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?
Laplace’s method for integration
approximate integrationasymptoticslaplace-method
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1.) Yes, but the result is false. Correctly: $$ \sqrt{\frac{2\pi}M} \cdot e^{M g(x_0)} \approx \int \sqrt{-g''(x)}\cdot e^{Mg(x)}\,\mathrm{d}x $$ Which is something, but it is not trivial for me how an approximation for ${g(x_0)}$ follows. You have to be careful with the estimations.
2.) 3.) 4.) I think that it doesn't work for those cases. Here is why.
The reason the Laplace's method works is that $\lvert e^x\rvert$ is very small when $\mathbb{R}\ni x\to-\infty$. If $g(x)$ has a global maximum at $x_0$ then $e^{M g(x_0)}\cdot \int \frac{e^{M g(x)}}{e^{M g(x_0)}}$ uses the property that the integrand is small everywhere except near the global maximum.
So the real, global maximum is required (however multivariate generalizations exist).
Note that in practice, it is harder to evaluate the integral. On that I advise statistical physics papers, for example: http://www3.eng.cam.ac.uk/research_db/publications/gc121
At the end you might find that if you have enough points to approximate the integral then you have enough points to take just the maximum. So the Laplace's method is superfluous.
I assume that $\ln$ denotes the principal value of the logarithm. $I$ can be evaluated in closed form in terms of two ${_1 \hspace {-1px} F_1}$ functions, but getting the asymptotics for those is not easy. Instead, we can apply the steepest descent method to the integral directly.
Assume $\epsilon > 0$ and $\lambda > 0$. The stationary points of $\phi(m) = -m^2 - \ln(\lambda + i \epsilon - 2 m)/2$ are at $$m_{1, 2} = \frac {\lambda + i \epsilon \pm \sqrt {(\lambda + i \epsilon)^2 - 4}} 4.$$ The level curves of $\operatorname {Re} \phi$ look like this (the values increase from blue to magenta to red):
Because of the branch cut, we cannot deform the contour $(-\infty, \infty)$ to go through $m_1$ while avoiding the red regions. Therefore, we need to take a contour that goes through $m_2$ in the direction from left to right. This corresponds to $m = m_2 + \xi \sqrt {-2/\phi''(m_2)}$ near the saddle point $m_2$. Then $$I \sim \sqrt {-\frac 2 {\phi''(m_2)}} \int_{\mathbb R} e^{N (\phi(m_2) - \xi^2)} d\xi = \sqrt {-\frac {2 \pi} {N \phi''(m_2)}} \, e^{N \phi(m_2)}, \quad N \to \infty.$$
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}):