Let $(\mathbf M'.\hat{\mathbf n})$ and $f(R,\theta)$ be a continuous function of $R$ and let $f(0,\theta)=0$. Then how shall we prove the following improper integral converges:
$\displaystyle\lim \limits_{a \to 0} \int^{2 \pi}_0\int^b_a (\mathbf M'.\hat{\mathbf n})(\hat{\mathbf r})\ \sqrt{{f_x}^2+{f_y}^2+1}\ \dfrac{1}{R^2+[f(R,\theta)]^2}\ R\ dR\ d\theta \tag 1$
Edit:
It has been said in an answer to use the following inequality:
$0 \leq \frac{R}{R^2 + f^2} \leq \left( \frac{R}{R^2} = \frac{1}{R} \right)$
However I get my integral as diverging:
$\displaystyle\lim \limits_{a \to 0} \int^b_a \dfrac{dR}{R}=-\lim \limits_{a \to 0} \left[ \ln |R| \right]^b_a=\lim \limits_{a \to 0} \left( \ln \dfrac{ |a|}{|b|} \right)=\ln (0)=??$
As I am getting $\ln (0)$, there might be something wrong in my above calculation.
Please show where am I wrong and also how the above improper integral $(1)$ converges.
Let
$$ I = \int^{2 \pi}_0 \int^b_0 \dfrac{(\mathbf M'.\hat{\mathbf n})(\hat{\mathbf r}) \sqrt{{f_x}^2+{f_y}^2+1}}{R^2+[f(R,\theta)]^2} R \,\mathrm{d}R \, \mathrm{d}\theta $$
We may take $(\mathbf M'.\hat{\mathbf n})(\hat{\mathbf r}) = 1$ (because it has no $R$ dependence -- so we may bound it by its min and max on the unit sphere). Since we require $f$ is continuously differentiable in $x$ and $y$, at the coordinate singularity $(r = x = y = 0)$, we must have $f_x = f_y = 0$ and for any $\varepsilon > 0$, thought of as small, we may find a disk centered at $(x,y) = (0,0)$ of radius $D$, on which ($f$ is nearly flat) $$ 1 \leq \sqrt{{f_x}^2+{f_y}^2+1} < 1 + \varepsilon \text{.} $$ On this disk, necessarily, $|f(R,\theta)| < \varepsilon D$. So the integrand of $I$ lies in the interval $$ \left[ \frac{R}{R^2 + \varepsilon^2 D^2}, \frac{(1+\varepsilon)R}{R^2 + \varepsilon^2 D^2} \right) \text{.} $$
The integral on the interval $[0,D]$ of the lower bound is $\frac{1}{2} \ln(1 + 1/\varepsilon^2)$and of the upper bound is $\frac{1}{2} (1+\varepsilon) \ln(1 + 1/\varepsilon^2)$, both of which diverge to $\infty$ as $\varepsilon \rightarrow 0$. So $I$ diverges to $\infty$.
As long as $f$ has to be approximately flat in a neighborhood of the origin, i.e., has to have continuous partial derivatives near the coordinate singularity, this is unavoidable. This condition requires, on a neighborhood of the origin, $f(R,\theta) \sim 0$, $f_x^2 + f_y^2 \sim 0$, so your integral is morally $R/R^2$ in a neighborhood of the origin. To prevent this behaviour, you would need either $f_x^2 + f_y^2 \sim R^\alpha$ for $\alpha > 0$ or $f(R,\theta) \sim R^\alpha$ for $\alpha < 1$ (and not have inadvertently arranged for the one to cancel a similar change in the other). In either case, $f$ has a nondifferentiable point at the origin and you have arranged to get a different power of $R$ behaviour to prevent divergence.