Let $m, n >0$ fixed. I have this random variable $X$ whose pdf $f(x)$ is defined as:
$f(x)=\frac{\frac{x^{m - 1}}{(1+x)^{m+n}}}{B(m,n)}, x \in [1, \infty), f(x)= 0 $ elsewhere.
N.B. Yes it is really $[1, \infty),$ not $[0, \infty).$
We can see that $f(x) \ge 0, \int_{\mathbb{R}}f(x)dx=1,$ so clearly $f(x)$ is a PDF. But I'm wondering whether I should call $X$ to follow the beta distribution with parameters $m,n$, as normally Beta distribution has support on $[0,1]$ only, not on $[1, \infty)$. If the answer is yes, great! If we can't call $X$ to follow a Beta distribution, should we say $X$ follow a transformed beta distribution? Is there any name for $X?$ Clearly, $X=\frac{B}{1-B}, B \sim B(m,n).$ But this transformation is bijective, so should we also call $X$ to be a Beta random variable?
Thank you!
ADDENDUM: To deter any confusion that may arise after reading the answer(s), let me state that I made a bad mistake in defining $f(x)$ earlier: I defined it as:
$f(x)=\frac{\frac{x^{m/2 - 1}}{(1+x)^{m+n}}}{B(m,n)}, x \in [1, \infty), f(x)= 0 $ elsewhere.
which was not the intended definition.
I don't know why you seem to think that $f$ integrates to $1$. To rephrase your claimed density, it says
$$f(x \mid m,n) = \frac{\Gamma(m+n)}{\Gamma(m)\Gamma(n)} x^{m/2 - 1} (1+x)^{-(m+n)}. \tag{1}$$ We will deal with the question of the support later. Now consider the case $m = 2$: We get
$$f(x \mid 2, n) = \frac{\Gamma(2+n)}{\Gamma(2)\Gamma(n)} x^0 (1+x)^{-(2+n)} = \frac{n(n+1)}{x^{n+2}}. \tag{2}$$ Its antiderivative is
$$\int f(x \mid 2,n) \, dx = \frac{n}{(x+1)^{n+1}} + C. \tag{3}$$ So its integral on $[0, \infty)$ is $n$, and on $[1,\infty)$, it is $n/2^{n+1}$, neither of which is $1$ except when $n = 1$ in the first case and certainly not for any general $n$. Therefore, your claim that $f$ integrates to $1$ is false.
If the support is on $[0,\infty)$, then the correct constant of proportionality that makes $f$ a density is not $1/B(m,n)$ but instead $$\frac{\Gamma(m+n)}{\Gamma(\frac{m}{2})\Gamma(\frac{m}{2}+n)} = \frac{1}{B(\frac{m}{2}, \frac{m}{2} + n)}.$$ You may verify this yourself with a suitable choice of $m, n$.
If the support is on $[1,\infty)$, then the constant includes a hypergeometric factor.
As for your claim that $X = B/(1-B)$ where $B$ is beta with general parameters $m, n$, the actual PDF of the transformed variable is
$$f_X(x) = \frac{1}{B(m,n)} x^{m-1} (1+x)^{-(m+n)} \mathbb 1 (x > 0). \tag{4}$$ That is to say, the support of such a variable is on $(0,\infty)$, which should be obvious from the transformation itself, since $$\lim_{B \to 0^+} \frac{B}{1-B} = 0$$ and in particular, $B = 1/3$ implies $X = (1/3)/(1-1/3) = 1/2 < 1$.
Returning to your original question about the name of such a distribution, if the support is on $(0,\infty)$, then $X = B/(1-B)$ as illustrated in $(4)$ is called Beta prime, or inverted beta, although I avoid the latter terminology to avoid confusion with "inverse beta", which is presumably $Y = 1/B$, with density
$$f_Y(y) = \frac{1}{B(m,n)} y^{-(m+n)} (y-1)^{n-1} \mathbb 1 (y > 1). \tag{5}$$
If we wish to retain the shape of $(4)$ but restrict the support to $[1,\infty)$, then this would be a left-truncated Beta prime distribution.