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.
 A: $$\int_{-\infty}^{a}f(x)dx$$
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.
A: In section 179: Application to infinite integrals of the rules for substitution and integration by parts in A course in pure mathematics by G. H. Hardy the substitution of improper integrals is explicitly stated. It is based upon the substitution rule together with integration by parts for definite integrals. We find

Transformation by substitution: Suppose that
\begin{align*}
\int_a^{\infty}\phi(x)\,dx\tag{1}
\end{align*}
is convergent. Further suppose that, for any value of $\xi$ greater than $a$, we have, as in $\S$161,
\begin{align*}
\int_a^{\xi}\phi(x)\,dx=\int_b^{\tau}\phi\{f(t)\}f^{\prime}(t)\,dt,\tag{2}
\end{align*}
where $a=f(b),\xi=f(\tau)$. Finally suppose that the functional relation $x=f(t)$ is such that $x\to \infty$ as $t\to \infty$. Then, maing $\tau$ and so $\xi$ tend to $\infty$ in (2), we see that the integral
\begin{align*}
\int_{b}^{\infty}\phi\{f(t)\}f^{\prime}(t)\,dt\tag{3}
\end{align*}
is convergent and equal to the integral (1).
On the other hand it may happen that $\xi\to\infty$ as $\tau \to -\infty$ or as $\tau \to c$. In the first case we obtain
\begin{align*}
\color{blue}{\int_{a}^{\infty}\phi(x)\,dx}&=\lim_{\tau\to-\infty}\int_b^{\tau}\phi\{f(t)\}f^{\prime}(t)\,dt\\
&=-\lim_{\tau\to -\infty}\int_{\tau}^b\phi\{f(t)\}f^{\prime}(t)\,dt\color{blue}{=-\int_{-\infty}^b\phi\{f(t)\}f^{\prime}(t)\,dt}.
\\
\cdots\qquad\qquad&
\end{align*}

which is the relevant part to answer your question. We can use the same arguments as above to derive
\begin{align*}
\int_{-\infty}^a\phi(x)\,dx=-\int_{b}^\infty\phi\{f(t)\}f^{\prime}(t)\,dt
\end{align*}
and with $f(t)=2a-t, f^{\prime}(t)=-1$ and $b=f(a)$ we obtain
\begin{align*}
\color{blue}{\int_{-\infty}^a\phi(x)\,dx=\int_{a}^\infty\phi(2a-t)\,dt}
\end{align*}
