<s>[INST] The closed form of $\int_{0}^{\infty} \frac{\cos x}{\left(x^{2}+1\right)^{n}} d x$ We are going to deal with the integral
$$
I_{n}=\int_{0}^{\infty} \frac{\cos x}{\left(x^{2}+1\right)^{n}} d x
$$
by differentiation on its related integral
$$
J(a)=\int_{0}^{\infty} \frac{\cos x}{x^{2}+a} d x,
$$
where $a>0$
Using contour integration along anti-clockwise direction of the path
$$\gamma=\gamma_{1} \cup \gamma_{2} \textrm{ where }  \gamma_{1}(t)=t+i 0(-R \leq t \leq R)  \textrm{ and } \gamma_{2}(t)=R e^{i t}  (0<t<\pi) $$
$$
\begin{aligned}
J(a) &=\frac{1}{2} \int_{-\infty}^{\infty} \frac{\cos x}{x^{2}+a} d x \\
&=\frac{1}{2} \operatorname{Re} \int_{\gamma} \frac{e^{i z}}{z^{2}+a} d z \\
&=\frac{1}{2} \operatorname{Re}\left[2 \pi i \operatorname{Res}\left(\frac{e^{i z}}{z^{2}+a}, \sqrt{a} i\right)\right] \\
&=\frac{1}{2} \operatorname{Re}\left[2 \pi i \frac{e^{i(\sqrt{a} i)}}{2 \cdot \sqrt{a} i}\right] \\
&=\frac{1}{2} \operatorname{Re}\left[\frac{\pi}{\sqrt{a}} e^{-\sqrt{a}}\right] \\
&=\frac{\pi}{2 \sqrt{a}} e^{-\sqrt{a}}
\end{aligned}
$$
Differentiating both sides w.r.t. $a$ by $(n-1)$ times yields
$$
\left.I_{n}=\frac{\pi}{(n-1)!}J^{(n-1)}(1)=\frac{(-1)^{n-1} \pi}{2(n-1) !} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{e^{-\sqrt{a}}}{\sqrt{a}}\right)\right|_{a=1}
$$
Noting that
$$
\frac{d}{d a}\left(e^{-\sqrt{a}}\right)=-\frac{1}{2 \sqrt{a}} e^{-\sqrt{a}},
$$
we have $$
I_{n}=\left.\frac{(-1)^{n} \pi}{(n-1) !} \cdot \frac{d^{n}}{d a^{n}}\left(e^{-\sqrt{a}}\right)\right|_{a=1}
$$
By the formula, we can find
$$
I_{1}=\frac{\pi}{2e};I_{2}=\frac{\pi}{2 e};
I_{3}=\frac{7 \pi}{16 e} ;
I_{4}=\frac{37 \pi}{96e} ;
I_{5}=\frac{133 \pi}{384 e};\cdots\ 
I_{10}=\frac{12994393 \pi}{53084160 e}
$$
Suggestions on how to deal with the last derivative is highly appreciated.
 [/INST] From this answer of mine: $$I_{n+1}=\frac{\pi/e}{2n!}\sum_{k=0}^n\frac{2^{-n-k}(n+k)!}{k!(n-k)!}=\frac{\pi/e}{2^{2n+1}n!}\sum_{k=0}^n\frac{2^k(2n-k)!}{k!(n-k)!}.$$
</s>