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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?

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    $\begingroup$ The main contribution comes from $x \lesssim 1$. Change $x \mapsto 1 - x$. $\endgroup$ – Felix Marin Oct 22 '20 at 23:45
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}):

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  • $\begingroup$ I believe we can apply Watson's lemma here to get more terms. en.wikipedia.org/wiki/Watson%27s_lemma $\endgroup$ – Alexandru Ionut Oct 23 '20 at 3:01
  • $\begingroup$ @AlexandruIonut Yes. Of course. It can be improved. I just got the "leading term". Thanks. $\endgroup$ – Felix Marin Oct 23 '20 at 3:05
  • $\begingroup$ Thanks! I split the final integral into 2 parts and used Watson's lemma on both and got the next term to be $-\frac{\sqrt{2}}{2}\frac{e^\lambda}{\lambda^{-3}}$ $\endgroup$ – logistress Oct 23 '20 at 12:08

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