$\text{Evaluate:} \lim_{b \to 1^+} \int_1^b \frac{dx}{\sqrt{x(x-1)(b-x)}}$

I'm trying to solve the following problem

$$\text{Evaluate:} \lim_{b \to 1^+} \int_1^b \frac{dx}{\sqrt{x(x-1)(b-x)}}$$

I'm tempted to think it's 0 because the bounds would be about equal, but that's not the correct answer. But I don't know what integration techniques to use, so any hints are greatly appreciated.

I don't know how to solve for the indefinite integral itself.

• Start with $x=\cos^2y+b\sin^2y$ – lab bhattacharjee Jul 16 '15 at 12:56
• I simplified the integral to $\lim_{b \to 1^+} \int_{x=1}^{x=b} \frac{2}{\sqrt{\cos^2{y}+b\sin^2{y}}}$ How do I change the bounds? – Vishwa Iyer Jul 16 '15 at 13:02
• @labbhattacharjee Just as a general question, is it possible to solve the integral using partial fractions? Or is that only applicable if there is no root present? – Gummy bears Jul 16 '15 at 13:05
• @VishwaIyer Is that $y^2$ or is it $ycos^2y$? – Gummy bears Jul 16 '15 at 13:06
• It's $\cos^2{y}$. I was having trouble with the formatting. – Vishwa Iyer Jul 16 '15 at 13:07

We can solve this using complex analysis: The function $x(x-1)(b-x)$ has a holomorphic square root $f$ on $\mathbb{C}\setminus\left((-\infty,0]\cup[1,b]\right)$ because every loop in this domain circles two zeroes of $x(x-1)(b-x)$ - more concretely, for $(r,\phi)\in (0,\infty)\times (-\pi,\pi)$, set $g(re^{i\phi}) = \sqrt{r}e^\frac{i\phi}{2}$. Then $g$ is the holomorphic extension of $\sqrt\cdot$ to $\mathbb{C}\setminus(-\infty,0]$, and $f(z)$ is the unique holomorphic extension of $ig(z)g(z-1)g(z-b)$ to the domain given above. Indeed, we can check directly that for $x\in (0,1)$, we have $\lim_{\epsilon\to 0} f(x\pm i\epsilon) = i\sqrt{x(1-x)(b-x)}$, so that we can extend $f$ to this interval. On the other hand, for $x\in [1,b]$, we have $\lim_{\epsilon\to 0} f(x\pm i\epsilon) = \mp\sqrt{x(x-1)(b-x)}$. Thus, we have $\int_1^b \frac{\mathrm dx}{\sqrt{x(x-1)(b-x)}} = \frac{1}{2}\lim_{\epsilon\to 0}\int_{1-\epsilon}^{b-\epsilon} f(z)\mathrm dz + \int_{b+\epsilon}^{1+\epsilon} f(z)\mathrm dz$, and in the limit we can close this path by vertical edges to get $\int_1^b \frac{\mathrm dx}{\sqrt{x(x-1)(b-x)}} = \frac{1}{2} \oint_\gamma f(z)\mathrm d z$ where $\gamma$ is any path which circles the "slit" $[1,b]$ once in counterclockwise direction. For $b$ small, we can for example take the circle of radius $\frac{1}{2}$ around $1$. Then the path of integration no longer depends on $b$, and by dominated convergence we can exchange the limit with integration to see that $\lim_{b\to 0}\int_1^b \frac{\mathrm dx}{\sqrt{x(x-1)(b-x)}} = \frac{1}{2}\oint_\gamma \frac{\mathrm d z}{g(z)i(z-1)}$ and by the Cauchy integral formula this is $\frac{2\pi i}{2ig(1)} = \pi$.

• This is impressive. – Axel Jul 17 '15 at 7:25

First you can notice that the $\frac{1}{\sqrt{x}}$ term is between $1$ and $\frac{1}{\sqrt{b}}$, so there is no need to bother about it. Indeed this approximation will yield a relative mistake of at most $\sqrt{b}$.

Indeed,

$$\frac{1}{\sqrt{b}}I = \frac{1}{\sqrt{b}} \int_1^b \frac{dx}{\sqrt{(x-1)(b-x)}} \leq \int_1^b \frac{dx}{\sqrt{x(x-1)(b-x)}} \leq \int_1^b \frac{dx}{\sqrt{(x-1)(b-x)}} = I .$$

Then you can see by an affine change of variable that the remaining integral $I$ doesn't depend on $b$.

All you need to do then is a simple change of variable to show that the answer is $\pi$. Can you try or do you want me to spell it out?

• i think you will not reglect the first term. Instead you approximate it as one. Which is indeed a very good approximation. Just to be totally correct...Otherwise nice (+1) – tired Jul 16 '15 at 13:27
• Ok, I'll edit to explain. – Axel Jul 16 '15 at 13:28
• What change of variable do you use? – Vishwa Iyer Jul 16 '15 at 14:07
• $x = 1 + (b-1)u$ then $u=\sin^2(t)$. – Axel Jul 16 '15 at 14:25

Hint: Let $~b=1+\epsilon,~$ where $~\epsilon\to0^+,~$ and $~x=1+t,~$ where $~t\in(0,\epsilon).~$ Then let $~t=\epsilon~u.$