$n$-th derivative of Beta function We pretty much know nothing about the high order derivatives of the Beta function. Well, we known for the example some recursive formulae for $\Gamma^{(n)}(1)$ as well as $\Gamma^{(n)}\left(\frac{1}{2}\right)$ but that is all. Yesterday , I encountered the following integral:
$$\int_{0}^{\infty} \frac{\log^n x}{\left ( 1+x^2 \right )^m}\, {\rm d}x$$
As it can easily be seen , we have that:
$$\int_{0}^{\infty} \frac{\log x}{\left ( 1+x^2 \right )^m}\, {\rm d}x ={\rm B}^{(1)} \left ( \frac{1}{2}, m- \frac{1}{2} \right ) \implies  \int_{0}^{\infty} \frac{\log^n x}{\left ( 1+x^2 \right )^m}\, {\rm d}x = {\rm B}^{(n)} \left ( \frac{1}{2}, m- \frac{1}{2} \right )$$
The question is whether or not we can express that last $n$-th derivative in a closed form - most likely in an inductive form. I guess it will be difficult but with mathematics you never know. 
 A: Let us denote with 
$$f(n, m)=\int_0^\infty \frac{\log^n x}{(1+x^2)^m}\, {\rm d}x$$
It is quite straigtforward to note that:
$$f(n, 1)=\int_{0}^{\infty} \frac{\log^n x}{1+x^2}\, {\rm d}x = \left\{\begin{matrix}
0 &,  &{\rm odd} \\ 
 2n! \beta(n+1)&, &{\rm even} 
\end{matrix}\right.$$
Here $\beta$ denotes the Beta dirichlet function.
On the other hand 
$$f(1,m)= \int_{0}^{\infty}\frac{\log x}{\left ( 1+x^2 \right )^m}= {\rm B}^{(1)} \left ( \frac{1}{2}, m - \frac{1}{2} \right )= \frac{\Gamma \left ( \frac{1}{2} \right ) \Gamma \left ( m- \frac{1}{2} \right )  \bigg[ \psi^{(0)}\left ( \frac{1}{2}\right) - \psi^{(0)} \left ( m- \frac{1}{2} \right )  \bigg]}{m^2 \Gamma(m)}$$
Using gammatester's comment for the more general result, that is for $f(n,m)$ using Liebniz's general rule we have that:
$$f(n, m)= \int_{0}^{\infty} \frac{\log^n x}{\left ( 1+x^2 \right )^m}\, {\rm d}x = {\rm B}^{(n)} \left ( \frac{1}{2}, m- \frac{1}{2} \right )= \frac{1}{m^2  \Gamma(m)}\sum_{k=0}^{n} \binom{n}{k} \Gamma^{(k)} \left ( \frac{1}{2} \right )\Gamma^{(n-k)} \left ( m- \frac{1}{2} \right )$$
Of course for $\Gamma^{(n)} \left(\frac{1}{2} \right)$ we have the following reduction formula:
$$\Gamma^{(n+1)}\left ( \frac{1}{2} \right )= -\left ( \gamma +2 \log 2 \right ) \Gamma^{(n)} \left ( \frac{1}{2} \right ) +n ! \sum_{k=1}^{n}\frac{(-1)^{k+1}}{\left ( n-k \right )!} \left ( 2^{k+1}-1 \right )\zeta(k+1) \Gamma^{(n-k)}\left ( \frac{1}{2} \right )$$
