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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?

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up vote 11 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: $$ U(1) = \int_0^1 \frac{du}{\sqrt{u(1-u)}} = \pi $$

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+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! –  Chris's sis 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
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