Euler's infinite product for the sine function
$$\displaystyle \sin( x) = x \prod_{k=1}^\infty \left( 1 – \frac{x^2}{\pi^2k^2} \right)$$
http://en.wikipedia.org/wiki/Basel_problem
We know that $\sin( x)$ satisfies $y''+y=0$ differential equation.
$$\displaystyle \frac{\sin'( x)}{\sin( x)} = \frac{1}{x}-2x \sum_{k=1}^\infty \frac{1}{\pi^2k^2-x^2}$$
$$\displaystyle \sin'( x) = \sin( x)\left(\frac{1}{x}-2x \sum_{k=1}^\infty \frac{1}{\pi^2k^2-x^2} \right)$$
$$\displaystyle \sin''( x) = \sin'( x) \left(\frac{1}{x}-2x \sum_{k=1}^\infty \frac{1}{\pi^2k^2-x^2}\right)+ \sin( x) \left(-\frac{1}{x^2}-2\sum_{k=1}^\infty \frac{1}{\pi^2k^2-x^2}-4x^2 \sum_{k=1}^\infty \frac{1}{(\pi^2k^2-x^2)^2}\right)$$
$$\displaystyle \sin''( x) = \left(\frac{\sin( x)}{x}-2x \sin( x) \sum_{k=1}^\infty \frac{1}{\pi^2k^2-x^2}\right) \left(\frac{1}{x}-2x \sum_{k=1}^\infty \frac{1}{\pi^2k^2-x^2}\right)+ \sin{x} \left(-\frac{1}{x^2}-2\sum_{k=1}^\infty \frac{1}{\pi^2k^2-x^2}-4x^2 \sum_{k=1}^\infty \frac{1}{(\pi^2k^2-x^2)^2}\right)$$
$$\displaystyle \sin''( x) = \sin x \left(+4x^2\left(\sum_{k=1}^\infty \frac{1}{\pi^2k^2-x^2} \right)^2-6 \sum_{k=1}^\infty \frac{1}{\pi^2k^2-x^2} -4x^2\sum_{k=1}^\infty \frac{1}{(\pi^2k^2-x^2)^2} \right)$$
If $\sin( x)$ satisfies $y''+y=0$ differential equation.
Then $$ 4x^2\left(\sum_{k=1}^\infty \frac{1}{\pi^2k^2-x^2} \right)^2-6 \sum_{k=1}^\infty \frac{1}{\pi^2k^2-x^2} -4x^2\sum_{k=1}^\infty \frac{1}{(\pi^2k^2-x^2)^2}=-1$$ must be equal
I am stuck to prove the relation in another way. How can I prove that the last relation is equal to $-1$ ?
Note:
If
$x=0$
Easily we can see that
$$ -6 \sum_{k=1}^\infty \frac{1}{\pi^2k^2} =-1$$
$$ \sum_{k=1}^\infty \frac{1}{k^2} =\frac{\pi^2}{6}$$
This result famous basel problem result.
Thanks for answers
Best Answer
The first sum is a known sum which I will not prove here:
$$\sum_{k=1}^{\infty} \frac{1}{\pi^2 k^2-x^2} = \frac{1}{2 x} \left ( \frac{1}{x}-\cot{x}\right)$$
The second sum, on the other hand, I could not find in a reference. You can, however, evaluate it using residues. That is,
$$\sum_{k=-\infty}^{\infty} \frac{1}{(\pi^2 k^2-x^2)^2} = -\sum_{\pm} \text{Res}_{z=\pm x/\pi} \frac{\pi \cot{\pi z}}{(\pi^2 z^2-x^2)^2}$$
I will spare you the residue calculation here; needless to say, the result for the sum is
$$\sum_{k=-\infty}^{\infty} \frac{1}{(\pi^2 k^2-x^2)^2} = \frac{\cot^2{x}}{2 x^2} + \frac{\cot{x}}{2 x^3}+ \frac{1}{2 x^2}$$
which means that
$$\sum_{k=1}^{\infty} \frac{1}{(\pi^2 k^2-x^2)^2} = \frac{\cot^2{x}}{4 x^2} + \frac{\cot{x}}{4 x^3}+ \frac{1}{4 x^2} - \frac{1}{2 x^4}$$
I also leave the algebra to the reader in plugging these expressions into the equation the OP has provided. In the end, yes, the relation is true.