Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I would like to find the limit of $$ \int_1^a \frac{\mathrm dt}{\sqrt{t(t-1)(a-t)}}$$

when $$ a\rightarrow1^+$$ It seems that $$ \int_1^a \frac{\mathrm dt}{\sqrt{t(t-1)(a-t)}}\sim_{a\rightarrow 1^+} \pi$$

What bothers me is that $a$ is in the integrand and I cannot find an equivalent of $$\frac{1}{\sqrt{t(t-1)(a-t)}}$$ when $a\rightarrow1^+$

Moreover the integral $$ \int \frac{\mathrm dt}{\sqrt{t(t-1)(a-t)}}$$ "cannot be computed", is not simple.

Do you have any idea?

share|cite|improve this question
up vote 12 down vote accepted

Don't say "cannot be computed". It is an elliptic integral: $$ U(a):=\int_{1}^{a} \frac{dt}{\sqrt{t (t - 1) (a - t)}} = \frac{2 K \Bigl(\sqrt{\frac{a - 1}{a}}\Bigr)}{\sqrt{a}} $$ and of course $2K(0) = \pi$.

Now, of course, the question is to show $K(0)=\pi/2$ directly, without using knowledge of elliptic integrals. For that I chose a change of variables: $u=(t-1)/(a-1)$ to make this an integral from $0$ to $1$ $$ U(a) = \int_0^1\frac{du}{\sqrt{u(1-u)(1+(a-1)u)}} $$ Now the limit at $a=1$ is clear: $$ U(1) = \int_0^1 \frac{du}{\sqrt{u(1-u)}} = \pi $$

share|cite|improve this answer
+1. I was writing with the solution with the same change of variable. – user17762 Jun 2 '12 at 16:35
@GEdgar: it's simply awesome! – user 1618033 Jun 2 '12 at 16:36
Thank you very much for this nice answer! I would never have thought of such a substitution! – Chon Jun 2 '12 at 16:44
Your edit brought this to the top of the list, and I thought, "I know how to do that." When I opened the page, I saw my deleted answer (deleted because it was too similar). Bummer. – robjohn Jul 2 '13 at 17:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.