Showing $\lim_{\nu\to\infty}\ln(\coth(\frac\nu2))\to2e^{-\nu}$ and $ \lim_{\nu\to0}\ln(\coth(\frac\nu2))\to-\ln(\frac\nu2)$

Question

I am struggling to derive a couple limits I have come across in a paper I am reading. Both involve the natural log of the hyperbolic cotangent. The paper seems to be saying the two terms trend the same, or are equivalent, as nu gets large, or small.

$$ \lim_{\nu \rightarrow \infty }\ln\left(\coth\left(\frac{\nu }{2}\right)\right)\rightarrow 2e^{-\nu} $$

$$ \lim_{\nu \rightarrow 0}\ln\left(\coth\left(\frac{\nu }{2}\right)\right)\rightarrow -\ln\left(\frac{\nu }{2}\right) $$

I should mention these are not explicitly stated as limit formulas in the paper. They are evaluating a boundary term after integrating by parts. The final expression is obviously convenient for their purposes.

This is the paper for anyone wishing more context. See top, right of page 2.

"Kramers-Kronig, Bode, and the meaning of zero" (arXiv link) by John Bechhoefer

Thanks

Answer 1

These results are called asymptotics, so I will use some notation that is normal for that kind of analysis. For the first one, we know that $\coth x \to 1$ when $x>>1$ so rewrite the log expression as

$$\log\left(\frac{e^{\frac{\nu}{2}}+e^{-\frac{\nu}{2}}}{e^{\frac{\nu}{2}}-e^{-\frac{\nu}{2}}}\right) = \log\left(\frac{e^{\frac{\nu}{2}}-e^{-\frac{\nu}{2}}+2e^{-\frac{\nu}{2}}}{e^{\frac{\nu}{2}}-e^{-\frac{\nu}{2}}}\right) = \log\left(1+\frac{2e^{-\frac{\nu}{2}}}{e^{\frac{\nu}{2}}-e^{-\frac{\nu}{2}}}\right)$$

Since $\nu$ is very large, in the denominator we have $e^{\frac{\nu}{2}}-e^{-\frac{\nu}{2}} \approx e^{\frac{\nu}{2}}$. Continuing on we get

$$\log\left(1+\frac{2e^{-\frac{\nu}{2}}}{e^{\frac{\nu}{2}}-e^{-\frac{\nu}{2}}}\right) \approx \log\left(1+2e^{-\nu}\right) = 2e^{-\nu}+O\left(e^{-2\nu}\right)$$

by Taylor series. The $O$ means the largest extra term that differentiates the true values of the two expressions is only as big as $e^{-2\nu}$ times a constant.

For the second one, use small angle approximation

$$\log \coth\left(\frac{\nu}{2}\right) = -\log \tanh\left(\frac{\nu}{2}\right) \approx -\log \frac{\nu}{2}$$

since $\tanh x \approx x$ for $x<<1$

Answer 2

Hint:

For the first limit, rewrite the argument of the log as $$ \coth(\frac{\nu }{2})=\frac{1+\mathrm e ^{-\nu}}{1-\mathrm e ^{-\nu}},$$ and use that near $u=0$, one has $$\ln(1+u)=u+o(u),\qquad \ln(1-u)=-u+o(u).$$

For the second limit, add the expansion at order $1$ of $\;\mathrm e^{-\nu}=1-\nu+o(\nu)$.

Answer 3

For the second one, use $$ \mathop {\lim }\limits_{x \to 0} x\coth x = 1. $$ For the first one, use $$ \coth x = \frac{{\cosh x}}{{\sinh x}} = \frac{{1 + e^{ - 2x} }}{{1 - e^{ - 2x} }} = (1 + e^{ - 2x} )(1 + e^{ - 2x} +\! \mathcal{O}(e^{ - 4x} )) = 1 + 2e^{ - 2x} +\!\mathcal{ O}(e^{ - 4x} ) $$ as $x\to+\infty$ and $$ \mathop {\lim }\limits_{x \to 0} \frac{{\log (1 + x)}}{x}. $$