# Infinite summation : $\displaystyle S= \sum_{r=0}^{\infty} \frac{\cos (2xr)}{4r^2 - 1}$

So I ran into this problem on Quora:

$$\displaystyle S = \sum_{r=0}^{\infty} \frac{\cos (2xr)}{4r^2 - 1}$$

Initially I thought it cannot be resolved as telescoping method failed. But I tried something else.

MY ATTEMPT:

Let $$\displaystyle S = \mathbb{R} \left [ \sum_{r=0}^{\infty} \frac {e^{-\iota 2rx}}{4r^2 -1} \right ] \equiv \mathbb{R} [J], \$$ where$$\ \mathbb{R} \$$ shows real part. So let us find $$J.$$

Now, $$\displaystyle J(x) = \sum_{r=0}^{\infty} \frac {e^{-\iota 2rx}}{4r^2 -1} \implies J'(x) = \sum_{r=0}^{\infty} \frac {-\iota (2r)e^{-\iota 2rx}}{4r^2 -1} \$$ and hence $$\displaystyle J''(x) = -\sum_{r=0}^{\infty} \frac {(4r^2)e^{-\iota 2rx}}{4r^2 -1}.$$

Therefore some maipulations and we get: $$\displaystyle J'' + J = - \sum_{r=0}^{\infty} e^{-\iota 2rx} \implies J'' + J = \iota \sin(x) e^{-\iota x}.$$

We can solve this by variation of parameters:

Initial soltuions of the equation $$\ \displaystyle J'' + J = 0 \$$ are $$\displaystyle e^{\iota x} , e^{-\iota x}$$. And let $$\ \displaystyle J= C_0 e^{\iota x} \phi_1 + C_1 e^{-\iota x} \phi_2 \$$ satisfy $$\ \displaystyle J'' + J = \iota \sin(x) e^{-\iota x}. \$$

Therefrore, $$\displaystyle e^{\iota x} \phi_1' + e^{-\iota x} \phi_2' = 0 \\ e^{\iota x} \phi_1' - e^{-\iota x} \phi_2' = \sin(x) e^{-\iota x}$$

Solving for $$\ \displaystyle \phi_1 , \phi_2\$$ we get:

$$\displaystyle J = C_0 e^{\iota x}\left ( \frac{1}{6} \cos (x) + \frac{1}{3} \sin ^2(x) e^{\iota x} \right ) + \frac{C_1}{2} e^{-\iota x} \cos (x)$$

But the issue is $$\ \displaystyle C_0 , C_1 .\$$ I was only able to find that $$\ \displaystyle C_0 + 3C_1 = 3 ,\$$ using the condition :$$\displaystyle J(x=0) = J(x=\pi) = J(x=-\pi) = \frac{1}{2} = \frac{C_0}{6} + \frac{C_1}{2}$$

I was only able to conclude:

$$\displaystyle S = \mathbb{R} \left[ C_0 e^{\iota x}\left ( \frac{1}{6} \cos (x) + \frac{1}{3} \sin ^2(x) e^{\iota x} \right ) + \frac{C_1}{2} e^{-\iota x} \cos (x) \right] \\ \text{where} \ C_0 + 3C_1 = 3$$

I want anyone guide me further. Or suggest some another method to solve this.

• It is a known identity $\left|\sin x\right| = \frac{2}{\pi}-\sum_{k = 1}^\infty \frac{4}{\pi(4k^2-1)}\cos(2k x)$ (quite known in this site). The RHS is the Fourier series of LHS. Commented Mar 26, 2022 at 14:35
• @RiverLi Thank you sharing such a beautiful Identity. This just give the answer rightaway. Commented Mar 26, 2022 at 17:07

$$2\sum_{r=0}^\infty\dfrac{e^{2irx}}{4r^2-1}=\sum_{r=0}^\infty\dfrac{e^{2irx}}{2r-1}-\sum_{r=0}^\infty\dfrac{e^{2irx}}{2r+1}$$

Now $$\sum_{r=0}^\infty\dfrac{e^{2irx}}{2r-1}=-1+e^{ix}\sum_{r=1}^\infty\dfrac{e^{ix(2r-1)}}{2r-1}$$ $$=-1+e^{ix}\ln\dfrac{1+e^{ix}}{1-e^{ix}}$$ $$=-1+e^{ix}\left(i\pi+\ln\dfrac{e^{ix/2}+e^{-ix/2}}{e^{ix/2}-e^{-ix/2}}\right)\text{ using }\ln(1-y)=-\sum_{n=1}^\infty\dfrac{y^r}r$$

$$=-1+e^{ix}\left(i\pi-\dfrac{i\pi}2+\ln\cot\dfrac x2\right)$$

The real part will be $$-1+\cos x\cdot\ln\cot\dfrac x2-\dfrac\pi2\cdot\sin x$$

Similarly, calculate $$\sum_{r=0}^\infty\dfrac{e^{2irx}}{2r+1}$$

• Wow, such a great method. Thank you for sharing. Will try this. Commented Mar 11, 2022 at 14:59
• Nice and easy solution Commented Mar 11, 2022 at 15:13
• Using this I got my answer as $\ \displaystyle S = -\frac{1}{2} - \frac{\pi \sin x}{4} . \$ Seems correct as $\ \displaystyle S(x=0) = -\frac{1}{2} \$ Commented Mar 11, 2022 at 15:40

Below is presented one of the way - how the problem can be solved via the complex integration. This solution I also posted on Quora.

Let's consider

$$\displaystyle S_0=\sum_{r=1}^\infty\frac{\cos 2rx}{4r^2-1}=\frac{1}{2}\Big(\Re\sum_{r=-\infty}^\infty\frac{e^{2irx}}{4r^2-1}+1\Big)=\frac{1}{2}(\Re S+1)\tag*{}$$

To evaluate $$S=\sum_{r=-\infty}^\infty\frac{e^{2irx}}{4r^2-1}$$ we introduce the function $$g(z)=\frac{2\pi i}{e^{2\pi iz}-1}$$, which has the residues $$\operatorname{Res}g(z)=1$$ at the points $$s=0,\pm1,\pm2 ...$$

Let's consider $$\displaystyle I=\oint \frac{2\pi i}{e^{2\pi iz}-1}\frac{e^{2izx}}{4z^2-1}dz\tag*{}$$

where we impose the condition $$x\in[0;\pi]$$, and the integration is along a contour that encloses the axis $$X$$ (red contour on the picture below). It also encloses all the poles of the function (because they lie on the axis $$X$$).

Then we can deform the contour - making a circle of radius $$R\to\infty$$. Due to the requirement $$x\in[0;\pi]\,\,\, |\frac{1}{e^{2\pi iz}-1}\frac{e^{2izx}}{4z^2-1}|<|\frac{1}{4z^2-1}|\sim \frac{1}{R^2}$$ at $$R\to\infty$$, and, therefore, $$I\to0$$ at $$R\to\infty$$.

It means that $$\displaystyle I=\oint=\sum\operatorname{Res}\frac{2\pi i}{e^{2\pi iz}-1}\frac{e^{2izx}}{4z^2-1}=0\tag*{}$$ On the other hand,

$$\displaystyle \sum\operatorname{Res}\frac{2\pi i}{e^{2\pi iz}-1}\frac{e^{2izx}}{4z^2-1}=\sum\operatorname{Res}_{z=\pm1/2}\frac{2\pi i}{e^{2\pi iz}-1}\frac{e^{2izx}}{4z^2-1}+S\tag*{}$$

Therefore, $$\displaystyle S=-\operatorname{Res}_{z=\pm1/2}\frac{2\pi i}{e^{2\pi iz}-1}\frac{e^{2izx}}{4z^2-1}\tag*{}$$

The residues evaluation is straightforward:

$$\displaystyle S=-2\pi i\Big(\frac{e^{ix}}{4(e^{\pi i}-1)}+\frac{e^{-ix}}{(-4)(e^{-\pi i}-1)}\Big)=-\frac{\pi}{2}\sin x\tag*{}$$ and the sum $$S_0$$

$$\displaystyle S_0=\frac{1}{2}(\Re S+1)=\frac{1}{2}(1-\frac{\pi}{2}\sin x); \,\,x\in[0;\pi]\tag*{}$$

Then $$\sum_{r=0}^\infty\frac{\cos 2rx}{4r^2-1}=S_0-1=-\frac{1}{2}(1+\frac{\pi}{2}\sin x); \,\,x\in[0;\pi]$$

We can rewrite the answer for $$x\in[-\pi;\pi]$$ as

$$\displaystyle \boxed{\,\,\sum_{r=0}^\infty\frac{\cos 2rx}{4r^2-1}=-\frac{1}{2}\big(1+\frac{\pi}{2}|\sin x|\big); \,\,x\in[-\pi;\pi]\,\,}\tag*{}$$

• This looks cool, but will take me some time to understand it. Commented Mar 11, 2022 at 15:04
• Thanks alot for sharing it. Commented Mar 11, 2022 at 15:04

Putting it as a comment may be too long, since this sum seems to be not a trivial one. Thus I provide a closed form here, obtained by CAS (which is large even without a step-by-step solution):

$$\sum _{r=1}^{\infty} \frac{\cos (2 r x)}{4 r^2-1}=\\-\frac{1}{4} e^{-i x} \left(-2 e^{i x}+e^{2 i x} \tanh ^{-1}\left(e^{-i x}\right)-\tanh ^{-1}\left(e^{-i x}\right)-e^{2 i x} \tanh ^{-1}\left(e^{i x}\right)+\tanh ^{-1}\left(e^{i x}\right)\right)$$

The sum up to $$k$$ involves the Lerch transcendent $$\Phi$$ and looks as follows:

$$\sum _{r=1}^k \frac{\cos (2 r x)}{4 r^2-1}=\\-\frac{1}{8 (2 k+1)}e^{-4 i x} \left(-2 k \left(e^{-2 i x}\right)^{k-1} \Phi \left(e^{-2 i x},1,k+\frac{3}{2}\right)-\left(e^{-2 i x}\right)^{k-1} \Phi \left(e^{-2 i x},1,k+\frac{3}{2}\right)+2 k \left(e^{-2 i x}\right)^k \Phi \left(e^{-2 i x},1,k+\frac{3}{2}\right)+\left(e^{-2 i x}\right)^k \Phi \left(e^{-2 i x},1,k+\frac{3}{2}\right)-e^{6 i x} \left(e^{2 i x}\right)^k \Phi \left(e^{2 i x},1,k+\frac{3}{2}\right)+e^{8 i x} \left(e^{2 i x}\right)^k \Phi \left(e^{2 i x},1,k+\frac{3}{2}\right)-2 k e^{6 i x} \left(e^{2 i x}\right)^k \Phi \left(e^{2 i x},1,k+\frac{3}{2}\right)+2 k e^{8 i x} \left(e^{2 i x}\right)^k \Phi \left(e^{2 i x},1,k+\frac{3}{2}\right)+2 \left(e^{-2 i x}\right)^{k-1}+2 e^{6 i x} \left(e^{2 i x}\right)^k-8 k e^{4 i x}-4 k e^{3 i x} \tanh ^{-1}\left(e^{-i x}\right)+4 k e^{5 i x} \tanh ^{-1}\left(e^{-i x}\right)+4 k e^{3 i x} \tanh ^{-1}\left(e^{i x}\right)-4 k e^{5 i x} \tanh ^{-1}\left(e^{i x}\right)-4 e^{4 i x}-2 e^{3 i x} \tanh ^{-1}\left(e^{-i x}\right)+2 e^{5 i x} \tanh ^{-1}\left(e^{-i x}\right)+2 e^{3 i x} \tanh ^{-1}\left(e^{i x}\right)-2 e^{5 i x} \tanh ^{-1}\left(e^{i x}\right)\right)$$

To resolve this in detail (step by step), a closer look at the Lerch transcendent might give insights into the sum.

• Yes, exactly I saw the solution for infinite sum on wolfram alpha. But I do not know how to get it. Commented Mar 11, 2022 at 14:51
• Thanks for sharing 'Lerch transcendents'. Will look into it. Commented Mar 11, 2022 at 14:53
• Yes - the Lerch transcendent might be the direction to investigate. My first idea would be to take a closer look at the $k$-sum and look how it behaves by inserting very large values for $k$. Commented Mar 11, 2022 at 14:53
• Sure, would definitly try it. Commented Mar 11, 2022 at 14:54