∫(0 to ∞) sin x^p dx

∫[0~∞] sin(x^p)dx (t=x^p, p>1)
=(1/p)∫[0~∞] t^(1/p -1) sint dt=(1/p)∫[0~∞] x^(1/p -1) sinx dx
=1/[pΓ(1-1/p)] ∫[0~∞] t^(-1/p)exp(-xt)sinx dt dx
=1/[pΓ(1-1/p)] ∫[0~∞] t^(-1/p)/(1+t^2) dt (L{sinx}= 1/(s^2+1) )
=1/[pΓ(1-1/p)] ∫[0~π/2] (cosx/sinx)^(1/p) dx
=1/[2pΓ(1-1/p)] B(1-a, a) (a= 0.5(1+1/p) )
=Γ(1/p]/[2pΓ(1/p)Γ(1-1/p)]* π/sin(aπ)
=Γ(1/p)*sin(π/p) /(2πp)* π/cos(π/2p)
=Γ(1/p)sin(π/2p)/p

我這題第一個印象是走扇形路徑積分

不過我沒有去算

不曉得扇形可不可以積出來