I am studying z transform and I couldn't get how to derive these two formulas
$$Z[ nu_{n}] =-z \frac{d}{dz} Z(u_{n}) $$
$$Z\left[ \frac{1}{n} u_{n}\right] =- \int_0^z z^{-1}Z(u_{n}) $$
These formula are somewhat similar to the corresponding ones of Laplace transforms which also involves derivatives and integrals. Any idea on how to prove it?
Thanks.
Look, the first one is pretty straightforward to prove:
For a discrete time domain function $f(k)$, its Z-Transform is defined as:
$$\mathcal{Z}\{f(k)\}=\sum_{k=-\infty}^{+\infty}f(k)z^{-k}$$
We can differentiate on both sides:
\begin{align} \frac{d}{dz}\mathcal{Z}\{f(k)\}&=\frac{d}{dz}\sum_{k=-\infty}^{+\infty}f(k)z^{-k}\\ &=\sum_{k=-\infty}^{+\infty}\frac{d}{dz}f(k)z^{-k}\\ &=\sum_{k=-\infty}^{+\infty}f(k)\frac{d}{dz}z^{-k}\\ &=\sum_{k=-\infty}^{+\infty}-kf(k)z^{-k-1}\\ &=-z^{-1}\sum_{k=-\infty}^{+\infty}kf(k)z^{-k}\\ &=-z^{-1}\mathcal{Z}\{kf(k)\}\\ \end{align}
So, finally:
$$\mathcal{Z}\{kf(k)\}=-z\frac{d}{dz}\mathcal{Z}\{f(k)\}$$
The integral property, on the other hand, is a little harder to prove with such a direct method, but let's go:
\begin{align} \int\mathcal{Z}\{f(k)\}dz&=\int\left[\sum_{k=-\infty}^{+\infty}f(k)z^{-k}\right]dz\\ &=\sum_{k=-\infty}^{+\infty}\int f(k)z^{-k}dz\\ &=\sum_{k=-\infty}^{+\infty}f(k)\int z^{-k}dz\\ &=\sum_{k=-\infty}^{+\infty}\frac{f(k)}{1-k}z^{1-k}\\ &=-z\sum_{k=-\infty}^{+\infty}\frac{f(k)}{k-1}z^{-k}\\ \end{align}
Defining $p=k-1$, then $k=p+1$ and the sum can be rewritten as:
\begin{align} &=-z\sum_{p=-\infty}^{+\infty}\frac{f(p+1)}{p}z^{-p-1}\\ &=-\sum_{p=-\infty}^{+\infty}\frac{f(p+1)}{p}z^{-p}\\ \end{align}
Applying the convolution theorem:
\begin{align} -\sum_{p=-\infty}^{+\infty}\frac{f(p+1)}{p}z^{-p}&=-\left[\sum_{p=-\infty}^{+\infty}f(p+1)z^{-p} * \sum_{p=-\infty}^{+\infty}\frac{1}{p}z^{-p}\right]\\ \end{align}
Now, applying the time-shift property of the Z-transform:
\begin{align} -\left[\sum_{p=-\infty}^{+\infty}f(p+1)z^{-p} * \sum_{p=-\infty}^{+\infty}\frac{1}{p}z^{-p}\right]&=-\left[z\sum_{p=-\infty}^{+\infty}f(p)z^{-p} * \sum_{p=-\infty}^{+\infty}\frac{1}{p}z^{-p}\right]\\ &=-z\left[\sum_{p=-\infty}^{+\infty}f(p)z^{-p} * \sum_{p=-\infty}^{+\infty}\frac{1}{p}z^{-p}\right]\\ &=-z\left[\sum_{p=-\infty}^{+\infty}\frac{f(p)}{p}z^{-p}\right]\\ \end{align}
Now, reverting the substitution of $p=k-1$ and $k=p+1$, we get:
\begin{align} &=-z\left[\sum_{k=-\infty}^{+\infty}\frac{f(k-1)}{k-1}z^{-k+1}\right]\\ &=-z\left[z\sum_{k=-\infty}^{+\infty}\frac{f(k-1)}{k-1}z^{-k}\right]\\ &=-z\left[\sum_{k=-\infty}^{+\infty}\frac{f(k)}{k}z^{-k}\right]\\ &=-z\mathcal{Z}\left\{\frac{f(k)}{k}\right\} \end{align}
Which means that:
$$\int\mathcal{Z}\{f(k)\}dz=-z\mathcal{Z}\left\{\frac{f(k)}{k}\right\}$$
And, finally:
$$\mathcal{Z}\left\{\frac{f(k)}{k}\right\}=-\frac{1}{z}\int\mathcal{Z}\{f(k)\}dz$$
Which actually differs from the property you posted on your question... If the integral is in $z$, then it makes a difference to have $z^{-1}$ inside or outside the integral.
Hope I was able to help.