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$$\sum _{ n=-\infty }^{ \infty }{ \frac { 1 }{ { \left( j\omega +j\frac { 2\pi n }{ T } \right) }^{ 2 } } }$$

final answer is $$-\frac { { T }^{ 2 } }{ 2 } \left( \frac { 1 }{ 1-cos(\omega T) } \right)$$

Where$$ j=\sqrt { -1 }$$

I tried to take out $j$ which gives me negative sign and then I tried to solve the parentheses but things are getting complicated. Is there any mathematical trick to get the final answer as shown?

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Sheldon
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  • You should have used $i=\sqrt{-1}$ which I think is a more common form of "iota" and you may not need to explicitly state , then, $i=\sqrt{-1}$ – RE60K Aug 06 '14 at 16:59
  • Especially since, after getting rid of $i^2$ and replacing $\omega$ by $\omega T$, the question reduces to show that, for every real number $\omega$ not a multiple of $2\pi$, $$\sum _{ n=-\infty }^\infty\frac1{\left(\omega +2\pi n\right)^2}= \frac1{1-\cos\omega}.$$ And this notationally simpler version could ring a bell... – Did Aug 06 '14 at 17:10

1 Answers1

1

$$\sum _{ n=-\infty }^{ \infty }{ \frac { 1 }{ { \left( j\omega +j\frac { 2\pi n }{ T } \right) }^{ 2 } } }=\sum _{ n=-\infty }^{ \infty }{ \frac { 1 }{ { j^2\left( \omega +\frac { 2\pi n }{ T } \right) }^{ 2 } } }=-\sum _{ n=-\infty }^{ \infty }{ \frac { 1 }{ { \left( \omega +\frac { 2\pi n }{ T } \right) }^{ 2 } } }=$$

$$-\sum _{ n=-\infty }^{ \infty }{ \frac { 1 }{ { \left( \omega +\frac { 2\pi n }{ T } \right) }^{ 2 } } }=\frac{d}{d\omega } (\sum _{ n=-\infty }^{ \infty }{ \frac { 1 }{ { \left( \omega +\frac { 2\pi n }{ T } \right) } } })$$

now we need to find out

$$\sum _{ n=-\infty }^{ \infty }{ \frac { 1 }{ { \left( \omega +\frac { 2\pi n }{ T } \right) } } }$$

There is a series representation as partial fraction expansion where just translated reciprocal functions are summed up, such that the poles of the cotangent function and the reciprocal functions match. Reference link
Please check note 12.

$$\pi \cot(\pi x)=\sum _{ n=-\infty }^{ \infty }{ \frac { 1 }{ { \left( x+n \right) } } }$$

There is also a proof in here how to get that series expansion.

Now we can use this identity and get :

$$\sum _{ n=-\infty }^{ \infty }{ \frac { 1 }{ { \left( \omega +\frac { 2\pi n }{ T } \right) } } }=\frac { T }{ 2\pi }\sum _{ n=-\infty }^{ \infty }{ \frac { 1 }{ { \left( \frac {\omega T }{ 2\pi } +n \right) } } }=$$

$$=\pi . \frac { T }{ 2\pi }\cot(\pi \frac {\omega T }{ 2\pi })=\frac { T }{ 2 }\cot(\frac {\omega T }{ 2 })$$

Now just you need to do final step.

$$\sum _{ n=-\infty }^{ \infty }{ \frac { 1 }{ { \left( j\omega +j\frac { 2\pi n }{ T } \right) }^{ 2 } } }=\frac{d}{d\omega } (\sum _{ n=-\infty }^{ \infty }{ \frac { 1 }{ { \left( \omega +\frac { 2\pi n }{ T } \right) } } })=\frac{d}{d\omega }(\frac { T }{ 2 }\cot(\frac {\omega T }{ 2 }))$$

Can you finalize it to get the answer that you wrote? Please let me know if you need help.

EDIT: I also found a nice proof for series expansion of $ \cot(x)$ in here. It was shared by @Raymond Manzoni

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