My question is if it is possible to find a compactly supported smooth function $\varphi:\mathbf{R}\to \mathbf{R}$ s.t. the following integration $\int_{\mathbf{R}}\varphi(t)e^{itx}e^{tx}dt$ stays bounded for $x\in \mathbf{R}$.
Some more thoughts: for example, if we take $\varphi$ to be supported on the interval $[1,2]$, and take $x$ to be some large number, then the term $e^{tx}$ hehaves like $e^x$, which might(?) be able pulled out from the integration to obtain $e^{x}\int_{\mathbf{R}}\varphi(t)e^{itx}dt$, thus in order for the last expression to be bounded, we need $\int_{\mathbf{R}}\varphi(t)e^{itx}dt$ to have exponential decay! This is not possible as if it is true, we know $\varphi$ will be real analytic from the Fourier inverse formula, which will then be identically zero from the fact that it has compact support.
I'm now really hoping that before pulling out the term $e^{tx}$, the term $e^{itx}$ will generate enough cancellation!