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[hide] Question 09
Decide whether the following improper integral converges or diverges: $${\begin{aligned}\int _{1}^{\infty }{\frac {x^{2}+1}{x^{3}+2}}\mathrm {d} x.\end{aligned}}$$ (Note that the value of this integral is not requested.)

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1

For a more elementary solution that doesn't use the limit comparison test, consider the following rewriting of the integrand: $${\frac {x^{2}+1}{x^{3}+2}}={\frac {x^{2}}{x^{3}+2}}+{\frac {1}{x^{3}+2}}={\frac {1}{x+{\frac {2}{x^{2}}}}}+{\frac {1}{x^{3}+2}}.$$ Can you use this to show that the integrand is bounded by ${\frac {K}{x}}$ for some $K$? Alternatively, the second solution applies the limit comparison test.

[show] Hint 2

Note first that, since the denominator does not vanish on the interval $[1,\infty )$, the integral is improper because one of the integration bounds is infinity (as opposed to there being a singular point in the domain of integration). Now the integrand can be approximated as follows for large $x$: $${\frac {x^{2}+1}{x^{3}+2}}\approx {\frac {x^{2}}{x^{3}}}={\frac {1}{x}}.$$ Can you use this approximation together with the Limit Comparison test?

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[show] Solution 1

Following the hint, we have $${\frac {x^{2}+1}{x^{3}+2}}={\frac {1}{x+{\frac {2}{x^{2}}}}}+{\frac {1}{x^{3}+2}}.$$ Given that the first term in the sum above looks close to ${\frac {1}{x}}$, and we know that $\int _{1}^{\infty }{\frac {1}{x}}\;\mathrm {d} x$ diverges, we should try to prove that the integral above diverges as well. To do this, let us try to bound the integrand below by a function that diverges. Since $x\in [1,\infty ),$ we have $${\frac {1}{x+{\frac {2}{x^{2}}}}}+{\frac {1}{x^{3}+2}}>{\frac {1}{x+{\frac {2}{x^{2}}}}}.$$ If we could say next that ${\frac {1}{x+{\frac {2}{x^{2}}}}}>{\frac {1}{x}}$, we would be done. But, as we can see by substituting some values for $x$ this is not true! However, if we can show that ${\frac {1}{x+{\frac {2}{x^{2}}}}}>{\frac {K}{x}}$, for some $K>0$, we are still ok, since the integral $\int _{1}^{\infty }{\frac {K}{x}}\;\mathrm {d} x$ diverges as well. To achieve this, we should make sure that ${\frac {2}{x^{2}}}$ is not too large, so that the ratio ${\frac {1}{x+{\frac {2}{x^{2}}}}}$ is not much smaller than ${\frac {1}{x}}$. Since $x\geq 1$, we have ${\frac {2}{x^{2}}}\leq {\frac {2}{1}}<2x,$ so we find $${\frac {1}{x+{\frac {1}{x^{2}}}}}\geq {\frac {1}{x+2x}}.$$ Thus we have shown that the integrand ${\frac {x^{2}+1}{x^{3}+2}}$ is bounded below by ${\frac {K}{x}}$, with $K={\frac {1}{3}}$, on the domain of integration. Since the integral of the lower bound, $$\int _{1}^{\infty }{\frac {1}{3x}}\;\mathrm {d} x$$ diverges, so does the integral $$\int _{1}^{\infty }{\frac {x^{2}+1}{x^{3}+2}}\;\mathrm {d} x$$

[show] Solution 2

Here is a solution using the limit comparison test, Theorem 1.12.22 in CLP2 . We expect the integrand to behave like ${\frac {1}{x}}$, whose integral from 1 to $\infty$ diverges, so we apply the limit comparison test with $f(x)={\frac {x^{2}+1}{x^{3}+2}}$ and $g(x)={\frac {1}{x}}$ (notation as in the theorem statement in CLP). We compute then $$\lim _{x\to \infty }{\frac {f(x)}{g(x)}}=\lim _{x\to \infty }{\frac {\frac {x^{2}+1}{x^{3}+2}}{\frac {1}{x}}}=\lim _{x\to \infty }{\frac {x^{2}+1}{x^{3}+2}}\cdot x=\lim _{x\to \infty }{\frac {x^{3}+x}{x^{3}+2}}=\lim _{x\to \infty }{\frac {1+1/x^{2}}{1+2/x^{3}}}={\frac {1+0}{1+0}}.$$ Since the above limit exists and is non-zero, the improper integral $\int _{1}^{\infty }f(x)\mathrm {d} x$ diverges by the limit comparison test (b).