I need to obtain an approximation to an integral of the form:
$$I = \int_0^1 \mathrm e^{M \sum_i f_i(x_i)} \mathrm \delta\left(\sum_i x_i - 1\right) d\mathbf x$$
where $M$ is a large real number. Here the functions $f_1(x_1), f_2(x_2), \dots, f_N(x_N)$ are $N$ real functions, sufficiently smooth.
I suspect I can apply a variant of the Laplace method, but I am not sure how to proceed due to the presence of the Dirac delta. Any suggestions appreciated. Thanks.
Note: See Laplace method on a simplex for a more general version of this problem, where the integrand is not factorized.
This is not a complete answer, only my attempt to solve the problem.
Let $F_k(v)$ be the Fourier transform of $\mathrm e^{M f_k(x_k)}$, restricted to $[0,1]$,
$$F_k(v)=\int_0^1 \mathrm{e}^{-2\pi\mathrm{i}x_k v_k + Mf_k(x_k)}\mathrm{d}x_k \qquad (1)$$
By the convolution theorem
$$I = \int_{-\infty}^{\infty}\prod_{k}F_{k}\left(v\right)\mathrm{e}^{2\pi\mathrm{i}v}\mathrm{d}v$$
So far so good. To approximate $F_k(v)$, I would like to try to use the Laplace method. I'm not sure about the correctness of this approach here, but let's move forward. We make a second-order Taylor expansion:
$$f_k(x_k) \approx f_k(x_k^*) + \frac{1}{2} f_k^{''}(x_k^*)(x_k-x_k^*)^2$$
Assume that $0<x_k^*<1$ and $f_k^{''}(x_k)<0$. In the limit $M\rightarrow\infty$ we can approximate:
$$\begin{aligned} F_k(v_k) &\approx \mathrm{e}^{M f_k(x_k^*)} \int_{-\infty}^{\infty} \mathrm{e}^{-2\pi\mathrm{i} x_k v_k + \frac{M}{2} f_k^{''}(x_k^*)(x_k-x_k^*)^2}\mathrm{d}x_{k} \\ & = \sqrt{\frac{2\pi}{M |f_k^{''}(x_k^*)|}} \exp\left(-2\pi\mathrm{i}x_k^*v+Mf_k(x_k^*)+\frac{2\pi^2v^2}{Mf_k^{''}(x_k^*)}\right) \end{aligned}$$
Hence
$$\begin{aligned} I &\approx \sqrt{\frac{2\pi}{M\prod_{s}\left|f_{s}^{\prime\prime}\left(x_{s}^{*}\right)\right|}}\mathrm{e}^{M\sum_{s}f_{s}\left(x_{s}^{*}\right)}\int_{-\infty}^{\infty}\exp\left\{ -2\pi\mathrm{i}v\left(\sum_k x_k^*-1\right)+\frac{2\pi^{2}v^{2}}{M}\sum_k\frac{1}{f_k^{''}(x_k^*)}\right\} \mathrm{d}v \\ &=\frac{1}{\sqrt{\left(\prod_k|f_k^{''}(x_k^*)|\right)\left(\sum_k\frac{1}{|f_k^{''}(x_k^*)|}\right)}}\exp\left\{ M\sum_kf_k(x_k^*)+\frac{M}{2} \frac{\left(\sum_k x_k^*-1\right)^2}{\sum_k\frac{1}{f_k^{''}(x_k^*)}}\right\} \end{aligned}$$
The problem with this "solution" is that I'm not sure if it is right. How can I check?