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

I'm supposed to be showing that the integral $$\int_0^\infty \frac 1 {(1+x)\sqrt x}\,\mathrm d x$$ 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?

• Just a side note, if you do parts by integrating one of the functions in the product, and then do parts again, but this time by differentiating the thing you just integrated, you will always just get the integral that you started with. Commented Feb 10, 2018 at 19:29
• Yup, seems so. Maybe integration by parts isn't the right approach here, since there no matter how I choose $f^\prime$ and $g$ I always end up with an expression with a natural logarithm in it. Commented Feb 10, 2018 at 19:31
• @TheSodesa where your estimate holds, you have bounded below by a convergent integral. Commented Feb 10, 2018 at 19:53
• This question is a mess: it asks about an improper integral on a finite closed interval, but the question in the body refers to another integral on the whole real line...of another function! This is too messy. Focusing in one single question, without all the rest, would make this way clearer. Commented Feb 10, 2018 at 20:00
• Your “It can easily be verified” is in fact false. Commented Feb 10, 2018 at 20:05

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*}

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.

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

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

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.

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

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

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

• this is mostly on OP, but this wasn't really their question Commented Feb 10, 2018 at 19:36
• The question is the divergence of $\int_0^1{dx\over x(x+1)}$. Isn’t it? At least it is the title of the question. Commented Feb 10, 2018 at 19:39
• not if the first line of the body is to be believed; op used a incorrect estimate to "reduce" the problem to the one in the title Commented Feb 10, 2018 at 19:40