What is the relationship between two definitions of Turing-computability of a partial function? When one wants to know that whether a partial function $f \colon \Sigma^{*} \supsetneq \mathrm{dom}(f) \rightarrow \Sigma^{*}$ is Turing-computable, there are two methods that I think they are both useful.


We can define a total function $\bar{f}$ such that
$$ \bar{f}(x) = \left\{
\begin{aligned} 
&amp;f(x),&amp;x \in \mathrm{dom}(f) \\
&amp;\bot,&amp;x \not\in \mathrm{dom}(f) \\
\end{aligned}
\right.$$
Thus, $f$ is Turing-computable if $\bar{f}$ is Turing-computable by a TM using $\bar{\Sigma} = \Sigma \cup \{ \bot \}$.
(Some answers use this definition. see [1] and [2])
Let $M$ be a TM, and We say $f$ is Turing-computable by $M$ if
$$M(x) = \left\{
\begin{aligned}
&amp;f(x),&amp;&amp; x \in \mathrm{dom}(f) \\
&amp;\bot,&amp;&amp;x \not\in \mathrm{dom}(f) \\
\end{aligned}
\right.$$
where $M(x) = \bot$ means that $M$ never halts on $x$. (This definition is introduced in my books.)


These two definitions are not equivalent, since $M$ can computes some functions whose domain is Turing-recognizable instead of Turing-decidable in sense of definition 2.

I want know that is the definition 2 more powerful? In another word, if $f$ is Turing-computable in sense of definition 1, can we prove that it is also Turing-computable in sense of definition 2?
Definition 1 is incorrect. It does not define the notion of partial computable function correctly. Definition 2 is correct.

Proposition: A partial function $f$ is computable according to your definition 1 if, and only if, it is a partial computable function whose domain $\mathrm{dom}(f)$ is a computably decidable set.

Proof.
Suppose $f$ is computable according to definition 1. Then for every $n \in \mathbb{N}$, we have $n \in \mathrm{dom}(f)$ if, and only if, $\bar{f}(n) \neq \bot$. So we can decide whether $n \in \mathrm{dom}(f)$.

Conversely, suppose $f$ is a partial computable function whose domain $\mathrm{dom}(f)$ is decidable. Then $\bar{f}$ is computable because we can compute it like this: given input $n$, if $n \in \mathrm{dom}(f)$ then output $\bot$, otherwise output $f(n)$. $\Box$