Convergence of an integral $\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$

Question

$\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx=\lim\limits_{t\to\infty}\int_1^{t} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$

Partial integration can't solve the integral $\int \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$.

What substitution (or other methods) would you suggest?

$x^2+1>x^2$ would help (I hope) – Claude Leibovici Dec 14 '15 at 11:17 — Claude Leibovici, Dec 14 '15 at 11:17

mathcounterexamples.net · Accepted Answer · 2015-12-14T11:25:15.880

2

For $x \in [1,\infty)$, you have $$0 \le \frac{1}{x\sqrt[3]{x^2+1}} \le \frac{1}{x^\frac{5}{3}}$$ and $\int_1^\infty \frac{dx}{x^\frac{5}{3}}$ is convergent.

edited Dec 14 '15 at 11:25

answered Dec 14 '15 at 11:19

mathcounterexamples.net

70,018

I think it's $x^(5/3)$ on the right. – coffeemath Dec 14 '15 at 11:22
@coffeemath You're right! $1+\frac{2}{3} = \frac{5}{3}$. Thanks. – mathcounterexamples.net Dec 14 '15 at 11:26

score 2 · Answer 2 · answered Dec 14 '15 at 11:23

2

$$0\leq \int_{1}^{+\infty}\frac{dx}{x\sqrt[3]{x^2+1}} = \int_{0}^{+\infty}\frac{du}{\sqrt[3]{e^{2u}+1}}\leq \int_{0}^{+\infty}e^{-2u/3}\,du=\frac{3}{2}.$$

answered Dec 14 '15 at 11:23

Jack D'Aurizio

353,855

coffeemath · Answer 3 · 2015-12-14T14:18:23.070

One can actually evaluate this integral "exactly" using a substitution and then partial fractions. Let $x^2+1=u^3$ so that the radical part becomes just $u.$ Also $2x\ dx = 3u^2\ du,$ and since the original integral doesn't have $2x\ dx$ in its numerator we can put it there by multiplication of numerator and denominator by $2x$ and then there will then be a $2x^2$ in the denominator, and from our substitution, $x^2=u^3-1.$ Finally when $x$ goes from $1$ to $\infty,$ our $u$ will range from $a=2^{(1/3)}$ to $\infty.$ So the starting integral has become $$\int_a^\infty\frac{3u^2\ du}{2(u^3-1)u}$$ Now since the denominator factors into $2(u-1)(u^2+u+1)u,$ partial fractions can be applied to the $u$ integral. An involved task, but could be done.

Edit: I should have noticed that now one can cancel a $u$ from numerator and denominator, so the partial fractions to be done is simply $(3/2)[u]/[u^3-1].$

Domenico Vuono · Answer 4 · 2015-12-14T17:29:11.837

0

The integral $$\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$$ converges.

Indeed $$\sum_{n=1}^\infty {\frac{1}{n\sqrt[3]{n^2+1}}} $$ converges. Indeed $$ {\frac{1}{n\sqrt[3]{n^2+1}}}\sim_{\infty} {\frac{1}{n\sqrt[3]{n^2}}} $$ that converges

edited Dec 14 '15 at 17:29

answered Dec 14 '15 at 14:15

Domenico Vuono

2,672

Why that argument works? It is not difficult to exhibit some functions $f$ such that the convergence/divergence of $\int_{0}^{+\infty}f(x),dx$ does not imply the convergence/divergence of $\sum_{n\geq 0}f(n)$ and vice-versa. – Jack D'Aurizio Dec 14 '15 at 17:43
Now I have a doubt... Why not?? – Domenico Vuono Dec 14 '15 at 17:57
You may take a Riemann- or Lebesgue integrable function over $\mathbb{R}^+$, re-define it over $\mathbb{N}$ and have that the value of the integral is still the same as before while $\sum_{n\geq 0}f(n)$ is divergent. – Jack D'Aurizio Dec 14 '15 at 18:02
For brevity: you should mention that some kind of monotonicity is used here. – Jack D'Aurizio Dec 14 '15 at 18:03
I'm attending the first year at university.... I believed to use the Integral test – Domenico Vuono Dec 14 '15 at 18:09
That involves some kind of monotonicity of the integrand function, as already mentioned. – Jack D'Aurizio Dec 14 '15 at 18:18
Ok${}{}{}{}{}{}$ – Domenico Vuono Dec 14 '15 at 18:20

Convergence of an integral $\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$

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