How to prove the integral converges? 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.
 A: 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.
