How to show: $-\sum_{k=-\infty}^\infty \frac{i*k*(-1)^k}{1+k^2}e^{-i*k*x}=2\sum_{k=1}^\infty \frac{k(-1)^{k-1}}{1+k^2}sin(kx)$

Question

I am struggling to show below in a big question:

$-\sum_{k=-\infty}^\infty \frac{i*k*(-1)^k}{1+k^2}e^{-i*k*x}=2\sum_{k=1}^\infty \frac{k(-1)^{k-1}}{1+k^2}sin(kx)$

Tried to analyse with geometric series (no use).

Any help is appreciated

Answer 1

Hint 1

$$e^{-ikx} = -i\sin(kx) + \cos(kx)$$

Hint 2

The Sine function is an odd function, whilst Cosine is even function.

Let's proceed

Once you wrote the series split into sine and cosine part, you have this

$$-\sum_{k = -\infty}^{+\infty}\frac{k(-1)^k}{1+k^2}\sin(kx)- \sum_{k = -\infty}^{+\infty}i\frac{k(-1)^k}{1+k^2}\sin(kx)$$

Now, according to the parity of those functions, you can immediately see that the first series does contain an even function, and the second one does contain an odd function (respectively $k\sin(kx)$ which is even and $k\cos(kx)$ which is odd). So the second series is zero by definition.

You remain with

$$-\sum_{k = -\infty}^{+\infty}\frac{k(-1)^k}{1+k^2}\sin(kx)$$

and since it's even, you can write

$$-2\sum_{k = 0}^{+\infty}\frac{k(-1)^k}{1+k^2}\sin(kx)$$

The minus sign can be incorporated in the $(-1)^k$ term which will become $(-1)^{k-1}$ thence the result.

$$2\sum_{k = 0}^{+\infty}\frac{k(-1)^{k-1}}{1+k^2}\sin(kx)$$