<s>[INST] Improper integral $\int_0^\infty\frac{x^{2n+1}}{(1+x^2)^2(2+x^2)^N}dx,\ \ \ n\le N$ How can I evaluate this integral?
$$\int_0^\infty\frac{x^{2n+1}}{(1+x^2)^2(2+x^2)^N}dx,\ \ \ n\le N$$
Maybe there is a recurrence relation for the integral?
 [/INST] Let $I_{n,N}$ denote the integral in question, where $n$ and $N$ are nonnegative integers such that $n\le N$. With the change of variables $y=2+x^2$ and then using the binomial expansion of $(y-2)^n$, we get
\begin{equation}
    2I_{n,N}=\int_2^\infty\frac{(y-2)^n\,dy}{(y-1)^2y^N}
    =\sum_{j=0}^n\binom nj (-2)^{n-j}J_{N-j},
\end{equation}
where
\begin{equation}
    J_k:=\int_2^\infty\frac{y^{-k}\,dy}{(y-1)^2}. 
\end{equation}
Integrating by parts, we get
\begin{equation}
    J_k=2^{-k}-kM_{k+1},
\end{equation}
where
\begin{equation}
    M_k:=\int_2^\infty\frac{y^{-k}\,dy}{y-1}=\int_0^{1/2}(1-t)^{-1}t^{k-1}\,dt,
\end{equation}
using $t=1/y$. Further,
\begin{align*}
    M_k&=\int_0^{1/2}[(1-t)+t](1-t)^{-1}t^{k-1}\,dt \\ 
    &=\int_0^{1/2}t^{k-1}\,dt+\int_0^{1/2}(1-t)^{-1}t^k\,dt  \\ 
    &=\frac1{k2^k}+M_{k+1}. 
\end{align*}
So,
\begin{align*}
    2I_{n,N}&=\sum_{j=0}^n\binom nj (-2)^{n-j}2^{j-N} \\ 
&   -\sum_{j=0}^n\binom nj (-2)^{n-j}(N-j)M_{N-j+1} \\ 
&   =2^{-N}1(n=0)-\sum_{j=0}^n\binom nj (-2)^{n-j}(N-j)M_{N-j+1}.
\end{align*}
Thus,
\begin{align*}
    I_{n,N}&=2^{-N-1}1(n=0)+\sum_{j=0}^n\binom nj (-2)^{n-j-1}(N-j)M_{N-j+1}, \tag{1}
\end{align*}
and the $M_k$'s are given by the simple recurrence
\begin{equation}
    M_{k+1}=M_k-\frac1{k2^k} \tag{2}
\end{equation}
for natural $k$,
with $M_1=\ln2$.

From this, one can also get a (double) recurrence for $I_{n,N}$. Indeed, by (1),
\begin{equation}
    I_{n,N}=2^{-N-1}1(n=0)+\sum_{j=0}^\infty a_{j,n}L_{N-j}, 
\end{equation}
where
\begin{equation}
    a_{j,n}:=\binom nj (-2)^{n-j-1},\quad L_k:=kM_{k+1}.
\end{equation}
From $\binom nj=\binom{n-1}{j-1}+\binom{n-1}j$, we get
\begin{equation}
    a_{j,n}=a_{j-1,n-1}-2a_{j,n-1}. 
\end{equation}
So,
\begin{align*}
    I_{n,N}&=\sum_{j=0}^\infty a_{j-1,n-1}L_{N-j}-2\sum_{j=0}^\infty a_{j,n-1}L_{N-j} \\  
    &=\sum_{i=-1}^\infty a_{i,n-1}L_{N-1-i}-2\sum_{j=0}^\infty a_{j,n-1}L_{N-j} \\  
    &=I_{n-1,N-1}-2I_{n-1,N} \tag{3}
\end{align*}
if $1\le n\le N$, and (by (1)) $I_{0,N}=2^{-N-1}-NM_{N+1}/2$, with the $M_k$'s given by recurrence (2).
This conclusion has been verified for a few pairs $(n,N)$.
</s>