Do we have step by step proof for this expression which is concerned with the expectation of an exponential function with a variable (R) that satisfies uniform distribution. I have been trying to understand how it was derived but its seems obscure to me up till now:
$\frac{\lambda_1}{2}E_R (e^{-B_{1}R^2})E_R (e^{-D_{1}R^2}) = \\ \frac{\lambda_1 \pi}{2\sqrt{B_{1}D_{1}}(R_{max}-R_{min})^2}[\phi (2\sqrt{B_{1}}R_{max})-\phi (2\sqrt{B_{1}}R_{min})][\phi (2\sqrt{D_{1}}R_{max})-\phi (2\sqrt{D_{1}}R_{min})]$
Where $\phi$ is the cdf of standard normal distribution
R satisfies uniform distribution in $ [R_{min}, R_{max}]$ $(0\leq R_{min} < R_{max})$
Consider the general form of the following expectation, where $R$ has a uniform probability density funcion $f$ on the interval $(a,b)$: $$\begin{align}E[e^{-c\,R^2}]&=\int_{-\infty}^\infty e^{-c\,r^2}\,f(r)\,dr\\ \\ &=\int_{-\infty}^\infty e^{-c\,r^2}\frac{1}{b-a}[a<r<b]\,dr\\ \\ &=\frac{1}{b-a}\int_{a}^b e^{-c\,r^2}\,dr\\ \\ &=\frac{1}{b-a}\left(\int_{-\infty}^b e^{-c\,r^2}\,dr\ -\ \int_{-\infty}^a e^{-c\,r^2}\,dr\right)\tag{1} \end{align}$$ Now condider the definition of the standard normal CDF: $$\Phi(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^x e^{-\frac{1}{2}t^2}\,dt.$$ Letting $r=\frac{1}{\sqrt{2c}}\,t$, so $c\,r^2=\frac{1}{2}t^2$ and $dr=\frac{1}{\sqrt{2c}}dt$, we have: $$\begin{align}\int_{-\infty}^a e^{-c\,r^2}\,dr= \int_{-\infty}^{\sqrt{2c}\,a}e^{-\frac{1}{2}t^2}\,\frac{1}{\sqrt{2c}}dt=\sqrt{\frac{\pi}{c}}\,\Phi(\sqrt{2c}\,a) \end{align}$$ Substituting this, and the similar result involving $b$, into (1) gives the following: $$\begin{align}E[e^{-c\,R^2}]=\frac{1}{b-a}\sqrt{\frac{\pi}{c}}\bigg( \Phi(\sqrt{2c}\,b)\ -\ \Phi(\sqrt{2c}\,a)\bigg)\end{align}$$ Applying this formula to each of the expectations in your question will give the desired result. (Note that the factor of $\frac{\lambda_1}{2}$ is superfluous, as it appears on both sides of your equation.)