# Why does $\int_0^1 \frac {dx}{x^{0.5}(1-x)^2}$ diverge?

I solved this question and was sure about my answer that it converges but the answer was it diverges, with no explanation,
I still don't get what's my mistake, here's what I did:

I wanted to use the limit comparison test, $$\displaystyle \frac {1}{x^{0.5}(1-x)^2}=\frac {1}{x^{0.5}(1-2x+x^2)}=\frac {1}{x^{0.5}-2x^{1.5}+x^{2.5}}$$, at $$x\to 0$$ , this function behaves like $$\displaystyle \frac {1}{x^{0.5}}$$.
$$\displaystyle \lim_{x\to0} \frac {x^{0.5}}{x^{0.5}-2x^{1.5}+x^{2.5}}$$ then I divided by $$x^{0.5}$$ and got $$\displaystyle \lim_{x\to0} \frac {1}{1-2x^{}+x^2}=1$$.

And the integral $$\displaystyle \int_0^1 \frac{dx}{x^{0.5}}$$ converges, then my integral should converge too.

Now I'm not sure where is my mistake, I would appreciate any help.

Thanks in advance.

## 1 Answer

You are right about $$x=0$$. It's what happens at $$x=1$$ that makes your integral diverge.

To be more precise: your integral converges if and only if both integrals$$\int_0^{1/2}\frac1{\sqrt x(1-x)^2}\,\mathrm dx\quad\text{and}\quad\int_{1/2}^1\frac1{\sqrt x(1-x)^2}\,\mathrm dx$$converge. Your computations show indeed that the first one converges. But the second one diverges.

• ohh thanks, I didn't notice that, appreciate it! – Pwaol Jan 31 at 19:00
• If I take $\frac {1}{(1-x)^2}$ to compare the second one, then substitute $t=1-x$ in the integral, and get $\int^1_{1/2} \frac {1}{t^2}$, that would be a validate proof that the second integral diverges? – Pwaol Jan 31 at 19:12
• @Pwaol Yes, that is correct. – José Carlos Santos Jan 31 at 19:16