I want to approximate a integral given like $$\langle x\rangle=\int\limits_0^1 \mathrm dxx \int\limits_0^1\mathrm dy f(y,x)g(y)$$ And $f(y,x)$ is $$f(x,y)=\frac m{h(x)}\exp\Bigg(L\int\limits_0^x \mathrm dz \ln\bigg(1-\frac m{h(z)}\bigg)\Bigg)\Theta(u-x)$$ One good thing which could make problem doable is, $L$ is large. Given $L$ is large, then once change order of integration and use the Laplace's Method to approximate the integral. Functions $g \ge 0,h>0$ are smooth with well defined inverse and bounded. Minima of $h$ is $m\equiv\min\limits_{0\le z\le y} h(z)$ and $u=h^{-1}(m)$. If things get complicated one can choose $$h(z)=\frac1 {1+bz-\frac{z^2}{2}}\quad\mathrm{and}\quad g(y)=1.$$
With such a choice $$m=\begin{cases}h(y)&y\le b\\ h(b) & b<y \end{cases}\quad\mathrm{and}\quad u=\begin{cases}y&y\le b\\ b & b<y \end{cases}$$
An Attempt
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Integral can be divided into two parts $${\cal I}_1=\int\limits_0^1 \mathrm dxx \int\limits_0^b\mathrm dy f(y,x)g(y)$$ and $${\cal I}_2=\int\limits_0^1 \mathrm dxx \int\limits_b^1\mathrm dy f(y,x)g(y)$$ Firts let's evaluate ${\cal I}_2$ \begin{align} {\cal I}_2=&\int\limits_0^1 \mathrm dxx \int\limits_b^1\mathrm dy f(y,x)g(y)\\ =&\int\limits_0^1 \mathrm dxx \int\limits_b^1\mathrm dy g(y)\frac{h(b)}{h(x)}e^{L\int\limits_0^x\mathrm dz\ln\big(1-\frac{h(b)}{h(z)}\big)}\Theta(b-x)\\ =&\int\limits_b^1\mathrm dyg(y)\int\limits_b^1\mathrm dxx\frac{h(b)}{h(x)}e^{L\int\limits_0^x\mathrm dz\ln\big(1-\frac{h(b)}{h(z)}\big)}\\ \approx&\int\limits_b^1\mathrm dyg(y)be^{L\int\limits_0^b\mathrm dz\ln\big(1-\frac{h(b)}{h(z)}\big)} \end{align} To get last expression we have used Laplace method knowing function in exponential is decreasing function of $x$. I hope the approximation done above is fine!? Let's shoot for ${\cal I}_2$ \begin{align}{\cal I}_1=&\int\limits_0^1 \mathrm dxx \int\limits_0^b\mathrm dy f(y,x)g(y)\\ =&\int\limits_0^1 \mathrm dxx \int\limits_0^b\mathrm dy g(y)\frac{h(y)}{h(x)}e^{L\int\limits_0^x\mathrm dz\ln\big(1-\frac{h(y)}{h(z)}\big)}\Theta(y-x) \end{align} Now, we exchange the order of integral \begin{align}{\cal I}_1\stackrel{?}{=}&\int\limits_0^b \mathrm dyg(y)h(y) \int\limits_0^y\mathrm dx \frac x{h(x)}e^{L\int\limits_0^x\mathrm dz\ln\big(1-\frac{h(y)}{h(z)}\big)}\\ \approx&0 \end{align}