Is $f(t,\beta ,\mu , \theta )= e^{\theta(t-\mu)^{\beta}}$ a well known distribution on the unit interval? $\newcommand{\erf}{\operatorname{erf}}$
I have tried out to define a new  distribution lie in [0,1] using my special function which is montioned in this paper, The new special function is defined as:
$$I(a)=\int_{0}^{a}{(e^{-x²})}^{\erf(x)}dx, \tag{01}$$ take $t=\erf(x)$ then $x=\erf^{-1}(t)$ such that $ dx=\dfrac{\sqrt{\pi}}{2}e^{(\erf^{-1}(t))^2 }dt$ by substitution in $\int_{0}^{a}{(e^{-x²})}^{\erf(x)}dx$ yield to get : $$\dfrac{\sqrt{\pi}}{2}\int_{0}^{b}{(e^{(\erf^{-1}(t))^2-t(\erf^{-1}(t))^2})}dt, \tag{02}$$ using this approximation for small $t$ we have :$\erf^{-1}(t) \sim t\dfrac{\sqrt{\pi}}{2}$ , Replace this approixmation gives the following :$$\dfrac{\sqrt{\pi}}{2}\int_{0}^{b}{e^{\dfrac{\pi}{4}(t^2-t^3)}}dt .\tag{03}$$
Now for the latter equality I want to define a new distribution since we are now in [0,1], because the precedent variable change $t$ vary from $0$ to $1$ because $ \erf^{-1}(\infty)=1 $, Then let us define a Unit distribution with $3$ parametre : $\beta$ and $\mu$ and $\theta $ such that:
$$ f(t,\beta ,\mu , \theta )=e^{\theta(t-\mu)^{\beta}} \text{and} f(t,\beta ,\mu , \theta )=0,  \text{for}: t >1 \tag{04}$$ , Now my question here is this a known distribution and if it is what is it ?
Edit: I have edited parameters for generalisation to be clear
 A: The distribution seems to be close to a truncated form of the generalized Gaussian distribution which has the same formula (except it has support on entire $\mathbb{R}$ and some parameters are in a different place, for instance  $\theta = -1/\alpha^\beta$ )
$$f(x) \propto e^{-(\vert x-\mu \vert /\alpha)^\beta} $$

One difference is that the function uses the absolute value of $x-\mu $. But , this might be necessary for your distribution as well. In the cases $\mu>0$ and $\beta \notin \mathbb{Z}$ then you the term $(x-\mu)^\beta$ is a non-integer power of a negative number which is problematic. So I guess that your distribution should use the restriction $\mu<0$ in which case the difference with the absolute sign is not relevant.
A: $f$, in your formulation, is not a probability density function as it does not integrate to 1 on the interval $t\in(0,1)$ for all values of the parameters. You would need to normalize it, but then the expression would not be that nice as it will depend on the parameters through some special functions (actually the cdf of the Generalized Normal distribution). 
One alternative is to use the logit transformation applied to the Generalized Normal distribution, which leads to
$$f(t;\alpha,\beta,\mu) = {\frac {\beta }{2\alpha \Gamma (1/\beta )}} \frac{1}{t(1-t)}\;e^\left\{-\left(\left\vert\mbox{logit}(t)-\mu \right\vert/\alpha \right)^{\beta }\right\}, $$
$t\in(0,1)$, $\mu\in{\mathbb R}$, $\alpha,\beta>0$.
You can call it the Generalized Logit-Normal distribution, which generalizes the logit normal distribution:
https://en.wikipedia.org/wiki/Logit-normal_distribution
