I am reading on Meyer wavelets and have trouble understanding the proof of the orthogonality of the shifted wavelets. The definitions are as follows:
$\eta \in L^2(\mathbb{R})$ is a scaling function that satisfies the following:
$H_r := \overline{\mathrm{span}}\{\eta\left(\frac{\cdot-n}{2^r}\right) | ~r \in \mathbb{Z},~ n \in \mathbb{N}\}$
$H_r \subset H_{r+1}$
$\hat{\eta}$ is continous and $\sum_{l\in\mathbb{Z}} |\hat{\eta}(p+2\pi l)|^2$ has no zeroes.
Then $\varphi$ is defined via
$$\hat{\varphi}(p) = \hat{\eta}(p)~\left(\sum_{l\in\mathbb{Z}} |\hat{\eta}(p+2\pi l)|^2\right)^{-\frac{1}{2}}$$
and the claim
$$\int_{-\infty}^\infty \varphi(x-m)^* \varphi(x-n)\, \mathrm{d}x = \delta_{mn}~~ \forall m,n\in \mathbb{Z}$$
is made. I have trouble proving the last statement. This is how far I got: For simplicity assume wlog $m=0$. Then the statement translates in the Fourier domain to (up to some normalization factors)
$$\int e^{ipn}\hat{\varphi}(p)^*\hat{\varphi}(p) \, \mathrm{d}p= \int e^{ipn} \frac{\hat{\eta}(p)^* \hat{\eta}(p)}{\sum_{l\in\mathbb{Z}} |\hat{\eta}(p+2\pi l)|^2} \, \mathrm{d}p = \delta_{n0}$$
and now we have a $2\pi$-periodic, real, positive and even denominator and $\hat{\eta}(p)^* \hat{\eta}(p) = \hat{\eta}(-p)~ \hat{\eta}(p)$ is a positive, even and real functions, due to the realness of $\eta$. So I can see, that the imaginary part of the integral vanishes as an integral over an odd function. But I can't see, why the real part should vanish too. Which of of the additional properties of $\eta$ do I have to use and how?
edit
now following @Dunham s advice I get the following,
$$ = \sum_{j \in \mathbb{Z}} \int_0^{2 \pi} \frac{\cos(p n)}{g(p)} \hat{\eta}(p+2 j \pi)^* \hat{\eta} (p+2 j \pi) \mathbb{d}p$$
$$ = 2 \sum_{j \in \mathbb{N}} \int_0^{2 \pi} \frac{\cos(p n)}{g(p)} \hat{\eta}(-p-2 j \pi) \hat{\eta} (p+2 j \pi) \mathbb{d}p $$
where
$$ g(p) = \sum_{l\in\mathbb{Z}} |\hat{\eta}(p+2\pi l)|^2$$
and $e^{ipn}$ was split in its real and imaginary part and the imaginary integral already vanishes.
However I still can't see, why it vanishes. Obviously in this situation every summand must individually vanish, but why can't $\frac{\hat{\eta}(-p-2 j \pi) \hat{\eta} (p+2 j \pi)}{g(p)}$ have most of its contribution to the integral, where cos is positive and almost none, where it is negative?
edit adding the final answer, in case somebody else stumbles upon this...
$$ ... = \int_0^{2\pi} \cos(np) \frac{\sum_{l\in\mathbb{Z}} |\hat{\eta}(p+2\pi l)|^2}{\sum_{l\in\mathbb{Z}} |\hat{\eta}(p+2\pi l)|^2} \mathrm{d}p = \delta_{n0} $$
Cheers, Jan Lukas