Dirac-Delta function representation - infinite sum involving trigonometric identities

Question

Proof of the identity: $$\delta (x-x') = \sum_{n=0}^{\infty} \Big\{ \cos[n \pi(x-x')] - \cos[n \pi(x+x')] \Big\} \tag{1}$$

I can intuitively tell that this function is $\infty$ for $x=x'$, and zero everywhere else, but I am looking for a mathematical proof. Especially that the integral from $-\infty$ to $\infty$ equals one.

PS: This was the 1st problem in a 4th semester exam for physics students...the solution shouldn't be too difficult i guess :)

Edit: For $x \neq x'$ we have: $$ \delta (x-x') = \sum_{n=0}^{\infty} \Big\{ \frac{e^{i n \pi (x-x')}+e^{-i n \pi (x-x')}}{2}-\frac{e^{i n \pi (x+x')}+e^{-i n \pi (x+x')}}{2} \Big\} \tag{2}$$

Which we can sum up for $x \neq x'$: $$ \frac{1}{2} \Big\{ \frac{1}{1-e^{i \pi (x-x')}} + \frac{1}{1-e^{-i \pi (x-x')}} - \frac{1}{1-e^{i \pi (x+x')}} - \frac{1}{1-e^{-i \pi (x+x')}} \Big\} \tag{3}$$

If we write that out, we can find that it equals zero.

For $x=x'$, the 1st cosine equals 1, so we sum up 1 from $n=0$ to $\infty$, and we get $ \infty $. But is the integral of this function really 1?

Answer 1

Hint: Transform the sums using eulers fromula

$$\cos(x)=\frac{e^{ix}+e^{-ix}}{2i}$$

The sum should become a geometric sum.

You can also rewrite $$\cos[n\pi(x-x')]=\Re\left[e^{in\pi(x-x')}\right]$$

Answer 2

Hint: We will assume that $x$ and $x^{\prime}$ are real (as opposed to complex) variables. OP's distribution $$ \sum_{n\in \mathbb{N}_0} \left\{ \cos[n \pi(x-x')] - \cos[n \pi(x+x')] \right\} ~=~ \frac{1}{2}\sum_{n\in \mathbb{Z}} \left\{ \exp[in \pi(x-x')] - \exp[in \pi(x+x')] \right\} ~=~III_2(x-x')-III_2(x+x')$$

is a linear combination of Dirac combs/shah distributions. It is not equal to the Dirac delta distribution $\delta(x-x')$ as OP's first formula (v4) suggests.