What's the Fourier transform of $f(x)/x$? What's the Fourier transform of $\frac{f(x)}{x}$? Is it even defined? Because I was thinking of the time dependent Schrödinger's equation for an electric potential:
$$ \frac{-\hbar^2}{2m} \nabla^2 \Psi + \frac{kq_1 q_2}{r} \Psi = i\hbar \frac{\partial \Psi}{\partial t} $$

So, I have to take the transform to solve it, but how do I take the transform of the potential part?
I think this might help. 

Suppose $\Psi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\phi(k)e^{ikx}dk$.

So $\Psi(x)$ is the Fourier Transform of $\phi(k)$.

Take the derivative of either side with respect to $x$. 

$\frac{\partial \Psi}{\partial x}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty ik\phi(k)e^{ikx}dk$

So we can get the transform of multiplying by a power of k if we just do the transform of $\phi$ then take the derivative of the transform. 

This suggests integrating the transform with respect to $x$ to divide and get the transform of $\phi(k)/k$. 

There are some restrictions on the method. The function which is to be integrated with respect to $k$ has to converge, so multiplying or dividing by powers of $k$ will only work for certain values of $\phi(k)$.
I think this might help. 

Suppose $\Psi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\phi(k)e^{ikx}dk$.

So $\Psi(x)$ is the Fourier Transform of $\phi(k)$.

Take the derivative of either side with respect to $x$. 

$\frac{\partial \Psi}{\partial x}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty ik\phi(k)e^{ikx}dk$

So we can get the transform of multiplying by a power of k if we just do the transform of $\phi$ then take the derivative of the transform. 

This suggests integrating the transform with respect to $x$ to divide and get the transform of $\phi(k)/k$. 

There are some restrictions on the method. The function which is to be integrated with respect to $k$ has to converge, so multiplying or dividing by powers of $k$ will only work for certain values of $\phi(k)$.