Proving Euler-Maclaurin approximation for even functions using Poisson summation

Question

I'd like to ask for help for the following from my Advanced Mathematics for Physics class (6th semester):

If $f(x)$ is a sufficiently regular and even function integrable in all $\mathbb{R}$, use Poisson's summation formula to prove the following approximation: $$ \sum_{n=0}^{\infty} f(n) = \int_0^{\infty} f(x)\ dx + \frac{1}{2} f(0) - \frac{1}{12} f'(0) + \frac{1}{720} f'''(0) + \cdots $$

We studied Poisson's summation formula as $\sum_{n \in \mathbb{Z}} f(n) = \sum_{n \in \mathbb{Z}} \hat{f}(n)$, where $ \hat{f}(\alpha) = \int_{\mathbb{R}} f(x) e^{-2 \pi i x \alpha}\,dx $ is the Fourier transform of $f$. The best I've been able to do is show that $$ \begin{align} \sum_{n \in \mathbb{Z}} f(n) = \sum_{n \in \mathbb{Z}} \hat{f}(n) \implies \sum_{n=0}^{\infty} f(n) = \int_0^{\infty} f(x)\ dx + \frac{1}{2} f(0) + \sum_{n=1}^{\infty} \int_{\mathbb{R}} f(x) e^{-2 \pi i x n} \, dx \end{align} \tag{because $f$ and $\hat{f}$ are both even } $$ and I tried to expand the integrand of the last term into its Maclaurin series but since I'd be taking the integral over the reals of a polynomial, I cannot prove the integral converges. I've tried reading proofs of the Euler-Maclaurin formula online but those I could find don't make use of the Poisson formula and use Bernoulli numbers extensively (I know they're important for this result but I'm not very familiar with them and my teacher expects us to solve this problem without using them).

Any help would be greatly appreciated.

(This is my first question on the site, I think I'm following the question guidelines but I apologize if I missed anything)

Answer 1

Finally got a solution. Posting it in case anyone's interested.

First, notice that because $ \int_0^{\infty} f(x) \, dx $ converges, $ x \to \infty \implies f^{(n)}(x) \to 0 $ for $ n = 0,1,\cdots, \infty $. Let us also remember the sine and cosine functions are bounded

Let's do the last integral by parts:

$ \begin{align} \int_{\mathbb{R}} f(x) e^{-2 \pi i x n} \, dx &= 2 \int_0^{\infty} f(x) \cos{(2 \pi x n)} \, dx \tag{because of parity} \\ &= 2 \left[ \left( \frac{f(x) \sin{(2 \pi n x)}}{2 \pi n} \right)_0^{\infty} - \int_0^{\infty} \frac{f'(x) \sin{(2 \pi n x)}}{2 \pi n} \, dx \right] \\ &= \cdots \\ &\approx 2 \left[ - \frac{f'(0)}{(2 \pi n)^2} + \frac{f'''(0)}{(2 \pi n)^4} \right] \end{align} $

The rest is easy.