Let $\lambda>-1;{\color{red}{\mu}}\in\mathbb{R};0<T\leq\infty, g(0)\not=0$. Watson's lemma is basically the ${\color{red}{\mu}}=0$ version of the following equation: \begin{equation} \int_0^{T}t^{\lambda+i{\color{red}{\mu}}}g(t)e^{-xt}\mathrm{d}t =\sum_{k=0}^{N}\frac{g^{(k)}(0)}{k!}\frac{\Gamma(\lambda+i{\color{red}{\mu}}+k+1)}{x^{\lambda+i{\color{red}{\mu}}+k+1}} +O\left(x^{-\lambda-N-2}\right),\quad\text{ as } x\to+\infty\tag{1} \end{equation}
I verified (1) by slightly modify the proof for Watson's lemma in wikipedia.org.
question (1) where can I find a reference for the proof of (1)?
question (2) how do I apply the original Watson's lemma to the following integral?
\begin{equation} \int_0^{1}(1-t)^{\nu+i{\color{red}{\mu}}}e^{-xt}\mathrm{d}t ,\quad\text{ as } x\to+\infty\tag{2} \end{equation}
If I set $\lambda=0,g(t)=(1-t)^{\nu+i{\color{red}{\mu}}}$ and use the binomial expansion formula \begin{equation} (1-t)^{\nu+i{\color{red}{\mu}}}=1-(\nu+i{\color{red}{\mu}})t+O(t^2) \qquad \text{ as } t\to 0^+\tag{3} \end{equation} then I obtain: \begin{equation} \int_0^{1}(1-t)^{\nu+i{\color{red}{\mu}}}e^{-xt}\mathrm{d}t =\frac{1}{x}-(\nu+i{\color{red}{\mu}})\frac{1}{x^2}+O(x^{-3}) \qquad \text{ as } x\to+\infty\tag{4} \end{equation}
If ${\color{red}{\mu}}>x$, then $|\nu+i{\color{red}{\mu}}|/x>1$. Thus (4) is divergent. This is the problem I encountered. How to make it convergent?
Thanks- mike