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Check whether $$\int_{-\infty}^{\infty}\frac1{\sqrt{x^{10}+2}}\ dx$$ converges or diverges.

I used the Limit Comparison Test with the function $\frac1{x^5}$: $$\lim_{x\to\infty}\frac{\sqrt{x^{10}+2}}{x^5}=1$$ Also, $$\int_{-\infty}^{\infty}\frac1{x^5}\ dx=\infty$$ Therefore, it means that:$$\int_{-\infty}^{\infty}\frac1{\sqrt{x^{10}+2}}\ dx=\infty$$ However, the latter integral converges. What is wrong with this approach?

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  • $\begingroup$ $x^{-5}$ has bad behaviour at $0$, whereas $\frac{1}{\sqrt{x^{10} + 2}}$ does not. You need to take this into account. $\endgroup$ – AlohaSine Feb 18 '17 at 22:21
  • $\begingroup$ @TrevorNorton Actually my textbook states that it is used for integrals. $\endgroup$ – user372003 Feb 18 '17 at 22:22
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    $\begingroup$ The limit comparison test for improper integrals states that if $f$ and $g$ are continuous on the domain.......here we do not have that. @Denis $\endgroup$ – Ahmed S. Attaalla Feb 19 '17 at 0:27
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Hint The equation for $\int_{-\infty}^{\infty} \frac{dx}{x^5}$ is incorrect, but all of the trouble with this integral is near the origin in the sense that, for example, $$\int_1^{\infty} \frac{dx}{x^5} = \frac{1}{4} < \infty .$$

Additional hint This suggests writing $$\int_{-\infty}^{\infty} \frac{dx}{\sqrt{x^{10} + 2}} = \int_{-\infty}^{-1} \frac{dx}{\sqrt{x^{10} + 2}} + \int_{-1}^{1} \frac{dx}{\sqrt{x^{10} + 2}} + \int_{1}^{\infty} \frac{dx}{\sqrt{x^{10} + 2}}.$$

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Note that integrals can diverge at a finite point, while summations diverge at infinity for the most part. This is why the limit comparison test doesn't make sense for integrals. What you have proven, though, is that there exists $X,c$ such that

$$\int_X^\infty\frac1{\sqrt{x^{10}+2}}\ dx<c\int_X^\infty\frac1{x^5}\ dx$$

Unfortunately, $X>0$, so this argument doesn't extend to your integral, but you can note that

$$\int_0^\infty\frac1{\sqrt{x^{10}+2}}\ dx=\int_0^X\frac1{\sqrt{x^{10}+2}}\ dx+\int_X^\infty\frac1{\sqrt{x^{10}+2}}\ dx$$

Leaving the only remaining part to showing the integral is finite over $[0,X]$, and then doubling to reach $\mathbb R$.

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