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$$\int_2^\infty \frac{4x^3+3x^2-x}{5x^5-2x^4+x^2-2}\ln x\, dx$$

How should I approach this? can I look at $$\int_2^\infty \frac{\ln x}{x^2}\,dx \text{?}$$

Michael Hardy
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newhere
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1 Answers1

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The integrand function is bounded between $\frac{4\log x}{5 x^2}$ and $\frac{84\log x}{65 x^2}$ on the interval $(2,+\infty)$, hence the integral is converging since $$ \int_{2}^{+\infty}\frac{\log x}{x^2}\,dx = \frac{1+\log 2}{2}.$$

Jack D'Aurizio
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    How do you find the bounded functions? – newhere Aug 09 '16 at 18:19
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    @newhere: by just studying $\frac{4x^5+3x^4-x^3}{5x^5-2x^4+x^2-2}$ over $(2,+\infty)$, but that constants are not that important. The integrand function is asymptotically equivalent to $\frac{\log x}{x^2}$ that is an integrable function, so the original integral is convergent. – Jack D'Aurizio Aug 09 '16 at 18:22
  • Great answer! I must press further though... How did you get the upper bound $\frac{84\log x}{65 x^2}$? I don't see an obvious way to show this, though I must be missing something blatant. I can't see how you got those coefficients – Brevan Ellefsen Aug 09 '16 at 18:32
  • @BrevanEllefsen: the previous function in my comment is decreasing on $(2,+\infty)$, the constant $\frac{84}{65}$ just comes from evaluating it at $x=2$. – Jack D'Aurizio Aug 09 '16 at 18:36
  • @JackD'Aurizio ah, that makes perfect sense. Thank you for the clarification! – Brevan Ellefsen Aug 09 '16 at 18:37