1
$\begingroup$

$$\int_{0}^{\infty} \frac{\sqrt[3]x}{1+x}dx$$

I have read an example on book and they did the following:

$$\frac{\sqrt[3]x}{1+x^2}<\frac{\sqrt[3]x}{x}=\frac{1}{x^{\frac{2}{3}}}$$

and we know that the integral of $\frac{1}{x^{\alpha}}$ converges for $\alpha>1$

So $$\int_{0}^{\infty} \frac{\sqrt[3]x}{1+x}dx$$ converges, but does not this holds just for $\int_{c}^{\infty}\frac{1}{x^{\alpha}}$ where $c>0$ and $\alpha>1$?

Because $\int_{0}^{\infty}\frac{1}{x^2}$ diverges?

$\endgroup$
  • 1
    $\begingroup$ Your answer is incorrect; $\int_0^\infty \frac{\sqrt[3]{x}}{1+x} dx$ does diverge after all. In fact $\int_c^\infty \frac{\sqrt[3]{x}}{1+x} dx$ diverges for any $c > 0$. $\endgroup$ – Drew N Aug 4 '16 at 8:41
  • $\begingroup$ But $2/3<1$.... $\endgroup$ – Ahmed S. Attaalla Aug 4 '16 at 16:19
1
$\begingroup$

But does not this holds just for $\int_{c}^{\infty}\frac{dx}{x^{\alpha}}$ where $c>0$ and $\alpha>1$?

Yes. Here is how you obtain it. Assume $c>0$ and $\alpha>1$. Then $$ \int_{c}^{\infty}\frac{dx}{x^{\alpha}}=\lim_{M \to \infty}\int_{c}^M\frac{dx}{x^{\alpha}}=\lim_{M \to \infty}\left[ \frac{ x^{1-\alpha}}{1-\alpha}\right]_{c}^M=0-\frac{ c^{1-\alpha}}{1-\alpha}<\infty. $$

Observe that $\int_{c}^{\infty}\frac{dx}{x^2}$ ($c>0$) is convergent.

$\endgroup$
1
$\begingroup$

As you said one has, for any $\epsilon>0$ $$\int_0^\epsilon x^\alpha<\infty\iff a>-1$$ and $$\int_\epsilon^\infty x^\alpha<\infty \iff \alpha<-1 $$ But in your case near $0$ you have no problems as you add $+1$ in the denominator and so the function is bounded on $[0,\epsilon]$ and so the integral stays bounded also.

As $x+1\leq 2 x\iff\frac{1}{2x}\leq\frac{1}{1+x}$ for all $x$, we have a lower bound $$\int_1^\infty\frac{\sqrt[3]x}{1+x}dx\geq\int_1^\infty\frac{\sqrt[3]x}{2x}dx=\frac{1}{2}\int_1^\infty x^{-2/3} dx,$$ which diverges by the first case as $-2/3\geq-1$.

As the integrand is positive on $[0,1]$, no cancellation can take place. Thus the whole integral diverges.

Actually one can also bound $1 \leq x+1\leq2$ on $[0,1]$ to obtain the estimate $$\infty>\int_0^1\frac{\sqrt[3]x}{2}dx\leq \int_0^1\frac{\sqrt[3]x}{1+x}dx\leq \int_0^1\frac{\sqrt[3]x}{1}dx<\infty.$$ by the first case, as $1/3>-1$.

$\endgroup$
  • $\begingroup$ So what you are saying is that we looked at $\int_{0}^{c}\frac{1}{x^{\frac{5}{3}}}+\int_{c}^{\infty}\frac{1}{x^{\frac{5}{3}}}$ so the second term converges because $\frac{5}{3}>1$ but the first term converges too? $\endgroup$ – gbox Aug 4 '16 at 8:40
  • $\begingroup$ see my edit, I've added some clarifications $\endgroup$ – b00n heT Aug 4 '16 at 15:00
0
$\begingroup$

Your function $\frac{\sqrt[3]x}{1+x}\sim \frac{\sqrt[3]x}{x}=\frac{1}{x^{2/3}}$ as $x\rightarrow +\infty$. We now that the integral $\int_{c}^{\infty}\frac{1}{x^{\alpha}}$ converges for $a>1$ (with $c>0$). Since $2/3 < 1$, your integral diverges

$\endgroup$
0
$\begingroup$

$0<\int_0^1 f(x)dx<1$ and $\int_1^\infty f(x)dx>\int_1^\infty \frac{1}{2x^{\frac{2}{3}}}dx=\infty$.

Therefore, $\int_0^\infty f(x) dx$ diverges.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.