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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?

1 Answers 1

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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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    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.2013-07-02