Proving ∫0∞xne−txsin⁡xxdx=sin⁡nθ(1+t2)n2(n−1)! where θ=arcsin⁡11+t2 Now I didnt

Overall we want to prove that
${\displaystyle \frac{\sin \left(n\theta \right)}{{(1+t^2)}^{\frac{n}{2}}}(n-1)\ne \frac{(n-1)!}{{(1+t^2)}^n}\left(\frac{i}{2}[{(t-i)}^n-{(t+i)}^n]\right)}$
which boils down to showing that
${\displaystyle \sin \left(n\theta \right)=\frac{1}{{(1+t^2)}^{\frac{n}{2}}}\left(\frac{i}{2}[{(t-i)}^n-{(t+i)}^n]\right)}$
Hence the variable ${\displaystyle \theta }$ is defined in terms of an ${\displaystyle \arcsin }$ function we may recall the logarithmic definition of the inverse sine function aswell as the exponential definition of the sine function given by
${\displaystyle \arcsin \left(z\right)=-i\log (iz+\sqrt{1-z^2})}$ (1)
${\displaystyle \sin \left(z\right)=\frac{1}{2i}(e^{iz}-e^{-iz})}$ (2)
We are interested in ${\displaystyle \sin \left(n\theta \right)}$ where ${\displaystyle \theta =\arcsin \left(\frac{1}{\sqrt{1+t^2}}\right)}$ therefore we can deduce that
${\displaystyle \theta =\arcsin \left(\frac{1}{\sqrt{1+t^2}}\right)=-i\log (\frac{i}{\sqrt{1+t^2}}+\sqrt{1-\frac{1}{1+t^2}})=-i\log \left(\frac{1}{\sqrt{1+t^2}}[i+t]\right)}$
This leads us to

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