I was trying to write a formula for antidifference operator $\Delta^{-1}=(e^D-1)^{-1}$ using Fourier transforms. I obtained the formal formula:
$$\Delta^n[f](x)=\frac1{2 \pi }\int_{-\infty }^{+\infty } e^{-iwx}(e^{-i w}-1)^n\int_{-\infty }^{+\infty } e^{i t w} f(t) \, dt \, dw$$
For $n>0$ it works well in computer algebra systems, giving for instance, at $n=1$ the difference operator. But if we put $n=-1$ so to obtain antidifference, the computer algebra systems fail.
For instance, when taking $f(x)=x$, the expected result would be $\Delta^{-1}[f](x)=\frac{x^2}2-\frac{x}2+\frac1{12}$. But the computer algebras fail, returning
$$\mathcal{F}_w^{-1}\left[-\frac{i \sqrt{2 \pi } \delta '(w)}{-1+e^{-i w}}\right](x)$$
For $f(x)=1$ the expected result is $x-1/2$ while the output is
$$\mathcal{F}_w^{-1}\left[\frac{\sqrt{2 \pi } \delta (w)}{-1+e^{-i w}}\right](x)$$
The formula fails for any function I feed in.
So, I wonder, what the issue is here? Is it insufficient power of the computer algebra system, or these transforms are problematic themselves? The formula fails even if I fix $x=0$ so to obtain the constant term only.