# Convergence/divergence of $\int_{0}^{\infty} \frac{\sqrtx}{1+x}dx$

$$\int_{0}^{\infty} \frac{\sqrtx}{1+x}dx$$

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

$$\frac{\sqrtx}{1+x^2}<\frac{\sqrtx}{x}=\frac{1}{x^{\frac{2}{3}}}$$

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

So $$\int_{0}^{\infty} \frac{\sqrtx}{1+x}dx$$ 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?

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

## 4 Answers

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.

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 $$\int_1^\infty\frac{\sqrtx}{1+x}dx\geq\int_1^\infty\frac{\sqrtx}{2x}dx=\frac{1}{2}\int_1^\infty x^{-2/3} dx,$$ 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 $$\infty>\int_0^1\frac{\sqrtx}{2}dx\leq \int_0^1\frac{\sqrtx}{1+x}dx\leq \int_0^1\frac{\sqrtx}{1}dx<\infty.$$ by the first case, as $1/3>-1$.

• 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? – gbox Aug 4 '16 at 8:40
• see my edit, I've added some clarifications – b00n heT Aug 4 '16 at 15:00

Your function $\frac{\sqrtx}{1+x}\sim \frac{\sqrtx}{x}=\frac{1}{x^{2/3}}$ 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

$0<\int_0^1 f(x)dx<1$ 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.