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Is there a way to see that

$$\int_V^{\alpha} \frac{dE}{\sqrt{(\alpha - E)(E - V)}} = \pi$$

should be true without doing any work?

Is there a quick way to brute-force this without long computations?

Hoping people can explain maybe physics or maths thinking that makes this so obvious.

Thanks!

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    $\begingroup$ linear change of variables, suggest making the interval of integration from $-1$ to $1$ $\endgroup$
    – Will Jagy
    Jun 23, 2016 at 1:03
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    $\begingroup$ Hint: let $x=\frac{2}{\alpha-V}E-\frac{\alpha+V}{\alpha-V}$ and you will get the answer. $\endgroup$
    – xpaul
    Jun 23, 2016 at 1:15
  • $\begingroup$ Will's hint is a good one and can be used to transform the integral into $\int_{-1}^1\frac{{\rm d}x}{\sqrt{1-x^2}}$. To evaluate this one, take a look at this. $\endgroup$
    – Winther
    Jun 23, 2016 at 1:27

1 Answer 1

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As per the suggestion of Will Jagy given in the comments above, this all comes down to how we perform the change of variables of $E$. The easiest way to do this, is that we can perform a sequence of linear transformations

$$ [V,a] \;\; \to \;\; [0, a-V] \;\; \to \;\; [0,2] \;\; \to \;\;[-1,1] $$

by performing the following operations on $E$:

$$ E \;\; \to \;\; E-V \;\; \to \;\; \frac{2}{a-V} \cdot (E-V) \;\; \to \;\; \frac{2(E-V)}{a-V} - 1 \;\; = \;\; \frac{2E-a-V}{a-V} \;\; =\;\; u. $$

We can immediately see that $E = \frac{1}{2}[(a-V)u + a +V]$ and $du = \frac{2}{a-V}dE$. The hardest part is in the substitution in the integrand. Observe that

\begin{eqnarray*} \frac{1}{\sqrt{(a-E)(E-V)}} & = & \frac{1}{\sqrt{\left [a - \frac{1}{2}(a-V)u - \frac{1}{2}a - \frac{1}{2}V \right ]\cdot \left [\frac{1}{2}(a-V)u + \frac{1}{2}a + \frac{1}{2}V - V \right ] }} \\ & = & \frac{1}{\sqrt{\left [\frac{1}{2}(a-V) - \frac{1}{2}(a-V)u \right ] \cdot \left [\frac{1}{2}(a-V) + \frac{1}{2}(a-V)u \right ]}} \\ & = & \frac{2}{a-V} \cdot \frac{1}{\sqrt{1-u^2}}. \end{eqnarray*}

We therefore see that the change of variables in the original integral is given by

$$ \int_V^a \frac{dE}{\sqrt{(a-E)(E-V)}} \;\; =\;\; \int_{-1}^1 \frac{a-V}{2} \cdot \frac{2}{a-V} \frac{du}{\sqrt{1-u^2}} \;\; =\;\; \int_{-1}^1 \frac{du}{\sqrt{1-u^2}}. $$

Since I started writing this, it looks like xpaul chimed in with the same suggestion.

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