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If the critical point of the phase is in the domain of $\psi$, use $|a|$ in place of $\lambda$. Then (if $a>0$) you have a phase $\Phi(x)=x^2+bx/a$, and $|\Phi''(x)|\geq 2$, so you get decay of order $|a|^{-1/2}$. See p. 328 in Stein and Shakarchi vol. 4 for the estimate. Alternatively, you can complete the square and do a (linear) change of variables. ...

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The reason we can get that (twisted) Poisson summation formula in the first place is that in the primitive case you can interpolate the character to a smooth real function via Gauss sums. In the imprimitive case this is not the case anymore, and you can't get a function nice enough to anything that resembles a Poisson formula to hold. Of course nothing is ...

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Not sure if this is true, but me may write that \begin{eqnarray} \frac{\hat{f}(\tau,\xi)}{\tau+|\xi|^4+\varepsilon|\xi|^2+i} &=& \frac{\tau+|\xi|^4+i}{\tau+|\xi|^4+\varepsilon|\xi|^2+i} \frac{\hat{f}(\tau,\xi)}{\tau+|\xi|^4+i} \\ &:=& m_{\varepsilon}(\tau, \xi) \hat{F}(\tau, \xi) \end{eqnarray} and from the case $\varepsilon = 0$, you ...

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