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

Keaton Good

Keaton Good

Answered

2022-01-23

Proving 0xnetxsinxxdx=sinnθ(1+t2)n2(n1)! where θ=arcsin11+t2
Now I didnt

Answer & Explanation

plusmarcacw

plusmarcacw

Expert

2022-01-24Added 10 answers

Overall we want to prove that
sin(nθ)(1+t2)n2(n1)(n1)!(1+t2)n(i2[(ti)n(t+i)n])
which boils down to showing that
sin(nθ)=1(1+t2)n2(i2[(ti)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)=ilog(iz+1z2) (1)
sin(z)=12i(eizeiz) (2)
We are interested in sin(nθ) where θ=arcsin(11+t2) therefore we can deduce that
θ=arcsin(11+t2)=ilog(i1+t2+111+t2)=ilog(11+t2[i+t])
This leads us to
sin(nθ)=12i(eθeθ)=12i[(t+i)n(1+t2)n2(1+t2)n2(t+i)n]=12i[(t+i)n(1+t2)n2(ti)n(1+t2)

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