# Determining the convergence of $\int_{1}^{2}\frac{1}{\sqrt{x}(2-x)}\, \mathrm{d}x$ in simple way.

Is there a (simple) way to determine its convergence without determining its value?

I know that the function $x\mapsto \left (\sqrt{x}(2-x) \right )^{-1}$ is continuous for all $x\in \mathbb{R}\setminus \lbrace 0,2\rbrace$ so

$$\int_{1}^{2}\frac{1}{\sqrt{x}(2-x)}\, \mathrm{d}x=\lim_{y\to 2^{+}}\int_{1}^{y}\frac{1}{\sqrt{x}(2-x)}\, \mathrm{d}x.$$

The integral to left side converges if both integrals to right side converge $$\int_{1}^{2}\frac{1}{\sqrt{x}(2-x)}\, \mathrm{d}x=\int_{1}^{a}\frac{1}{\sqrt{x}(2-x)}\, \mathrm{d}x+\int_{a}^{2}\frac{1}{\sqrt{x}(2-x)}\, \mathrm{d}x$$ where $a\in (1,2)$. Lets say, for example $a=1/2$. According to the WolframAlpha it says it doesn't converge. How?

If it were $\int_{0}^{1}\frac{1}{\sqrt{x}(2-x)}\, \mathrm{d}x$ I would say it converges since the integral $\int_{0}^{1}\frac{1}{\sqrt{x}}\, \mathrm{d}x$ converges when comparing it.

Note that I am already aware of this expression $\frac{1}{\sqrt{x}(2-x)}=\frac{1}{2\sqrt{x}}+\frac{\sqrt{x}}{2(2-x)}$ - but I don't want to use it.

Another question is, what would happen if it were $\int_{0}^{2}\frac{1}{\sqrt{x}(2-x)}\, \mathrm{d}x$? I would say, that it diverges since $\int_{0}^{1}\frac{1}{\sqrt{x}(2-x)}\, \mathrm{d}x + \int_{1}^{2}\frac{1}{\sqrt{x}(2-x)}\, \mathrm{d}x$ diverges even though $\int_{0}^{1}\frac{1}{\sqrt{x}(2-x)}\, \mathrm{d}x$ converges.

I am sorry for the long post. I would be really happy if you could enlighten me.

• Did you mean $$\frac{1}{\sqrt{x(2-x)}}$$ perhaps? With the $2-x$ outside the root, the integral is infinite. Commented Apr 30, 2014 at 20:22
• No Daniel. I meant $1/(\sqrt{x}(2-x))$ :) Commented Apr 30, 2014 at 20:26

$$\int_{1}^{2}\frac{\mathrm{d}x}{\sqrt{x}(2-x)}\geqslant \int_{1}^{2}\frac{\mathrm{d}x}{\sqrt{2}(2-x)}\ \stackrel{(u=2-x)}{=}\ \frac1{\sqrt2}\int_{0}^{1}\frac{\mathrm{d}u}{u}=\ldots$$

• Thank you for your answer. I like it. Is it correct that the integral $\int_{0}^{2}\frac{1}{\sqrt{x}(2-x)}\, \mathrm{d}x$ diverges because $\int_{0}^{1}\frac{1}{\sqrt{x}(2-x)}\, \mathrm{d}x + \int_{1}^{2}\frac{1}{\sqrt{x}(2-x)}\, \mathrm{d}x$ even though the integral $\int_{0}^{2}\frac{1}{\sqrt{x}(2-x)}\, \mathrm{d}x$ converges? Commented Apr 30, 2014 at 20:40
• Disagree: only the second one diverges.
– Did
Commented Apr 30, 2014 at 20:42
• Okay, thanks. I realized I made a mistake so I was about to edit my comment. Commented Apr 30, 2014 at 20:44

The integral $\int_1^2 1/\sqrt{x}(2-x)$ converges if and only if the integral $\int_1^2 1/(2-x)$ converges because $1/\sqrt{x}$ is bounded and continuous on $[1,2]$. For $\int_1^2 1/(2-x)$ you get that it is equal to $\int_0^1 1/(1-x)$ and the indefinite integral of $1/(1-x)$ is $-\ln (1 - x)$ which approaches $\infty$ as $x \to 1$, so the integral diverges and doesn't have a finite value.

There is problem at the right side (upper limit) when $x\to 2$, We know that $\sqrt x < 2 , x \in (1, 2)$. So $$\int_1^2 \frac{1}{\sqrt x (x-2)}dx > \int_1^2 \frac{1}{2(x-2)}dx$$

If we change, $-y \to x-2$, this converts to $$\int_{1}^0 \frac{1}{y}dy = -\int_0^1 \frac{1}{y} dy$$

We know this doesn't converge, just set $w = \frac 1 y$, the integral converts into $$\int_{\infty}^1 w \left(-\frac 1 {w}^2 \right ) dw = \int_1^\infty \frac 1 w dw \approx \log \infty$$ Which doesn't converge.