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Suppose a wave function $\psi (x)=C\exp(ip_0x/\hbar)\exp(-|x|/(2\Delta x)) $ is given and we are asked to find the momentum space wavefunction. Then we take the Fourier transform as follow:

\begin{align} \phi(p) &= \frac{C}{\sqrt{2\pi \hbar}}\int^{\infty}_{-\infty}\exp\left(\frac{ip_0x}{\hbar}\right)\exp\left(-\frac{\vert x\vert}{2\Delta x}\right)\exp\left(-\frac{ipx}{\hbar}\right)\,{\rm d}x \\ &= \frac{C}{\sqrt{2\pi \hbar}}\int_0^\infty\exp\left(\frac{xi\left(p_0-p-\hbar/2\Delta x\right)}{\hbar}\right)\,{\rm d}x+\int_{-\infty}^0\exp\left(\cdots\right)\\ \end{align} Now the wavefunction can be normalised to $0$, hence it is a legitimate wavefunction so we should be able to get momentum space wavefunction. But the first integral blows up as $x \rightarrow \infty$. If one argues that $(p_0-p-\hbar/2\Delta x)$ should be negative (which is anyway not true in general) then the integral on right is bound to blow up. Then what is wrong? Or is there a way to get around this problem?

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    $\begingroup$ You seemed to have gained a factor of $i$ on the $\Delta x$ term from nowhere. Fix this and the integral converges $\endgroup$ – By Symmetry Feb 1 '18 at 10:55
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    $\begingroup$ FWIW, if you use \exp instead of e^{}, you don't need to \Large your equations to make them legible. $\endgroup$ – Kyle Kanos Feb 1 '18 at 12:41

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