# How would I go about integrating an improper integral with an absolute value in the denominator?

$\int^1_{-1} \frac{1}{\sqrt{\lvert{2x-x^2}\rvert}} dx$

I'm getting stumped. The integral is improper because at 0 the function does not exist.

I am thinking of completing the square and doing a trig substitution. This doesn't seem right to me though, because of the absolute value function in the square root in the denominator. Anyway, once I complete the square, the denominator is $\sqrt{\lvert {1-(x-1)^2} \rvert}$

I'm confused how to proceed. I think I then split it up between (x-1)^2 -1 and 1-(x-1)^2 in two separate integrals. But I'm honestly very confused right now on how to proceed.

Since $2x-x^2<0$ on $(-1,0)$ it is $|2x-x^2|=x^2-2x$ on $(-1,0).$ Since $2x-x^2>0$ on $(0,1)$ it is $|2x-x^2|=x^2-2x$ on $(0,1).$ Thus

$$\int^1_{-1} \frac{1}{\sqrt{\lvert{2x-x^2}\rvert}} dx=\int^0_{-1} \frac{1}{\sqrt{x^2-2x}} dx+\int^1_{0} \frac{1}{\sqrt{2x-x^2}} dx.$$

Now,

$$\int\frac{1}{\sqrt{x^2-2x}}dx=\int\frac{1}{\sqrt{x}\sqrt{x-2}}dx\underbrace{=}_{x=t^2}\int \frac{2tdt}{t\sqrt{t^2-2}}= \int \frac{2dt}{\sqrt{t^2-2}}\\ =\ln(2(\sqrt{t^2-2}+t))=\ln(2(\sqrt{x-2}+\sqrt{x})).$$

Proceed in a similar way with the other integral.

$2x-x^2=x(2-x)$; this is positive when $0<x\le 1$ and negative when $-1\le x<0$, so you can rewrite your integral as

$$\lim_{a\to 0^-}\int_{-1}^a\frac{dx}{\sqrt{x^2-2x}}+\lim_{a\to 0^+}\int_a^1\frac{dx}{\sqrt{2x-x^2}}\;.$$

Now you can write $x^2-2x=(x-1)^2-1$ and $2x-x^2=1-(x-1)^2$ and proceed.

Write the integral as

$$\int_{-1}^1 dx \, |x|^{-1/2} |2-x|^{-1/2}$$

Over the integration interval, $2-x \gt 0$, so we may split the integral up as follows:

$$\int_{-1}^0 dx \, (-x)^{-1/2} (2-x)^{-1/2} + \int_0^1 dx \, x^{-1/2} (2-x)^{-1/2} =\int_0^1 dx \, x^{-1/2} \left [(2-x)^{-1/2} + (2+x)^{-1/2} \right ]$$

Sub $x=u^2$ and get

$$2 \int_0^1 du \, \left [(2-u^2)^{-1/2} + (2+u^2)^{-1/2} \right ]$$

These are standard integrals:

$$\int_0^1 du \, (2-u^2)^{-1/2} = \int_0^{\pi/4} d\theta = \frac{\pi}{4}$$ $$\int_0^1 du \, (2+u^2)^{-1/2} = \int_0^{\arctan{1/\sqrt{2}}} d\theta \, \sec{\theta} = \log{\left (\frac{1+\sqrt{3}}{\sqrt{2}} \right )}$$

The result is

$$\frac{\pi}{2} + \log{\left (2+\sqrt{3} \right )}$$

• Just a question, If the (2-x) part of the function was not greater than zero over the integration interval, how would one proceed? – universality Dec 3 '14 at 3:28