$I_n(t,a) = \int_0^\infty \frac{\cos(xt)}{\left(x^2 + a^2\right)^n}\:dx$ Spurred on by this, here I'm hoping to resolve the following integral:
\begin{equation}
I_n(a,t) = \int_0^\infty \frac{\cos(xt)}{\left(x^2 + a^2\right)^n}\:dx
\end{equation}
Where $a,t \in \mathbb{R}^+$ and $n \in \mathbb{N}$. To begin with we observe that:
\begin{equation}
I_n(a,t) =  \int_0^\infty \frac{\cos(xt)}{\left(a^2\left(\frac{x^2}{a^2} + 1\right)\right)^n}\:dx = \frac{1}{a^{2n}}  \int_0^\infty \frac{\cos(xt)}{\left(\left(\frac{x}{a}\right)^2 + 1\right)^n}\:dx
\end{equation}
Let $u = \frac{x}{a}$:
\begin{align}
I_n(a,t) &= \frac{1}{a^{2n}} \int_0^\infty \frac{\cos(uat)}{\left(u^2 + 1\right)^n}\cdot a\:du = a^{1 - 2n}\int_0^\infty \frac{\cos(uat)}{\left(u^2 + 1\right)^n}\:du  \\
&=a^{1 - 2n}I_n(1, at)
\end{align}
Thus, we need only resolve the following integral to solve $I_n(a,t)$:
\begin{equation}
J_n(s) = \int_0^\infty \frac{\cos(su)}{\left(u^2 + 1\right)^n}\:du
\end{equation}
Noting $I_n(a,t) = J_n(at)$. Here we will proceed by forming a differential equation for $J_n(s)$. To do so, we employ Leibniz's Integral Rule and differentiate under the curve twice w.r.t $s$:
\begin{align}
\frac{d^2J_n}{ds^2} &= \int_0^\infty \frac{-u^2\cos(su)}{\left(u^2 + 1\right)^n}\:du = -\int_0^\infty \frac{\left(u^2 + 1 - 1\right)\cos(su)}{\left(u^2 + 1\right)^n}\:du \nonumber \\
&=-\left[\int_0^\infty \frac{\cos(su)}{\left(u^2 + 1\right)^{n - 1}}\:du -  \int_0^\infty \frac{\cos(su)}{\left(u^2 + 1\right)^n}\:du\right] \nonumber \\
&=-\left[J_{n - 1}(s) - J_n(s) \right] = J_n(s) - J_{n - 1}(s)
\end{align}
Thus we form the recursive differential equation:
\begin{equation}
\frac{d^2J_n}{ds^2}- J_n(s) = -J_{n - 1}(s)
\end{equation}
In order for a solution to be obtained, the following is required: $I_1(s)$, $I_n(0)$, and $I_n'(0)$. Thankfully these are all easy to obtain. Starting with $I_1(s)$ we find:
\begin{equation}
I_n(s) = \frac{\pi}{2}e^{-s}
\end{equation}
For $I_n(0)$ we have:
\begin{equation}
I_n(0) = \int_0^\infty \frac{1}{\left(u^2 + 1\right)^n}\:du
\end{equation}
Using the subsitution $u = \tan(w)$ we obtain a solution in terms of the Beta (and by extension Gamma) function:
\begin{align}
I_n(0) &= \int_0^\frac{\pi}{2} \frac{1}{\left(\tan^2(w) + 1\right)^n}\cdot \sec^2(w)\:dw = \int_0^\frac{\pi}{2} \cos^{2n - 2}(w)\:dw \nonumber \\
&= \frac{1}{2}B\left( \frac{2n - 1}{2}, \frac{1}{2} \right) = \frac{1}{2}\frac{\Gamma\left(\frac{2n - 1}{2}\right)\Gamma\left( \frac{1}{2} \right)}{\Gamma\left(\frac{2n - 1}{2} + \frac{1}{2} \right)} = \frac{\sqrt{\pi}}{2}\frac{\Gamma\left(\frac{2n - 1}{2}\right)}{\Gamma(n)}
\end{align}
For $I_n'(0)$ we have:
\begin{equation}
I_n'(0) = \int_0^\infty \frac{-x\sin(x \cdot 0)}{\left(x^2 + 1\right)^n} = 0
\end{equation}
Now, and here is where I'm unsure about my process - for our recursive differential equation we take the Laplace Transform:
\begin{align}
\mathscr{L}_{s \rightarrow p}\left[ \frac{d^2J_n}{ds^2} \right] - \mathscr{L}_{s \rightarrow p}\left[J_n(s) \right] &= -\mathscr{L}_{s \rightarrow p}\left[ J_{n - 1}(s) \right] \nonumber \\
p^2 \overline{J}_n(p) - pJ_n(0) - J_n'(0) - \overline{J}_{n}(p) &= -\overline{J}_{n - 1}(p) \nonumber \\ 
\left(p^2 - 1\right)\overline{J}_n(p) &= pJ_n(0) -\overline{J}_{n - 1}(p)
\end{align}
Thus,
\begin{equation}
\overline{J}_n(p) = \frac{p}{p^2 - 1} J_n(0) - \frac{1}{p^2 - 1}\overline{J}_{n - 1}(p)
\end{equation}
We now take the Inverse Laplace Transform:
\begin{align}
\mathscr{L}_{p \rightarrow s}^{-1} \left[\overline{J}_n(p)\right] &= \mathscr{L}_{p \rightarrow s}^{-1} \left[\frac{p}{p^2 - 1}\right]J_n(0)  - \mathscr{L}_{p \rightarrow s}^{-1} \left[\frac{1}{p^2 - 1}\overline{J}_{n - 1}(p)\right] \nonumber \\
J_n(s) &= J_n(0)\cosh(s) - \int_0^s \sinh(s - a)J_{n - 1}(a)\:da \nonumber \\
 &= J_n(0)\cosh(s) - \int_0^s \left[\sinh(s)\cosh(a) - \sinh(a)\cosh(s)\right]J_{n - 1}(a)\:da \nonumber \\
&= J_n(0)\cosh(s) - \sinh(s)\int_0^s\cosh(a) J_{n - 1}(a)\:da \nonumber \\
&\quad+ \cosh(s)\int_0^2 \sinh(a)J_{n - 1}(a)\:da
\end{align}
Now whilst we have a recursive integral form that governs $J_n(s)$ I am unsure how to solve it!.
Does anyone have any pointers about how to move forward?

Another approach (I believe) is to employ the linear D-operator. Here if we define $D = \frac{d}{ds}$ then our governing differential equation is given by:
\begin{equation}
\left(D - 1\right)\left(D + 1\right)\left[ J_{n}(s)\right] = -J_{n - 1}(s)
\end{equation}
Thus,
\begin{equation}
J_n(s) = -\left(\left(D - 1\right)\left(D + 1\right)\right)^{-1}\left[ J_{n-1}(s)\right]
\end{equation}
Which is my reasoning is correct implies that
\begin{align}
J_n(s) &= (-1)^n \left(\left(D - 1\right)\left(D + 1\right)\right)^{-(n - 1)}\left[ J_1(s)\right] = (-1)^n \left(\left(D - 1\right)\left(D + 1\right)\right)^{-(n - 1)}\left[ \frac{\pi}{2}e^{-s}\right] \nonumber \\
&= (-1)^n \frac{\pi}{2} \left(\left(D - 1\right)\left(D + 1\right)\right)^{-(n - 1)}\left[ e^{-s}\right]
\end{align}
 A: Start with the result (link):
$$\int_{0}^{\infty }{\frac{\cos \left( su \right)}{\left( {{u}^{2}}+p \right)}du}=\frac{\pi {{e}^{-s\sqrt{p}}}}{2\sqrt{p}}$$
Differentiating both sides $n-1$ times (w.r.t  $p$)
$$\int_{0}^{\infty }{\frac{\left( n-1 \right)!{{\left( -1 \right)}^{n-1}}\cos \left( su \right)}{{{\left( {{u}^{2}}+p \right)}^{n}}}du}=\frac{{{d}^{n-1}}}{d{{p}^{n-1}}}\left( \frac{\pi {{e}^{-s\sqrt{p}}}}{2\sqrt{p}} \right)$$
Setting $p=1$
$$\int_{0}^{\infty }{\frac{\cos \left( su \right)}{{{\left( {{u}^{2}}+1 \right)}^{n}}}du}=\frac{1}{{{\left( -1 \right)}^{n-1}}\left( n-1 \right)!}{{\left[ \frac{{{d}^{n-1}}}{d{{p}^{n-1}}}\left( \frac{\pi {{e}^{-s\sqrt{p}}}}{2\sqrt{p}} \right) \right]}_{p=1}}$$
Note that the integral in question is indeed an integral representation(see equation 5 here) of the Modified Bessel Function of the Second Kind ${{K}_{n}}\left( s \right)$ which is a solution to the Modified Bessel Differential Equation. After some research in special functions text-books I have found that almost (if not all authors) use Complex analysis methods to evaluate it , that’s why  I strongly believe that  forming a differential equation to find the integral is not an accessible method!!! and by the way here is the value of the integral in terms of speatial functions :
$$\frac{\sqrt{\pi }{{2}^{\frac{1}{2}-n}}{{K}_{\frac{1}{2}-n}}\left( s \right)}{{{s}^{\frac{1}{2}-n}}\Gamma \left( n \right)}$$
