✔ 最佳答案
1. 令 x=t^2, then
原式=∫[0~∞] 2t^2/(t^6+1) dt
=∫[-∞~∞] x^2/(x^6+1) dx
= ∫_c x^2/(x^6+1) dx (c為上半平面之"外圍")
又x^6+1=(x-w1)(x-w2)...(x-w6)
w1=cos(π/6)+i sin(π/6)
w2=cos(π/2)+i sin(π/2)
w3=cos(5π/6)+i sin(5π/6) (w1~w3在上半面,w4~w6在下半面)
lim(x->w) x^2*(x-w)/(x^6+1) (w=w1, w2, w3)
= w^2 /(6w^5)=1/(6w^3)=-w^3/6 (因 x^6=-1)
故原式=2πi*(-1/6)[w1^3+w2^3+w3^3]
=(-πi/3)*[ i - i + i ]=π/3
2.
∫[0~2π] cos(nx)/[1-2acosx + a^2] dx
=∫[-π~π] cos(nx)/(1-2a cosx +a^2) dx (週期 2π)
=∫[-π~π] [cos(nx)+i sin(nx)]/(1-2a cosx+a^2) dx (sin(nx)為奇函數)
=∫[0~2π] [exp(ix)]^n /{1+a^2-a[ exp(ix)+exp(-ix)] } dx
=∫_c z^n/[1+a^2-a(z+1/z)] *(-i/z) dz (c為 x^2+y^2=1逆時針)
= i* ∫_c z^n/[az^2-(1+a^2)z + a] dz
= i*∫_c z^n/[(z-a)(az-1)] dz
= i*2πi*{Residue of z^n/[(z-a)(az-1)] of the pole z=a} (Note: 1/a>1)
= -2π* a^n /(a^2-1)
= 2π a^n/(1-a^2)
2010-06-16 22:59:05 補充：
原式=∫[0~∞] 2t^2/(t^6+1) dt=∫[-∞~∞] t^2/(t^6+1) dt
=∫[-∞~∞] x^2/(x^6+1) dx (定積分的積分變數可任意更換,dummy variable)
2010-06-18 19:30:11 補充：
2x^2/(x^6+1)是偶函數,故∫[-∞~∞] x^2/(x^6+1) dx
= 2*∫[0~∞] x^2/(x^6+1) dx = ∫[0~∞] 2x^2/(x6+1) dx
