Let $\lambda>-1;{\color{red}{\mu}}\in\mathbb{R};0 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
applying Watson's lemma to an integral
1
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asymptotics
laplace-transform
laplace-method