How would one show that the integral $\int_0^\infty \frac 1 {(1+x)\sqrt x}\,\mathrm d x$ either diverges or converges?

Question

I'm supposed to be showing that the integral \begin{equation} \int_0^\infty \frac 1 {(1+x)\sqrt x}\,\mathrm d x \end{equation} either converges or diverges, using the comparison test. It can easily be verified that $$ \frac 1 {(1+x)\sqrt x} \geq \frac 1 {(1+x)x} = \frac 1 x\frac 1 {(1+x)}, $$ so if I was able to show that the integral $$ \int_0^\infty \frac 1 x\frac 1 {(1+x)}\,\mathrm d x $$ diverges I would be in the clear. I know this integral can be broken into two parts like so: $$ \lim_{c\to0^+}\int_c^1 \frac 1 x\frac 1 {(1+x)}\,\mathrm d x + \lim_{c\to\infty}\int_1^c \frac 1 x\frac 1 {(1+x)}\,\mathrm d x\,. $$ If either of these diverges, so does the sum. Working with the left one using integration by parts: \begin{align} &&\lim_{c\to0^+} \int_c^1{ \frac 1 x \frac 1 {1+x} }{x} &= \lim_{c\to 0^+}\left[ \frac 1 x \ln(1+x)\right]_c^1 - \int_c^1{ \frac {-1}{x^2} \ln({1+x}) }{x}\\ && &= \lim_{c\to\infty}\left[ \frac 1 x \ln(1+x)\right]_c^1 + \int_c^1{ \frac {1}{x^2} \ln(1+x) }{x}\\ && &= \lim_{c\to\infty}\left[ \frac 1 x \ln(1+x)\right]_c^1 + \left[ \frac {-1} x \ln(1+x)\right]_c^1 - \int_c^1{ \frac {-1} x \frac 1 {1+x} }{x}\\ \iff && 0&=0\,. \end{align} This doesn't exactly work. Is there an easier way to do this?

Answer 1

Deal first with the interval $[0,1]$ and then separately with $[1,\infty);$ $$ \frac 1 2 \le \frac 1 {1+x} \le 1 \text{ if } 0\le x\le 1. $$ $$ \int_0^1 \frac 1 {2\sqrt x}\,\mathrm{d}x \le \int_0^1 \frac 1 {(1+x)\sqrt x}\,\mathrm dx \le \int_0^1 \frac 1 {\sqrt x} \, \mathrm{d} x $$

$$ \int_1^\infty \frac 1 {(1+x)\sqrt x} \, \mathrm{d}x \le \int_1^\infty \frac 1 {x^{3/2}} \,\mathrm{d}x $$

Answer 2

By letting $y=\sqrt{x}$, then \begin{align*} \int_0^\infty \frac 1 {(1+x)\sqrt{x}}\,dx=\int_0^\infty\frac{2}{1+y^2}\,dy, \end{align*} and \begin{align*} \int_0^\infty \frac{1}{1+y^2} \, dy = \lim_{M\rightarrow\infty} \tan^{-1} y \bigg|_{y=0}^{y=M} = \frac\pi 2<\infty. \end{align*}

Maybe some sort of comparison: \begin{align*} \int_0^\infty \frac 1 {1+y^2}\,dy=\int_0^1 \frac 1 {1+y^2} \, dy + \int_1^\infty \frac 1 {1+y^2} \, dy, \end{align*} and \begin{align*} \int_1^\infty \frac 1 {1+y^2}\,dy \leq \int_1^\infty \frac 1 {y^2} \, dy = 1 < \infty. \end{align*}

Answer 3

A simple way is to use these comparisons

$$\frac{1}{(1+x)\sqrt x}\le \frac1{\sqrt x}$$ in the interval $(0,1)$ and

$$\frac{1}{(1+x)\sqrt x}\le \frac1{ x^{3/2}}$$ in the interval $(1,+\infty)$ to conclude the convergence of the given integral.

Answer 4

Just by eyeballing we can tell that the integral $$ \int_0^1\frac{1}{\sqrt{x}(1+x)}\mathrm dx $$ converges; near $0$, $\frac{1}{1+x}\approx 1$ leaving you to worry about $\frac{1}{\sqrt{x}}$ which converges.

Similar analysis will show that the integral converges near infinity, as the integrand is asymptotically equivalent to $$ \frac{1}{\sqrt{x}(1+x)}\approx\frac{1}{x^{3/2}} $$

Your estimate is thus not correct. Indeed, your inequality is equivalent to saying that $$ \sqrt{x}\leq x $$ when $x\in [0,1]$ which is not true.

Answer 5

$${1\over x(x+1)}={1\over x}-{1\over x+1}$$

Now we know that $\int_0^1{dx\over x}$ diverges and that $\int_0^1{dx\over x+1}=\log{2}$ and we get the divergence of $\int_0^1{dx\over x(x+1)}$ but this proves nothing concerning $\int_0^1{dx\over (x+1)\sqrt{x}}$ because contrary to what is stated in the question between $0$ and $1$ one has

$${1\over (x+1)\sqrt{x}}\leq {1\over x(x+1)}$$