求Improper積分值

I will use complex analysis to solve this problem.
Consider the contour integral which constitutes in three parts : L_R the segment from 0 to R, A_R the arc from R to Re^(2πi/n) and L_R' the segment from Re^(2πi/n) to 0.
The integral on the arc A_R tends to 0 as R tends to infinity because the integral is estimated by the sup norm of f multiplied by the length of arc, and because n>=2
| ∫A_R dz / (1+ z^n) | <= R[(2π)/n](B/R^n) -> 0
Consider the poles of 1/(1+z^n) which are e^(kπi/n) (k=0,1,...n-1) and among these poles, only e^(πi/n) is in the interior of the contour ( for large R) and so we only need to calculate the residue of this simple pole.
Using derivative, the residue is (1/n)e^[(-n-1)πi/n]=(-1/n)e^(πi/n)
Now if we parametrizing L_R' by t e^(2πi/n) with 0<=t<=R, we find that
∫L_R' dz / (1+ z^n) = -e^(2πi/n) ∫L_R dz / (1+ z^n)
Taking limit as R->∞ , we have ∫[0~∞] dz / (1+ z^n) = ∫L_R' dz /(1+ z^n) + ∫L_R dz / (1+ z^n)
= [1 - e^(2πi/n)] ∫L_R dx/(1+ x^n)
(2πi)(-1/n)e^(πi/n)= [1 - e^(2πi/n)] ∫L_R dx / (1+ x^n)
[e^(πi/n) - e^(-πi/n)] / (2i) ∫L_R dx/(1+ x^n) = π/n
∫L_R dx / (1+ x^n) = (π/n) / sin(π/n)
∫[0~∞] dx / (1+ x^n) = (π/n) / sin(π/n)

去年我也做過這題數，因為功課有這題。