TITLE: Approximating $\int_0^t f(u) e^{-\alpha (t-u)} \mathrm{d} u$ for $\alpha \gg 1$
QUESTION [0 upvotes]: I am trying to obtain an approximation for the integral
\begin{equation}
 \int_0^t f(u) e^{-\alpha (t-u)} \mathrm{d} u
\end{equation}
when $\alpha \gg 1$. Intuitively, for smooth enough $f(u)$ the expression must be dominated by the endpoint $u=t$ since other contributions are exponentially vanishing. My question is that whether there is a general method (similar to saddle-point approach) for approximating this integral?

REPLY [0 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{t}\on{f}\pars{u}
\expo{-\alpha\pars{t - u}}\,\,\dd u} =
\int_{0}^{t}\on{f}\pars{t - u}
\expo{-\alpha u}\,\,\dd u
\\[5mm] \stackrel{\mrm{as}\ \alpha\ \to\ \infty}{\sim}
\,\,\,&
\int_{0}^{\infty}\on{f}\pars{t}
\expo{-\alpha u}\,\,\dd u =
\bbx{{1 \over \alpha}\,\on{f}\pars{t}} \\ &
\end{align}
{"set_name": "stack_exchange", "score": 0, "question_id": 3880657}