请问这个反常积分怎么算出收敛值?
2个回答
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2x^2-2x+1 = 2(x -1/2)^2 +1/2
let
x -1/2 = (1/2)tanu
dx= (1/2)(secu)^2 du
x=1, u=π/4
x=+∞ , u=π/2
∫(1->+∞) dx/[x.√(2x^2-2x+1)]
=∫(π/4->π/2) (1/2)(secu)^2 du/{ [(1/2+(1/2)tanu] .(1/√2)secu }
=√2.∫(π/4->π/2) secu /(1+tanu) du
=√2.∫(π/4->π/2) 1/(cosu+sinu) du
=∫(π/4->π/2) 1/sin(u+π/4) du
=∫(π/4->π/2) csc(u+π/4) du
=[ln|csc(u+π/4) -cot(u+π/4)| ]|(π/4->π/2)
=ln(√2+1)
let
x -1/2 = (1/2)tanu
dx= (1/2)(secu)^2 du
x=1, u=π/4
x=+∞ , u=π/2
∫(1->+∞) dx/[x.√(2x^2-2x+1)]
=∫(π/4->π/2) (1/2)(secu)^2 du/{ [(1/2+(1/2)tanu] .(1/√2)secu }
=√2.∫(π/4->π/2) secu /(1+tanu) du
=√2.∫(π/4->π/2) 1/(cosu+sinu) du
=∫(π/4->π/2) 1/sin(u+π/4) du
=∫(π/4->π/2) csc(u+π/4) du
=[ln|csc(u+π/4) -cot(u+π/4)| ]|(π/4->π/2)
=ln(√2+1)
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