Transforming an improper integral The function $f(x)$ is defined in the interval $(-\infty, a]$, where $a$ is some real number.
If I have an improper integral of the form
$$\int_{-\infty}^{a} f(x) dx \tag{1}$$
and I need to decide if it converges or diverges, can I always transform it to this one
$$\int_{a}^{\infty} f(2a-x) dx \tag{2}$$
and study the integral (2) instead of (1)?
I am saying that (1) is convergent if and only if (2) is convergent.
First of all, is this statement true?
Also, in fact... I am also claiming that:
$$\int_{-\infty}^{a} f(x) dx = \int_{a}^{\infty} f(2a-x) dx$$
I think it is true (just by geometric considerations) because the graphs of the two functions $f(x)$ and $g(x) = f(2a-x)$ are symmetric with respect to the line $x=a$.
How can we justify this statement more formally?
Why I am asking this? Because in my book all criteria for convergence/divergence (of improper integrals) are given only for integrals of the kind (2). So that made me thinking that... OK, I need to have some way to deal with integrals of the kind (1). And as a result I came up with this transformation.

$$\int_{-\infty}^{a}f(x)dx$$
Put $x=2a-u \implies dx=-du$.
$$-\int_{\infty}^{a}f(2a-u)du$$
According to properties of definite integrals,
$$\int_{a}^{\infty}f(2a-u)du$$
Which is second integral you are talking about.
Secondly, you shouldn't use same variable for different bounds for 2 or more integrals.
In the description you claimed that,
$$\int_{-\infty}^{a} f(x) dx = \int_{a}^{\infty} f(2a-x)dx$$
So it's not correct to use the same variable for both integrals with different bounds. Instead, do this.
$$\int_{-\infty}^{a} f(x) dx = \int_{a}^{\infty} f(2a-u)du$$
Doing one U-substitution(As i have done in the starting), says that it's true.
