Laplace's method for integration I am trying to compute $$I(\lambda) = \int_{0}^1 \frac{x}{\sqrt{1+x^4}}e^{\lambda x}\mathrm{d}x $$ for large, real, positive $\lambda$. I'm attempting this with Laplace's method as suggested, however I fail to see how this would work considering $f(x) = x $ has no maximum on the interior of the interval $(0,1)$ and $f''(x) = 0$, which would be problematic in the formula for Laplace's method. Am I missing a small substitution trick here, or is the question poorly written?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\on{I}\pars{\lambda} & \equiv
\bbox[5px,#ffd]{\int_{0}^{1}
{x \over \root{1 + x^{4}}}\expo{\lambda x}\,\dd x}
\label{1}\tag{1}
\\[5mm] \stackrel{x\ \mapsto\ 1 - x}{=}\,\,\,&
\int_{0}^{1}
{1 - x \over \root{1 + \pars{1 - x}^{4}}}
\expo{\lambda\pars{1 - x}}\,\dd x
\\[5mm] = &\
\expo{\lambda}\int_{0}^{1}
\expo{-\lambda x}{1 - x \over
\root{1 + \pars{1 - x}^{4}}}\,\dd x
\\[5mm] \stackrel{\color{red}{\mrm{as}\ \lambda\ \to\ \infty}}{\sim}
\,\,\,&
\expo{\lambda}\int_{0}^{\infty}
\expo{-\lambda x}{1 - 0 \over
\root{1 + \pars{1 - 0}^{4}}}\,\dd x
\\[5mm] = &\
\bbx{{\root{2} \over 2}\,{\expo{\lambda} \over \lambda}}\label{2}\tag{2} \\ &
\end{align}
The $\ds{\color{darkblue}{blue}}$ one is the exact expression (\ref{1}) while the $\ds{\color{red}{red}}$ one is the asymptotic expression (\ref{2}):

