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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

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Isn't you own calculation a proof? Or are you looking for another proof? –  Maesumi Mar 20 '13 at 12:50
    
@Maesumi : Yes I proved I know but I need to see another proof without using my way. –  Mathlover Mar 20 '13 at 12:52

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
    
Thanks for answer . Is it also possible to prove the relation with elementary methods (such as power series or just algebraic operators )? –  Mathlover Mar 20 '13 at 13:34
    
I am sure some way using those methods applies, and I thought about using power/Laurent series. But in the end, residues are really the most powerful - and simplest - way to go. –  Ron Gordon Mar 20 '13 at 13:35

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