Approximating $\int_0^t f(u) e^{-\alpha (t-u)} \mathrm{d} u$ for $\alpha \gg 1$

Question

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?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}