Does this limit exist and if so what is it's value?

Question

Let $r,\sigma,\phi$ be positive real variables in: $$ \lim_{\phi\to 0} \left[ \sqrt{r^2+\sigma^2-2 r\sigma\cos(\phi)} + r\cos(\phi)\ln\left(-r\cos(\phi)+\sigma+\sqrt{r^2+\sigma^2-2 r\sigma\cos(\phi)}\right) -r-r\cos(\phi)\ln(r)-r\cos(\phi)\ln(1-cos(\phi))\right] $$ Note. Arising from this question: Could this be called Renormalization? .
Can't proceed without knowing the outcome of this limit. Please help.

Bonus. It would be even nicer if someone can calculate the accompanying integral: $$ \int_0^{2\pi}\left[ \sqrt{r^2+\sigma^2-2 r\sigma\cos(\phi)} + r\cos(\phi)\ln\left(-r\cos(\phi)+\sigma+\sqrt{r^2+\sigma^2-2 r\sigma\cos(\phi)}\right) -r-r\cos(\phi)\ln(r)-r\cos(\phi)\ln(1-cos(\phi))\right]d\phi $$ Otherwise I have to do it numerically, which is feasible anyway.

Answer 1

Most terms in the expression are completely benign, allowing us to take a limit simply by substituting $\phi = 0$. The two problematic terms can be combined as $$\lim_{\phi\to 0} \; r\cos\phi \left(\; \ln\left( - r\cos\phi + s + \sqrt{r^2+s^2-2r s\cos\phi} \right) - \ln(1 - \cos\phi ) \;\right) \qquad (\star)$$ Here, we have the danger of an "$\infty-\infty$" indeterminate form, should the argument of each log vanish. Note that $\lim_{\phi\to 0} \sqrt{r^2+s^2-2 r s \cos\phi} = |r-s|$, so that the first log's argument approaches either the (positive and finite) value $2(s-r)$ for $s > r$, or else $0$ for $r\geq s$. In the first case, the first log is finite, and the limit is therefore dominated by the second log (whose argument goes to zero), so that $(\star)$ is $+\infty$. The second case is the interesting one, so we'll assume from now on that $r \geq s$.

Ignoring the $r\cos\phi$ factor, we can re-write $(\star)$ as $$\lim_{\phi\to 0} \; \ln\frac{-r\cos\phi + s + t}{1-\cos\phi} = \lim_{\phi\to 0}\left(\;r + \frac{-r + s + t}{1-\cos\phi}\;\right) \qquad (\star\star)$$ where $t \geq 0$ abbreviates $\sqrt{r^2+s^2-2 r s\cos\phi}$, and can be interpreted as the length of the third side of a triangle whose other edges ---of lengths $r$ and $s$--- bound an angle of measure $\phi$. With that interpretation in mind, we can observe that the area, $A$, of that triangle is given by Heron's formula: $$A^2 = \frac{1}{16}(r+s+t)(-r+s+t)(r-s+t)(r+s-t)$$ and also by $$A = \frac{1}{2} r s \sin\phi$$ so that $$\begin{align} \frac{-r+s+t}{1-\cos\phi} &= \frac{(-r+s+t)}{\sin^2\phi}(1+\cos\phi) \\ &= \frac{16A^2}{(r-s+t)(r+s-t)(r+s+t)} \frac{r^2s^2}{4A^2}(1+\cos\phi) \\[6pt] &= \frac{4r^2s^2(1+\cos\phi)}{(r-s+t)(r+s-t)(r+s+t)} \end{align}$$ Recalling that $t \to r-s$ as $\phi\to 0$, we see that $(\star\star)$ becomes $$\lim_{\phi\to 0} \ln\left( r + \frac{4r^2s^2(1+\cos\phi)}{(r-s+t)(r+s-t)(r+s+t)} \right) = \ln\left( r + \frac{4r^2s^2(2)}{(2r-2s)(2s)(2r)} \right) = \ln\frac{r^2}{r-s}$$

Thus, the full original limit (assuming $r \geq s$) becomes $$( r - s ) - r - r \cdot 1 \cdot \ln r + r \cdot 1 \cdot \ln\frac{r^2}{r-s} = - s + r\ln\frac{r}{r-s}$$ with appropriate consideration of the case $r=s$.

Answer 2

Since $\cos(0)=1$, we have:

$$ \lim_{\phi\to 0} \left[ \sqrt{r^2+\sigma^2-2 r\sigma\cos(\phi)} + r\cos(\phi)\ln\left(-r\cos(\phi)+\sigma+\sqrt{r^2+\sigma^2-2 r\sigma\cos(\phi)}\right) -r-r\cos(\phi)\ln(r)-r\cos(\phi)\ln(1-\cos(\phi))\right] = \left[ |r-\sigma| + r\ln\left(-r+\sigma+|r-\sigma|\right) -r-r\ln(r)-r\ln(0)\right]=+\infty\times r $$

Thus, if $r=0$ and $\sigma>0$ then:

$$\left[ \sqrt{r^2+\sigma^2-2 r\sigma\cos(\phi)} + r\cos(\phi)\ln\left(-r\cos(\phi)+\sigma+\sqrt{r^2+\sigma^2-2 r\sigma\cos(\phi)}\right) -r-r\cos(\phi)\ln(r)-r\cos(\phi)\ln(1-\cos(\phi))\right]=\sigma-\cos\phi,\,\forall\, \phi.$$

If $r=0$ and $\sigma<0$ then:

$$\left[ \sqrt{r^2+\sigma^2-2 r\sigma\cos(\phi)} + r\cos(\phi)\ln\left(-r\cos(\phi)+\sigma+\sqrt{r^2+\sigma^2-2 r\sigma\cos(\phi)}\right) -r-r\cos(\phi)\ln(r)-r\cos(\phi)\ln(1-\cos(\phi))\right]=-\sigma,\,\forall\, \phi.$$

And if $r\neq0$ then you have to compare $r$ to $\sigma$ and conclude.