γ=√x3+2(x2−3x)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:
γ≈∞√x3x2ϕ→∫∞3√x3x2ϕ=∫∞31x4ϕ−32
That converges for ϕ>5/4
Then
γ≈31(x2−3x)ϕ→∫∞31(x2−3x)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 x→3+, one has
√x3+2(x2−3x)ϕ=√x3+2xϕ(x−3)ϕ∼√293ϕ⋅1(x−3)ϕ this integrand is convergent near x=3+ iff ϕ<1.
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