As promised, here is a little derivation to give some insight into the effects of windows on DFT values done in reverse.
Suppose that I have a DFT from some signal.
\begin{equation}
\begin{aligned}
X[k] &= \sum\limits_{n=0}^{N-1} x[n] e^{-i\frac{2\pi}{N}kn}
\end{aligned}
\end{equation}
I look at it, decide it needs a little "smoothing", so I decide to average neighboring DFT values to make a new DFT bin. Let's call it $Y$. The weights I'm going to use are just a "twist" of the (rescaled) VonHann values. If you plug in $\omega=0$ this becomes clear.
Plug and chug.
\begin{equation}
\begin{aligned}
Y[k] &= ( -e^{i\omega}, e^{i\omega}+ e^{-i\omega},-e^{-i\omega} ) \cdot (X[k-1],X[k],X[k+1]) \\
&= -e^{i\omega} X[k-1] + ( e^{i\omega}+ e^{-i\omega} ) X[k] -e^{-i\omega} X[k+1] \\
&= \sum\limits_{n=0}^{N-1} \left[ -e^{i\omega} e^{i\frac{2\pi}{N}n} + e^{i\omega}+ e^{-i\omega} -e^{-i\omega} e^{-i\frac{2\pi}{N}n} \right] x[n] e^{-i\frac{2\pi}{N}kn} \\
&= \sum\limits_{n=0}^{N-1} 2 \left[ \cos ( \omega) - \cos \left( \frac{2\pi}{N}n + \omega \right) \right] x[n] e^{-i\frac{2\pi}{N}kn} \\
\end{aligned}
\end{equation}
Now, my weighted average has all the sudden become a window function. Again plug in $\omega=0$ and you can see it is the VonHann.
Time for one of those trig identities.
\begin{equation}
\begin{aligned}
\cos(A) - \cos(B) &= \cos\left( \frac{A+B}{2} + \frac{A-B}{2} \right) - \cos\left( \frac{A+B}{2} - \frac{A-B}{2} \right) \\
&= -2 \sin\left( \frac{A+B}{2} \right) \sin\left( \frac{A-B}{2} \right) \\
\end{aligned}
\end{equation}
Substitute it in:
\begin{equation}
\begin{aligned}
Y[k] &= \sum\limits_{n=0}^{N-1} 4 \sin \left( \frac{n}{N}\pi \right) \sin \left( \frac{n}{N}\pi + \frac{\omega}{2} \right)
x[n] e^{-i\frac{2\pi}{N}kn} \\
Y[k] &= \sum\limits_{n=0}^{N-1} w_{\omega}[n] x[n] e^{-i\frac{2\pi}{N}kn} \\
\end{aligned}
\end{equation}
When you plug in $\omega=0$ you see why the VonHann is also called the sine squared window.
So, here is a family of window functions whose base member is the VonHann.
\begin{equation}
w_{\omega}[n] = 4 \sin \left( \frac{n}{N}\pi \right) \sin \left( \frac{n}{N}\pi + \frac{\omega}{2} \right)
\end{equation}
Like I said, I'm not a window expert as I don't care for them much, but I don't think you'll find these on the list in the references. Trig equations are slippery though, it could be one of those in different form.
Still it is interesting.
Hope this helps.
For those who say it might be worthless, I say not so fast.
Let $\omega = \frac{2\pi}{N}$
\begin{equation}
w[n] = 4 \sin \left( \frac{n}{N}\pi \right) \sin \left( \frac{n+1}{N}\pi \right)
\end{equation}
Which means is it zeroes out the last sample as well as the first and shifts the center of the window to $(N-1)/2$, the center of the sample set, vs $N/2$ which is the center of the repeat frame.
Extra credit:
Which value of $\omega$ cancels the "The twisting caused by being off bin." which is point #2 in the reference in my first answer.
