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

Keaton Good

Answered question

2022-01-23

Proving

\int_{0}^{\infty} x^{n} e^{- t x} \frac{\sin x}{x} d x = \frac{\sin n θ}{{(1 + t^{2})}^{\frac{n}{2}}} (n - 1)!

where

θ = \arcsin \frac{1}{\sqrt{1 + t^{2}}}

Now I didnt

plusmarcacw

Beginner2022-01-24Added 10 answers

Overall we want to prove that

\frac{\sin (n θ)}{{(1 + t^{2})}^{\frac{n}{2}}} (n - 1) \neq \frac{(n - 1)!}{{(1 + t^{2})}^{n}} (\frac{i}{2} [{(t - i)}^{n} - {(t + i)}^{n}])

which boils down to showing that

\sin (n θ) = \frac{1}{{(1 + t^{2})}^{\frac{n}{2}}} (\frac{i}{2} [{(t - i)}^{n} - {(t + i)}^{n}])

Hence the variable

θ

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

\arcsin (z) = - i \log (i z + \sqrt{1 - z^{2}})

(1)

\sin (z) = \frac{1}{2 i} (e^{i z} - e^{- i z})

(2)
We are interested in

\sin (n θ)

where

θ = \arcsin (\frac{1}{\sqrt{1 + t^{2}}})

therefore we can deduce that

θ = \arcsin (\frac{1}{\sqrt{1 + t^{2}}}) = - i \log (\frac{i}{\sqrt{1 + t^{2}}} + \sqrt{1 - \frac{1}{1 + t^{2}}}) = - i \log (\frac{1}{\sqrt{1 + t^{2}}} [i + t])

This leads us to

\sin (n θ) = \frac{1}{2 i} (e^{\in θ} - e^{- \in θ}) = \frac{1}{2 i} [\frac{{(t + i)}^{n}}{{(1 + t^{2})}^{\frac{n}{2}}} - \frac{{(1 + t^{2})}^{\frac{n}{2}}}{{(t + i)}^{n}}] = \frac{1}{2 i} [\frac{{(t + i)}^{n}}{{(1 + t^{2})}^{\frac{n}{2}}} - \frac{{(t - i)}^{n}}{{(1 + t^{2})}^{}}

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