Improper integral $\int _3^{\infty }\:\frac{\sqrt{x^3+2}}{\left(x^2-3x\right)^\phi}$

Question

$$\gamma = \:\frac{\sqrt{x^3+2}}{\left(x^2-3x\right)^\phi}$$ $$\int _3^{\infty }\gamma$$ I have to find for whom the $\phi > 0$ the improper integral converges, but i'm not very familiar with this type of integral. I tried like that:

$$\gamma \approx_{\infty }\:\frac{\sqrt{x^3}}{x^{2\phi}}\rightarrow \int _3^{\infty }\:\frac{\sqrt{x^3}}{x^{2\phi}} = \int _3^{\infty }\:\frac{1}{x^{\frac{4\phi-3}{2}}}$$ That converges for $\color{red}{\phi > 5/4}$
Then

$$\gamma \approx_3 \frac{1}{\left(x^2-3x\right)^\phi} \rightarrow \int _3^{\infty }\:\frac{1}{\left(x^2-3x\right)^\phi}$$ But i don't know how to solve it, can you give me a clue as to how to continue? Thanks

Answer 1

Hint. As $x \to 3^+$, one has $$ \frac{\sqrt{x^3+2}}{\left(x^2-3x\right)^\phi}= \frac{\sqrt{x^3+2}}{x^\phi(x-3)^\phi}\sim \frac{\sqrt{29}}{3^\phi}\cdot \frac1{(x-3)^\phi} $$ this integrand is convergent near $x=3^+$ iff $\phi<1$.

Answer 2

You're on the right way:

We have at $\infty$

$$\frac{\sqrt{x^3+2}}{(x^2-3x)^\phi}\sim_\infty \frac{1}{x^{2\phi-\frac32}}$$ so to insure the convergence at $\infty$ we should have $2\phi-\frac32>1\iff\phi>\frac54$

Moreover, let $t=x-3$ so $(x^2-3x)^\phi=(t^2+6t+9-3t-9)^\phi=(3t+t^2)^\phi\sim_03^\phi t^\phi$ so the integral converges at $0$ if and only if $\phi<1$. Since we can't have $\phi>\frac54$ and $\phi<1$ at same time then the given integral diverges.