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I have the following integral: $$I=\int_0^\infty\frac{x}{x^5+5}dx$$ I have to use the comparison theorem to tell if it's convergent or divergent.

I have tried using $\frac{x}{x^5}=\frac{1}{x^4}$ as a function to compare I with, but $\int_0^\infty\frac{1}{x^4}dx$ is divergent and bigger than I: therefore I cannot make any conclusion about I. I could use $\frac{1}{x^5+5}$ as a function to compare, but I don't know how to integrate that, and it is only similar to I at $\infty$, not at $0$.

How should I tell if it's convergent or divergent?

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  • $\begingroup$ "and it is only similar to I at $\infty$, not at $0$." is not really correct. Try graphing both functions. $\endgroup$
    – ViktorStein
    Mar 9, 2020 at 21:38

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Comparison with $\frac{1}{x^4}$ is the right way of doing this. Your integral is only improper at its upper boundary, and so the convergence there does not depend on the lower boundary: you could just as well test the convergence of the integral:

$$\int_c^{\infty}\frac{x}{x^5+5}dx$$

for some $c>0$ (e.g. $c=1$) - for which the function $\frac{1}{x^4}$ can be used. Due to convergence of:

$$\int_c^{\infty}\frac{dx}{x^4}$$

the original integral also converges.

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  • $\begingroup$ You could explicitly use the function $$f(x) := \begin{cases} x, & \text{if } x \in [0,1], \\ x^{-4}, & \text{if } x > 1 \end{cases}$$ as a dominating functions, whose integral is convergent. $\endgroup$
    – ViktorStein
    Mar 9, 2020 at 21:42
  • $\begingroup$ But how is the integral of $\frac{1}{x^4}$ convergent? If you separate it into two and take the limits, the part with $\infty$ goes to $0$, yes, but the part with $0$ goes to infinity, no? $$\lim_{t → 0^+} \int_t^c \frac{1}{x^4} = \infty$$ $\endgroup$
    – Thibaut B.
    Mar 9, 2020 at 21:58
  • $\begingroup$ My point is that you don't need to test the convergence of the integral on the interval $[0,c]$: the integral $\int_0^c\frac{x}{x^5+5}dx$ is proper as the interval is closed and the function that is being integrated is continuous on it. If you need one single function that will dominate your function across the whole interval $[0,\infty)$, then, as @ViktorGlombik said, you can patch it up in two parts. $\endgroup$
    – user700480
    Mar 9, 2020 at 22:03

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