Evaluating $\lim_{x\to0}x^{-2}\left(\frac{1}{2}\ln\cos\phi+\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\frac{\sin^{2}(nx)}{(nx)^{2}}\sin^{2}(n\phi)\right)$

Question

I am stuck at evaluating the limit$$\lim_{x\to0}x^{-2}\left(\frac{1}{2}\ln\cos\phi+\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\frac{\sin^{2}(nx)}{(nx)^{2}}\sin^{2}(n\phi)\right)$$ My attempt:$$\sin^2(nx)=\frac{1-\cos (2nx)}{2}=\frac12\sum_{k=1}^{\infty}(-1)^{k+1}\frac{(2nx)^{2k}}{(2k)!}$$Now I first evaluate:$$\begin{align} &\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\frac{\sin^{2}(nx)}{(nx)^{2}}\sin^{2}(n\phi)\\&=\frac12\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}(-1)^{k+1}\frac{(2nx)^{2k}}{(2k)!}\frac{(-1)^{n-1}}{n}\frac{1}{(nx)^{2}}\sin^{2}(n\phi)\\&=\frac12 \sum_{n,k\ge 1}(-1)^{k+1}(-1)^{n-1}\frac{(2n)^{2k}}{n^3(2k)!} x^{2(k-1)}\sin ^2 (n\phi)\end{align}$$The last formula above should be something like:$$-\frac12\ln \cos \phi+cx^2+o(x^2)$$Where $c$ is a constant.But I find it hard to find what the sum is.

Any help would be appreciated.

Answer 1

Let the expression inside the limit be $F(x, \phi)$. We want to evaluate $L = \lim_{x\to0} \frac{F(x, \phi)}{x^2}$. Since $F(x, \phi)$ is an even function of $x$, its Taylor series around $x=0$ is of the form $F(x, \phi) = F(0, \phi) + \frac{F''(0, \phi)}{2}x^2 + O(x^4)$. The limit is then $L = \frac{1}{2}F''(0, \phi)$, provided that $F(0, \phi)=0$.

The Taylor expansion of the term in the sum for small $x$ is given by $$ \frac{\sin^2(nx)}{(nx)^2} = \frac{(nx - (nx)^3/6 + \dots)^2}{(nx)^2} = \frac{(nx)^2 - (nx)^4/3 + O(x^6)}{(nx)^2} = 1 - \frac{n^2 x^2}{3} + O(x^4) $$ Substitute this into the expression for $F(x, \phi)$ $$ F(x, \phi) = \frac{1}{2}\ln\cos\phi + \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\sin^{2}(n\phi)\left(1 - \frac{n^2x^2}{3} + O(x^4)\right) $$ Further separating the terms by powers of $x$ gives $$ F(x, \phi) = \underbrace{\left(\frac{1}{2}\ln\cos\phi + \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\sin^{2}(n\phi)\right)}_{F(0, \phi)} - \frac{x^2}{3}\underbrace{\sum_{n=1}^{\infty}(-1)^{n-1}n\sin^{2}(n\phi)}_{S} + O(x^4) $$

Since $\sin^2\theta = \frac{1-\cos(2\theta)}{2}$, we write $$ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\sin^{2}(n\phi) = \frac{1}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} - \frac{1}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\cos(2n\phi)}{n} $$ Using the fact that $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}y^n}{n}=\ln(1+y)$ and its real part for $y=e^{i2\phi}$:

• $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} = \ln(2)$
• $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\cos(2n\phi)}{n} = \text{Re}[\ln(1+e^{i2\phi})] = \ln|1+e^{i2\phi}| = \ln(2\cos\phi)$ Thus, the sum is $\frac{1}{2}(\ln 2 - \ln(2\cos\phi)) = -\frac{1}{2}\ln\cos\phi$. So, $F(0, \phi) = \frac{1}{2}\ln\cos\phi - \frac{1}{2}\ln\cos\phi = 0$.

From the initial expansion, the limit is: $$ L = -\frac{1}{3} S = -\frac{1}{3}\sum_{n=1}^{\infty}(-1)^{n-1}n\sin^{2}(n\phi) $$ To evaluate the sum $S$, we again use $\sin^2(n\phi) = \frac{1-\cos(2n\phi)}{2}$: $$ S = \frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1}n - \frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1}n\cos(2n\phi) $$ These divergent series are interpreted via Abel summation, using the series $\sum_{n=1}^{\infty}n z^n = \frac{z}{(1-z)^2}$:

• $\sum_{n=1}^{\infty}(-1)^{n-1}n = \lim_{y\to 1^-} \sum (-1)^{n-1}ny^n = \lim_{y\to 1^-} \frac{y}{(1+y)^2} = \frac{1}{4}$.
• $\sum_{n=1}^{\infty}(-1)^{n-1}n\cos(2n\phi) = \text{Re}\left[\sum_{n=1}^{\infty}(-1)^{n-1}n(e^{i2\phi})^n\right] = \text{Re}\left[\frac{e^{i2\phi}}{(1+e^{i2\phi})^2}\right]$. Since $1+e^{i2\phi} = 2\cos\phi\,e^{i\phi}$, the term is $\text{Re}\left[\frac{e^{i2\phi}}{(2\cos\phi\,e^{i\phi})^2}\right] = \text{Re}\left[\frac{e^{i2\phi}}{4\cos^2\phi\,e^{i2\phi}}\right] = \frac{1}{4\cos^2\phi}$.

Substituting these results back into the expression for $S$: $$ S = \frac{1}{2}\left(\frac{1}{4}\right) - \frac{1}{2}\left(\frac{1}{4\cos^2\phi}\right) = \frac{1}{8}\left(1 - \frac{1}{\cos^2\phi}\right) = \frac{1}{8}\frac{\cos^2\phi-1}{\cos^2\phi} = -\frac{\sin^2\phi}{8\cos^2\phi} = -\frac{1}{8}\tan^2\phi $$ The limit is $L = -\frac{1}{3}S$: $$ L = -\frac{1}{3} \left(-\frac{1}{8}\tan^2\phi\right) = \frac{1}{24}\tan^2\phi $$

Answer 2

We can also use complex integration to find full asymptotics at $x\to0$.

Let's denote $$S(\phi)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\frac{\sin^{2}(nx)}{(nx)^{2}}\sin^{2}(n\phi)$$ and consider the integral in the complex plane along the following rectangular contour $C$, where $N=1, 2, 3...\to\infty\,$ and $\,2(x+\phi)\leqslant \pi\quad\tag{a}$

$$$$

$$$$

$$I=-\frac1{2\pi i}\oint_C\frac\pi{\sin\pi z}\frac{\sin^2(zx)\sin^2(z\phi)}{x^2z^3}dz=-\underset{z=0,1,2,...}{\operatorname{Res}}\frac\pi{\sin\pi z}\frac{\sin^2(zx)\sin^2(z\phi)}{x^2z^3}=S(\phi)$$ On the other hand, it is starightforward to see the the integrals along the path $2,\,3,\,4$ tend to zero at $N\to\infty$ (if the condition (a) above is met). It means that $$S(\phi)=-\frac1{2\pi i}\int_{\infty}^{-\infty}\frac\pi{\sin(\pi i t)}\frac{\sin^2(itx)\sin^2(it\phi)}{x^2(it)^3}idt$$ $$=\frac1{8x^2}\int_{-\infty}^\infty\frac{\left(\cosh(2tx)-1\right)\left(\cosh(2t\phi)-1\right)}{t^3\sinh(\pi t)}dt$$ Decomposing $\cosh(2tx)-1$ near $x=0$ (i.e. dropping exponentially small terms) $$S(\phi)\sim\frac14\int_{-\infty}^\infty\frac{\cosh(2t\phi)-1}{t\sinh(\pi t)}dt+\sum_{k=2}^\infty\frac{4^{k-1}x^{2k-2}}{2(2k)!}\int_{-\infty}^\infty\frac{\cosh(2t\phi)-1}{\sinh(\pi t)}t^{2k-3}dt$$ I leave to those who are interested to show that $$\frac14\int_{-\infty}^\infty\frac{\cosh(2t\phi)-1}{t\sinh(\pi t)}dt=\frac12\int_{-\infty}^\infty\frac{\sinh^2(t\phi)}{t\sinh(\pi t)}dt=-\frac12\ln(\cos\phi)$$ (see, for example, here or here )

Using also for odd $n$ $$J(n)=\int_{-\infty}^\infty\frac{\cosh(2t\phi)-1}{\sinh(\pi t)}t^ndt$$ $$=\frac1{2^n}\frac{\partial^n}{\partial\phi^n}\int_{-\infty}^\infty\frac{e^{2t\phi}}{\sinh(\pi t)}dt-\frac1{2^n}\frac{\partial^n}{\partial\phi^n}\bigg|_{\phi=0}\int_{-\infty}^\infty\frac{e^{2t\phi}}{\sinh(\pi t)}dt$$ and the value of the integral (the proof, for example, by means of complex integration is straightforward) $$\int_{-\infty}^\infty\frac{e^{2t\phi}}{\sinh(\pi t)}dt=\tan\phi$$ after some transformations we get the answer $$S(\phi)\sim-\frac12\ln(\cos\phi)+\sum_{k=2}^\infty\frac{\tan^{(2k-3)}(\phi)-\tan^{(2k-3)}(0)}{(2k)!}x^{2k-2}$$ $$=-\frac12\ln(\cos\phi)+\left(\frac1{\cos^2\phi}-1\right)\frac{x^2}{4!}+\frac{6-4\cos^2\phi-2\cos^4\phi}{\cos^4\phi}\,\frac{x^4}{6!}+O\big(x^6\big)$$ Numeric check (WA) confirms the answer.