Proving $\int_0^\infty x^ne^{-tx}\frac{\sin x}xdx=\frac{\sin n\theta}{(1+t^2)^{n/2}}(n-1)!$, where $\theta=\arcsin\frac1{\sqrt{1+t^2}}$

Overall we want to prove that

$$\frac{\sin(n\theta)}{(1+t^2)^{\frac{n}{2}}} (n-1)!=\frac{(n-1)!}{(1+t^2)^n}\left(\frac{i}{2}\left[(t-i)^n-(t+i)^n\right]\right)$$

which boils down to showing that

$$\sin(n\theta)=\frac1{(1+t^2)^{\frac n2}}\left(\frac{i}{2}\left[(t-i)^n-(t+i)^n\right]\right)$$

Hence the variable $\theta$ is defined in terms of an $\arcsin$ function we may recall the logarithmic definition of the inverse sine function aswell as the exponential definition of the sine function given by

\begin{align*} \arcsin(z)&=-i\log(iz+\sqrt{1-z^2})\tag1\\ \sin(z)&=\frac1{2i}(e^{iz}-e^{-iz})\tag2 \end{align*}

We are interested in $\sin(n\theta)$ where $\theta=\arcsin\left(\frac1{\sqrt{1+t^2}}\right)$ therefore we can deduce that

$$\theta=\arcsin\left(\frac1{\sqrt{1+t^2}}\right)=-i\log\left(\frac i{\sqrt{1+t^2}}+\sqrt{1-\frac1{1+t^2}}\right)=-i\log\left(\frac1{\sqrt{1+t^2}}[i+t]\right)$$

This leads us to

$$\sin(n\theta)=\frac1{2i}(e^{in\theta}-e^{-in\theta})=\frac1{2i}\left[\frac{(t+i)^n}{(1+t^2)^{\frac n2}}-\frac{(1+t^2)^{\frac n2}}{(t+i)^n}\right]=\frac1{2i}\left[\frac{(t+i)^n}{(1+t^2)^{\frac n2}}-\frac{(t-i)^n}{(1+t^2)^{\frac n2}}\right]$$

$$\therefore~\sin(n\theta)~=~\frac1{(1+t^2)^{\frac n2}}\left(\frac i2[(t-i)^n-(t+i)^n]\right)$$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\int_{0}^{\infty} x^{n}\expo{-tx}{\sin\pars{x} \over x} \,\dd x = {\sin\pars{n\theta} \over \pars{1 + t^{2}}^{n/2}}\,\pars{n - 1}!}}$ where $\ds{\bbox[5px,#ffd]{\theta \equiv \arcsin\pars{1 \over \root{1 + t^{2}}}}}$

---

\begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty} x^{n}\expo{-tx}{\sin\pars{x} \over x} \,\dd x} = \Im\int_{0}^{\infty} x^{n - 1}\expo{-tx}\pars{\expo{\ic x} - 1}\,\dd x \\[5mm] = &\ \Im\int_{0}^{\infty} x^{n - 1}\, \bracks{\expo{\pars{-t + \ic}x} - \expo{-tx}}\,\dd x = \Im\int_{0}^{\infty} x^{n - 1}\, \sum_{k = 0}^{\infty}{\bracks{\pars{-t + \ic}x}^{k} - \pars{-tx}^{k}\over k!}\,\dd x \\[5mm] = &\ \Im\int_{0}^{\infty} x^{n - 1}\, \sum_{k = 0}^{\infty}\bracks{\pars{t - \ic}^{k} - t^{k}}\,{\pars{-x}^{k} \over k!}\dd x \\[5mm] = &\ \Im\braces{\Gamma\pars{n}\bracks{\pars{t - \ic}^{-n} - t^{-n}}} = \pars{n - 1}!\,\Im\bracks{\pars{t - \ic}^{-n}} \\[5mm] = &\ \pars{n - 1}!\,\Im\braces{\bracks{\root{t^{2} + 1} \exp\pars{\ic\,\arctan\pars{-\,{1 \over t}}}}^{-n}} \\[5mm] = &\ {n! \over \pars{1 + t^{2}}^{n/2}} \sin\pars{n\arctan\pars{1 \over t}} \end{align} Note that $\ds{\arctan\pars{1 \over t} = \theta}$

Then, \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty} x^{n}\expo{-tx}{\sin\pars{x} \over x} \,\dd x} = {\sin\pars{\theta} \over \pars{1 + t^{2}}^{n/2}}\, \pars{n - 1}! \end{align}