Keaton Good

2022-01-23

Proving ${\int }_{0}^{\mathrm{\infty }}{x}^{n}{e}^{-tx}\frac{\mathrm{sin}x}{x}dx=\frac{\mathrm{sin}n\theta }{{\left(1+{t}^{2}\right)}^{\frac{n}{2}}}\left(n-1\right)!$ where $\theta =\mathrm{arcsin}\frac{1}{\sqrt{1+{t}^{2}}}$
Now I didnt

plusmarcacw

Expert

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