Proving $\int_0^\infty x^ne^{-tx}\frac{\sin x}xdx=\frac{\sin n\theta}{(1+t^2)^{n/2}}(n-1)!$, where $\theta=\arcsin\frac1{\sqrt{1+t^2}}$ I have stumbled upon the PDF by Leo Goldmakher from University of Toronto, Canada, named Differentiation Under The Integral Sign (PDF link). In that pdf, he gave a theorem (his Theorem 1, at the end of the note) which states as below:

For any real number $t\geq0$ and any integer $n\geq 1$ we have
$$\int_0^\infty x^n e^{-tx}\frac{\sin{x}}{x} dx = \frac{\sin{n\theta}}{(1+t^2)^{\frac{n}{2}}} (n-1)!$$
where $\theta =\arcsin {\frac{1}{\sqrt{1+t^2}}}$

Now I didn't understand how to prove this theorem?
I know
$$\displaystyle\int_0^\infty x^n e^{-tx}\sin{x} \frac{dx}{x} =\frac{(n-1)!}{(1+t^2)^n}\left(\frac{i}{2}\left[(t-i)^n-(t+i)^n\right]\right)$$
The Author  says substituting the above into (1) and tidying up a bit leads to this theorem.
I didn't understand how did he arrive at this theorem?
 A: 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
$$\sin(n\theta)=\frac1{2i}(e^{in\theta}-e^{-in\theta})=\frac1{2i}\left[\frac{(t+i)^n}{(1+t^2)^{\frac n2}}-\frac{(1+t^2)^{\frac n2}}{(t+i)^n}\right]=\frac1{2i}\left[\frac{(t+i)^n}{(1+t^2)^{\frac n2}}-\frac{(t-i)^n}{(1+t^2)^{\frac n2}}\right]$$

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

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