I have the following exponential function
$$f(t)= e^{-(\theta/t)^{\beta}-c\frac{\Gamma\Big(\alpha,\big(\frac{\theta}{t}\big)^\beta\Big)}{\Gamma(\alpha)}} $$
where $c,\ \alpha,\ \theta,$ and $\beta$ are constants. $\Gamma(,)$ and $\Gamma()$ are the upper incomplete gamma function and the gamma function, respectively. They are given by
$$
\Gamma\bigg(\alpha,\Big(\frac{\theta}{t}\Big)^\beta\bigg)=\int_{(\theta/t)^\beta}^\infty t^{\alpha-1}e^{-t}dt\\ \Gamma(\alpha)=\int_0^\infty t^{\alpha-1}e^{-t}dt
$$
I need to take the logarithm of $f(t)$. EDIT: But I have the following questions:
1. How does one perform the logarithm on the fractional part?
2. If $f(t)$ is a distribution function such that $t_1,t_2,…,t_n$ be an independent random sample from the distribution, how can we perform the logarithm on $f(t)$.
Taking these questions into account, should the logarithm of $f(t)$ be
$$
\ln\Big(f(t)\Big)=-\sum_i^n(\theta/t_i)^\beta-nc\Bigg[\frac{\sum_i^n\Gamma\Big(\alpha,\big(\theta/t_i\big)^\beta\Big)}{\sum_i^n\Gamma_i(\alpha)}\Bigg]
$$
Or should it be
$$
\ln\Big(f(t)\Big)=-\sum_i^n(\theta/t_i)^\beta-nc-\sum_i^n\Gamma\Big(\alpha,\big(\theta/t_i\big)^\beta\Big)+\sum_i^n\Gamma_i(\alpha)
$$
Also, should $\Gamma(,)$ and $\Gamma()$ also be transformed to the logarithm. If so, applying logarithm yields
$$
\ln\Bigg(\Gamma\bigg(\alpha,\Big(\frac{\theta}{t}\Big)^\beta\bigg)\Bigg)=\int_{(\theta/t)^\beta}^\infty -(\alpha-1)t^2dt\\ \Gamma(\alpha)=\int_0^\infty -(\alpha-1)t^2dt
$$
Are these correct? Should $\int$ sign be transformed to $\prod$ sign?
Any help would be appreciated.
Thanks in advance.
Best Answer
It is always true that $$\ln(\exp(g))=g$$ no matter how complex an expression $g$ may be. There is no need to take the logarithm of $g$, or transform products into sums, or any other such thing: the logarithm undoes the exponential, no more, no less.
So $$f(t)= e^{-(\theta/t)^{\beta}-c\frac{\Gamma\Big(\alpha,\big(\frac{\theta}{t}\big)^\beta\Big)}{\Gamma(\alpha)}} $$
$$\ln(f(t))= -(\theta/t)^{\beta}-c\frac{\Gamma\Big(\alpha,\big(\frac{\theta}{t}\big)^\beta\Big)}{\Gamma(\alpha)} $$
Aside: in a complex context, it is true that $$\exp(x)=y$$ has solutions other than $x=\ln(y)$, those being $x=\ln(y)+ n i 2 \pi$, where $n$ is an integer and $i$ the imaginary unit, so some care is occasionally needed.