Is there an alternative method to prove that $$\int_{1}^{\infty}x \psi^{(n)}(x) \mathrm dx = (-1)^{n+1}\Big[(n-2)!\zeta(n-1) + (n-1)!\zeta(n)\Big]$$
Context and Solution
The original problem given to me was
How can I prove that $$\sum_{k=3}^{n} \int_{1}^{\infty} x \hspace{.05cm}\psi^{(n)}(x) \mathrm dx = \zeta(2) + (-1)^k(k-1)! \zeta(k)$$ where $\psi^{(n)}(x)$ is the Polygamma function of order $n$.
I wanted to show the sum of the inner integrals seem to telescope, giving the RHS. Using a computer algebra system, I computed the $f(n)$ for some integer values, where $f(n) = \int_{1}^{\infty}x \psi^{(n)}(x) \mathrm dx$. Note that $f(n)$ did not converge for any integer less than $3$.
\begin{align*} &f(3) = \zeta(2) + 2 \zeta(3)\\ &f(4) = -2\zeta(3) - 6\zeta(4)\\ &f(5) = 6\zeta(4) + 24 \zeta(5)\\ &f(6) = -24\zeta(5) - 120\zeta(6)\\ \end{align*}
Which led me to make the following statement
$$\int_{1}^{\infty}x \psi^{(n)}(x) \mathrm dx = (-1)^{n+1}\Big[(n-2)!\zeta(n-1) + (n-1)!\zeta(n)\Big]$$
The following is my proof, using the series expansion of $\psi^{(n)}(x)$. Is there an alternative proof that avoids using the expansion?
\begin{align*} f(n) &= \int_{1}^\infty x \psi^{(n)}(x) \, dx = \int_{1}^\infty x \left[ (-1)^{n+1} n! \sum_{m=0}^\infty \frac{1}{(x + m)^{n+1}} \right] dx\\ &= (-1)^{n+1} n! \sum_{m=0}^\infty \int_{1}^\infty \frac{x}{(x + m)^{n+1}} \, dx \end{align*}
The inner integral evaluates to $$\int_{1}^\infty \frac{x}{(x + m)^{n+1}} \, dx = \frac{1}{n-1} (m+1)^{-(n-1)} - \frac{m}{n} (m+1)^{-n}$$
Substituting back into the summation gives
\begin{align*} f(n) &= (-1)^{n+1} n! \sum_{m=0}^\infty \left[ \frac{1}{n-1} (m+1)^{-(n-1)} - \frac{m}{n} (m+1)^{-n} \right]\\ &= (-1)^{n+1} n! \left[ \frac{1}{n-1} \sum_{m=0}^\infty (m+1)^{-(n-1)} - \frac{1}{n} \sum_{m=0}^\infty \frac{m}{(m+1)^n} \right] \end{align*}
We know
$$\sum_{m=0}^\infty (m+1)^{-(n-1)} = \zeta(n-1)$$ $$\sum_{m=0}^\infty \frac{m}{(m+1)^n} = \sum_{m=1}^\infty \frac{m-1}{m^n} = \zeta(n) - \zeta(n-1)$$
Substituting these gives
\begin{align*} f(n)&= (-1)^{n+1} n! \left[ \frac{\zeta(n-1)}{n-1} - \frac{\zeta(n) - \zeta(n-1)}{n} \right]\\ &= (-1)^{n+1} n! \left[ \frac{ \zeta(n-1) - (n-1) \zeta(n)}{n(n-1)} \right]\\ &= (-1)^{n+1} \Big[ (n-2)!\zeta(n-1) + (n-1)!\zeta(n) \Big] \end{align*}
as desired.
If you have any alternative tricks, techniques, or input, it would be highly appreciated. Thanks.
