What is the $n$ th derivative of $\ln(x)/(1+x^2)\:?$

Question

I'm into something but I came across the problem of finding a closed form for

$$\left( \frac{\ln x}{1+x^2} \right)^{(n)}$$

where the little $(n)$ denotes the $n$ th derivative of the function. After looking in some softwares, I believe the closed form is a rational function with $\ln x$ as a factor.

Thank for any help.

Edit: I found a way to change the problem above to this one:

$$ \left( \frac{\partial }{\partial x}\frac{(1-x)}{(1+x^2)(1-(1-x)y)}\right)^{(n)}$$

maybe this is easier?

EDIT $2$: I said closed formula but forgot to mention something easy to handle, and a summation like $\sum_{k=0}^n \cdots$ is hard, because after finding the derivative I'll need to find it's maximum...

A good example of what I mean Finding the $2n+1$ th derivative of $\frac{y^{2n+1}xy}{1-x^2y^2}$ with respect to $x$.

I'm trying to establish the irrationality of a constant using Beuker's integrals, and using this approach you'll eventually need to find the $n$ th derivative of a function. In this case it can be either the first function or the second showed here. The $y$ in the second came from a substitution in the Beuker's integral.

Answer 1

Find the nth derivative of $ln(x)$ and $(1+x^2)^{(-1)}$ and then use Liebnitz's rule $(fg)^{(n)} = \sum_{k=0}^n \binom{n}{k}f^{(k)}g^{(n-k)}$.

Answer 2

If this can help,

$$(x^2+1)f(x)=\ln(x)$$

$$2xf(x)+(x^2+1)f'(x)=\frac1x$$

$$2f(x)+4xf'(x)+(x^2+1)f''(x)=-\frac1{x^2}$$

and more generally

$$(k+1)(k+2)f^{(k)}(x)+2(k+2)xf^{(k+1)}(x)+(x^2+1)f^{(k+2)}(x)=\frac{(-1)^{k+1}}{x^{k+2}}.$$

Answer 3

Use the Leibniz rule $\def\d{\frac{d^n}{dx^n}}$ $\def\cc#1{\left[#1 - \mathit{c.\!c.}\right]}$ $\def\lnn#1{\underline{\ln}_{#1}}$ $\def\Im{\operatorname{Im}}$ $\def\Re{\operatorname{Re}}$ $\def\im#1{\Im\left( #1 \right)}$

$$\begin{align} (fg)^{(n)} &=\sum_{k=0}^n \binom n k f^{(n-k)} g^{(k)}\\ &=f^{(n)}g + \sum_{k=1}^n \binom n k f^{(n-k)} g^{(k)} \end{align}$$ for the product. For $n\geqslant1$, the $n$-th derivative of $g=\ln$ is: $$ \d\ln x=(-1)^{n+1}\frac{(n-1)!}{x^n} $$ From that we also get the $n$-th derivative of $f(x)=(1+x^2)^{-1}$ as $$ \d\frac1{x^2+1} = (-1)^n n! \Im{\frac{1}{(x-i)^{n+1}}} $$ by noticing that $$ \frac1{x^2+1} = \frac{i}{2}\left(\frac{1}{x+i} - \frac{1}{x-i}\right) = \Im{\frac1{x-i}} $$ Using the imaginary part only works for real $x$, and in the remainder I'll use $\Im$ assuming $x\in\mathbb R$. For complex $x$ it's the same route, just more tedious to write down. Plugging everything into Leibniz' rule, the binomial coefficients cancel out almost entirely and we get

$$ \d\frac{\ln x}{x^2+1} =(-1)^n n! \im{\frac{1}{(x-i)^{n+1}} \left( \ln x - \sum_{k=1}^n \frac 1k \left(1-i/x \right)^k \right)} $$

Writing the sum as incomplete lower Natural logarithm $$ \lnn{n} x = -\sum_{k=1}^n \frac 1k (1-x)^k $$ we get for $x\in\mathbb R$: $$ \d\frac{\ln x}{x^2+1} =(-1)^n n! \Im \frac{\ln x + \lnn{n}(i/x)}{(x-i)^{n+1}} $$

Moreover, this gives an asymptotic formula for $|x|<1$ due to $\lnn{\infty}=\ln$:

$$ \d\frac{\ln x}{x^2+1} \approx \Re \frac{(-1)^n n! \pi}{2(x-i)^{n+1}} \qquad\text{for large }n $$