INFINITE SUM 1/(n^4+4x^4)

Today, we will proof the result that appears here, namely:

$\sum_{n=-\infty}^{\infty}\frac{1}{n^4+4x^4}=\frac{\pi }{4x^3}\left(\frac{\sinh\left(2 \pi x\right)+\sin\left(2 \pi x\right)}{\cosh\left(2 \pi x\right)-\cos\left(2 \pi x\right)}\right)}$

Lets start by evaluating an easier infinte sum that is related to our goal Sum.

$\begin{aligned} \sum_{n=1}^{\infty}\frac{1}{n^4-x^4}&=\sum_{n=1}^{\infty}\frac{1}{(n^2-x^2)(n^2+x^2)}\\ &=\frac{1}{2x^2}\sum_{n=1}^{\infty}\left(\frac{1}{n^2-x^2}-\frac{1}{n^2+x^2}\right)\\ &=\frac{1}{2x^2}\left( \frac{1}{2x^2}-\frac{\pi \cot(\pi x)}{2x}+\frac{1}{2x^2}-\frac{\pi \coth(\pi x)}{2x}\right)\\ &=\frac{1}{2x^4}-\frac{\pi \coth(\pi x)}{4x^3}-\frac{\pi \cot(\pi x)}{4x^3}\\ &=\frac{2-\pi x(\cot(\pi x)+\coth(\pi x))}{4x^4} \end{aligned}$

$\boxed{\sum_{n=1}^{\infty}\frac{1}{n^4-x^4}=\frac{2-\pi x(\cot(\pi x)+\coth(\pi x))}{4x^4}}$ (1)

If we let $x \mapsto \sqrt{2}xe^{\frac{i \pi}{4}}$ in (1) we get

$\begin{aligned} \sum_{n=1}^{\infty}\frac{1}{n^4-\left(\sqrt{2}xe^{\frac{i \pi}{4}}\right)^4}&=\frac{2-\pi \sqrt{2}xe^{\frac{i \pi}{4}}\left(\cot\left(\pi \sqrt{2}xe^{\frac{i \pi}{4}}\right)+\coth\left(\pi\sqrt{2}xe^{\frac{i \pi}{4}}\right)\right)}{4\left(\sqrt{2}xe^{\frac{i \pi}{4}}\right)^4}\\ \sum_{n=1}^{\infty}\frac{1}{n^4+4x^4}&=\frac{2-\pi \sqrt{2}xe^{\frac{i \pi}{4}}\left(\cot\left(\pi \sqrt{2}xe^{\frac{i \pi}{4}}\right)+\coth\left(\pi\sqrt{2}xe^{\frac{i \pi}{4}}\right)\right)}{-16x^4}\\ &=-\frac{1}{8x^4}+\frac{\pi (1+i)}{16x^3}\left(\cot\left(\pi x(1+i)}\right)+\coth\left(\pi x(1+i)\right)\right) \end{aligned}$

Applying (A.7) and (A.8) we obtain

$\begin{aligned} &=-\frac{1}{8x^4}+\frac{\pi (1+i)(1-i)}{16x^3}\left(\frac{\sinh\left(2 \pi x\right)+\sin\left(2 \pi x\right)}{\cosh\left(2 \pi x\right)-\cos\left(2 \pi x\right)}\right)\\ &=-\frac{1}{8x^4}+\frac{\pi }{8x^3}\left(\frac{\sinh\left(2 \pi x\right)+\sin\left(2 \pi x\right)}{\cosh\left(2 \pi x\right)-\cos\left(2 \pi x\right)}\right)\\ \end{aligned}$

$\boxed{\sum_{n=1}^{\infty}\frac{1}{n^4+4x^4}=-\frac{1}{8x^4}+\frac{\pi }{8x^3}\left(\frac{\sinh\left(2 \pi x\right)+\sin\left(2 \pi x\right)}{\cosh\left(2 \pi x\right)-\cos\left(2 \pi x\right)}\right)}$ (2)

Now, observe the following

$\begin{aligned} \sum_{n=-\infty}^{\infty}\frac{1}{n^4+4x^4}&=\sum_{n=-\infty}^{-1}\frac{1}{n^4+4x^4}+\frac{1}{4x^4}+\sum_{n=1}^{\infty}\frac{1}{n^4+4x^4} \qquad (n \mapsto -n \, \text{in the first sum})\\ &=\sum_{n=\infty}^{1}\frac{1}{n^4+4x^4}+\frac{1}{4x^4}+\sum_{n=1}^{\infty}\frac{1}{n^4+4x^4} \\ &=\frac{1}{4x^4}+2\sum_{n=1}^{\infty}\frac{1}{n^4+4x^4} \\ \end{aligned}$

Using result (2)

$\begin{aligned} &=\frac{1}{4x^4}-\frac{1}{4x^4}+\frac{\pi }{4x^3}\left(\frac{\sinh\left(2 \pi x\right)+\sin\left(2 \pi x\right)}{\cosh\left(2 \pi x\right)-\cos\left(2 \pi x\right)}\right)}\\ \end{aligned}$

$\boxed{\sum_{n=-\infty}^{\infty}\frac{1}{n^4+4x^4}=\frac{\pi }{4x^3}\left(\frac{\sinh\left(2 \pi x\right)+\sin\left(2 \pi x\right)}{\cosh\left(2 \pi x\right)-\cos\left(2 \pi x\right)}\right)}}$ (3)

Appendix

$\boxed{\cos(x-y)-\cos(x+y)=2\cos x \, \cos y }$ (A.1)

Proof:

Recall

$\cos(x-y)=\cos x \, \cos y+\sin x \, \sin y$ (A.2)

$\cos(x+y)=\cos x \, \cos y-\sin x \, \sin y$ (A.3)

(A.2) - (A.3)

$\begin{aligned} \cos(x-y)-\cos(x+y)&=\cos x \, \cos y+\sin x \, \sin y-\cos x \, \cos y+\sin x \, \sin y\\ &=2\cos x \, \cos y \qquad \blacksquare \end{aligned}$

$\boxed{\sin (x+y)-\sin (x-y)&=2\sin x \, \cos y}$ (A.4)

Proof:

Recall

$\sin (x+y)=\cos x \sin y+ \sin x \cos y$ (A.5)

$\sin (x-y)=\sin x \cos y -\cos x \sin y$ (A.6)

Then, (A.5) - (A.6)

$\begin{aligned} \sin (x+y)-\sin (x-y)&=\cos x \sin y+ \sin x \cos y-\sin x \cos y +\cos x \sin y\\ &=2\sin x \, \cos y \qquad \blacksquare \end{aligned}$

$\boxed{\cot(x+iy)=\frac{\sin (2x)-i\sinh (2y)}{\cosh(2y)-\cos(2x)}}$ (A.7)

$\begin{aligned} \cot(x+iy)&=\frac{\cos (x+iy)}{\sin (x+iy)}\\ &=\frac{2\cos (x+iy)\sin (x-iy)}{2\sin (x+iy)\sin (x-iy)}\\ &=\frac{\sin (x+iy+x-iy)-\sin (x+iy-x+iy)}{\cos(x+iy-x+iy)-\cos(x+iy+x-iy)}\\ &=\frac{\sin (2x)-\sin (2iy)}{\cos(2iy)-\cos(2x)}\\ &=\frac{\sin (2x)-i\sinh (2y)}{\cosh(2y)-\cos(2x)}\qquad \blacksquare \end{aligned}$

$\boxed{\coth(x+iy)=\,\frac{\sinh\left(2x\right)-i\sin\left(2y\right)}{\cosh\left(2x\right)-\cos\left(2y\right)}}$ (A.8)

Proof:

$\begin{aligned} \coth(x+iy)&=\frac{\cosh(x+iy)}{\sinh(x+iy)}\\ &=\frac{\cos\left(i(x+iy)\right)}{-i\sin\left(i(x+iy)\right)}\\ &=i\,\frac{\cos\left(ix-y\right)}{\sin\left(ix-y)\right)} \\ &=i\,\frac{\cos\left(-(y-ix)\right)}{\sin\left(-(y-ix)\right)} \\ &=i\,\frac{2\cos\left(y-ix\right)}{-2\sin\left(y-ix\right)} \cdot \frac{\sin\left(y+ix)\right)}{\sin\left(y+ix)\right)}\\ &=-i\,\frac{\sin\left(y+ix+y-ix\right)-\sin\left(y+ix-y+ix\right)}{\cos\left(y-ix-y-ix\right)-\cos\left(y-ix+y+ix\right)}\\ &=-i\,\frac{\sin\left(2y\right)-\sin\left(2ix\right)}{\cos\left(-2ix\right)-\cos\left(2y\right)}\\ &=-i\,\frac{\sin\left(2y\right)+i\sinh\left(2x\right)}{\cosh\left(2x\right)-\cos\left(2y\right)}\\ &=\,\frac{\sinh\left(2x\right)-i\sin\left(2y\right)}{\cosh\left(2x\right)-\cos\left(2y\right)}\qquad \blacksquare \end{aligned}$

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