infty$. So, $\int_{-\infty}^\infty x^3 \; \mathrm{d}x$ doesn't exist, but $\lim_{a \to \infty} \int_{-a}^a x^3 \; \mathrm{d}x = 0$.

This also answers your main question. If we have that $\int_{-\infty}^\infty f(x) \; \mathrm{d}x$ exists, then it is equal to $\lim_{a \to \infty} \int_{-a}^a f(x) \; \mathrm{d}x$. For each $a$, we have $\int_{-a}^a f(x) \; \mathrm{d}x = \int_{-a}^a f(-x) \; \mathrm{d}x$, and thus the limits agree.
To show that this limit is equal to $\int_{-\infty}^\infty f(-x) \; \mathrm{d}x$, we just need to show that this limit exists. Observe that, for some (or any) $a_0 \in \Bbb{R}$, we have $\int_{a_0}^a f(-x) \; \mathrm{d}x = \int_{-a}^{-a_0} f(x) \; \mathrm{d}x$, hence $\int_{a_0}^\infty f(-x) \; \mathrm{d}x = \int_{-\infty}^{-a_0} f(x) \; \mathrm{d}x$. We can similarly see that $\int_{-\infty}^{a_0} f(-x) \; \mathrm{d}x = \int_{-a_0}^\infty f(x) \; \mathrm{d}x$. Therefore $\int_{-\infty}^\infty f(-x) \; \mathrm{d}x$ exists, and so
$$\int_{-\infty}^\infty f(x) \; \mathrm{d}x = \lim_{a \to \infty} \int_{-a}^a f(x) \; \mathrm{d}x = \lim_{a \to \infty} \int_{-a}^a f(-x) \; \mathrm{d}x = \int_{-\infty}^\infty f(-x) \; \mathrm{d}x.$$
But, this all relies on $\int_{-\infty}^\infty f(x) \; \mathrm{d}x$ existing. It doesn't work for $f(x) = x^3$, for example!
