3
$\begingroup$

I'm studying the Casimir Effect at finite temperature. To calculate the Helmoltz free energy in the canonical ensemble I need to sum a particular series. In some scientific papers it is suggested to use a modified Poisson formula over positive integers only that appears to be:

$$\frac{1}{2}F(0) + \sum_{n=1}^{\infty} F(n) = \pi \tilde{F}(0) + 2\pi \sum_{k=1}^{\infty} \tilde{F}(2\pi k)$$

where

$$\tilde{F}(k):= \frac{1}{\pi} \int_0^\infty F(t) \cos(kt)dt.$$

I can't understand how to demonstrate the validity of this expression starting from the standard summation formula, that is:

$$\sum_{n \in \mathbb{Z}} F(2\pi n) = \frac{1}{2\pi} \sum_{k \in \mathbb{Z}} \hat{F}(k)$$

where

$$\hat{F}(k):= \int_{-\infty}^\infty F(x) \exp\{-i k x\}dx$$

is the standard Fourier transform.

In fact this formula is deduced using tempered distributions so $F$ (that is a "test" function) has to have good property of convergence: this means I can't simply define a non-continuous function null for $n<0$.

In my particular case $F(x) = (ax)^2 \log(1-e^{-ax})$. Sorry for my bad English ;)

$\endgroup$

1 Answer 1

0
$\begingroup$

How about applying the following identity $$\sum_{n\in \mathbb{Z}} f(n) = f(0) + \sum_{n=1}^\infty [f(n) + f(-n)]$$ to both sides of the Poisson formula $$\sum_{n\in\mathbb{Z}} F(2\pi n) = \frac{1}{2\pi} \sum_{k\in\mathbb{Z}} \hat F(k).$$

We obtain immediately $$ F(0) + \sum_{n=1}^\infty [F(2\pi n) + F(-2\pi n)] = \frac{1}{2\pi} \hat F(0) + \frac{1}{2\pi} \sum_{k=1}^\infty[ \hat F( k) + \hat F(- k)].$$

Now let us assume $F(x) = F(-x)$, i.e., $F$ is symmetric. We can then simplify this expression as $$ F(2\pi n) + F(-2\pi n) = 2 F(2\pi n)$$ and $$\hat F(k) + \hat F(-k) = 2 \int_{-\infty}^\infty\! dx\, F(x) \cos ( k x) = 4 \int_{0}^\infty\! dx\, F(x) \cos ( k x) = 4\pi \tilde F( k) .$$

Thus, we have $$ F(0) + 2\sum_{n=1}^\infty F(2\pi n) = \tilde F(0) + 2 \sum_{k=1}^\infty \tilde F(k),$$ so the authors using the first formula seem to have a different definition of the Fourier transform.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .