Saturday, 4 June 2016

calculus - Improper integral intinfty3:fracsqrtx3+2left(x23xright)phi



γ=x3+2(x23x)ϕ
3γ

I have to find for whom the ϕ>0 the improper integral converges, but i'm not very familiar with this type of integral. I tried like that:



γx3x2ϕ3x3x2ϕ=31x4ϕ32
That converges for ϕ>5/4
Then



γ31(x23x)ϕ31(x23x)ϕ
But i don't know how to solve it, can you give me a clue as to how to continue? Thanks


Answer



Hint. As x3+, one has
x3+2(x23x)ϕ=x3+2xϕ(x3)ϕ293ϕ1(x3)ϕ this integrand is convergent near x=3+ iff ϕ<1.


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