Renormalization, integrating out high momenta Wilson way In equation $(12.5)$ in Peskin and Schroeder, they write out the generating function but leave out all quadratic terms of the form $\phi\hat{\phi}$ arguing that they vanish 

since Fourier components of different wavelengths are orthogonal. 

But then my question is why doesn't the same argument apply to terms of the form $$\phi^3\hat\phi, $$
which they include?
Here, 
$$\hat{\phi}(k):= \begin{cases}
\phi(k)\hspace{1cm}\text{ for } b\Lambda\leq \lvert k\rvert < \Lambda,\\
0\hspace{1.55cm}\text{ otherwise,}
\end{cases}
$$
where $b<1$ is some fraction. 
But if I'm not totally wrong, if $\phi\hat{\phi}$ vanish then so should $\phi^2\hat{\phi}^2$, by the same argument, right? If not, why not? 
 A: I'm not being precise, but morally:
Imagine you were integrating out all modes above a frequency $b\Lambda$. Consider $\omega < b\Lambda < 3 \omega$.
A mode $\phi$ with frequency $\omega$ when cubed, will have some part of it as mode of frequency $3 \omega$, since: $\sin (3x) = 3 \sin (x) - 4 \sin^3 (x)$. (Easier to see that ${(e^{i \omega t})}^3 = e^{i 3 \omega t}$). So $\phi^3$ might contain frequencies above the "Wilsonian cutoff" $b \Lambda$ so one has to be careful about its inner product with $\hat{\phi}$ (remember, you still have the $\int d^d k$) -- it will not identically be zero -- so you cannot throw away those terms.

EDIT: Ah, I realize now that my notation could be confusing. I apologize. @JeffDror has a nice answer.
In essence, remember that those terms are still being integrated over some sets of momenta. Jeff has shown clearly how momentum conservation (which gives an overall $\delta$-function for the momenta being integrated over) shows that $\phi \hat{\phi}$ will vanish while you cannot say the same for higher momenta.
As for generalizing my argument, note that 
$$\int d^d x \phi(x) \phi(x) \longrightarrow \int d^d k_1 d^d k_2 \phi(k_1) \phi(k_2) \delta(k_1 + k_2) = \int d^d k \phi(k) \phi(-k)$$
(The $-k$ comes because of momentum conservation.)
When you consider a higher order term
$$\int d^d x \phi(x) \phi(x) \phi(x) \phi(x) \longrightarrow \\
 \int \; d^d k_1 \, d^d k_2 \, d^d k_3 \, d^d k_4 \; \phi(k_1) \phi(k_2) \phi(k_3) \phi(k_4)  \delta(k_1 + k_2 + k_3 + k_4) \\  \neq \int d^d k_1 d^d k_2 \phi^2(k_1) \phi^2(k_2) \delta(k_1 + k_2)$$
The last term might be zero by arguments similar to the one I made above. But notice that it is not equal to the thing you started off with. I hope that clears up the fog.
A: I find the whole notation here a bit confusing since we are talking about momenta modes while using real space notation (maybe I'm the only one...). To clarify what is going on we can switch to momentum space instead. Consider the quadratic term in the exponential:
\begin{align} 
\int d^4x \phi ^2 & = \int d^4x \int d^4k d^4k' \phi _k  \phi ^\ast  _{ k ' }  e ^{ i ( k - k '  ) x } \\ 
& = \int d^4k \phi _k \phi _k ^\ast \\ 
& = \int _0 ^{ b \Lambda } d^4k \phi _k \phi _{ k} ^\ast + \int _{ b \Lambda } ^{ \Lambda } d^4k \phi _k \phi _k ^\ast 
\end{align} 
Now we can clearly make the split that the authors want to make,
\begin{equation} 
= \int _0 ^\Lambda ( \phi _k \phi _k ^\ast + \hat{\phi} _k \hat{\phi} _k ^\ast ) 
\end{equation} 
Notice that the cross term has dropped out of the expression. 
Now to see why this is not the case for the quartic term just repeat the procedure above. I find,
\begin{equation} 
\int d^4x \phi ^4 = \int _0 ^{ b \Lambda } d^4k _1 d^4k _2 d^4k _3 \phi _1 \phi _2 \phi _3 \phi _{ 1 +  2 +  3} + \int _{b \Lambda } ^{ \Lambda } d^4k _1 d^4k _2 d^4k _3 \phi _1 \phi _2 \phi _3 \phi _{ 1 +  2 +  3} 
\end{equation} 
The problem is that we can't just split up the integrals again since while we can ``choose the range'' of $ \phi _1 ,\phi _2 , $ and $ \phi _3 $ the range of $ \phi _{ 1+2+ 3}$ is undetermined. We can for example have $ k _1 = b\Lambda / 2 , k _2 = b\Lambda / 2 , k _3 = b \Lambda /2 $ but $ k _{1+2+3} = 3 b \Lambda /2 $. Therefore we can't drop the cross terms for any the quartic interaction.
Note: I was sloppy here about conjugates but I'm sure you get the idea.
