This is the integral I am struggling with: $$ \int_{U}^{\alpha} \frac{dE}{\sqrt{(\alpha-E)(E-U)}} $$ The solution claims it is $\pi$. However using an integral calculator, the integral seems to be zero.
How to evaluate $\int_{U}^{\alpha} \frac{dE}{\sqrt{(\alpha-E)(E-U)}}$ with the constraint $E \leq U$
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What do you mean by the constraints $E\leq U $? $E$ is the dummy variable in the integral, so this doesn't make sense. Maybe you mean $ U\leq \alpha$? – Kenny Wong May 20 '23 at 09:39
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1Intuitively, consider an upper semicircle $y=\sqrt{(\alpha-E)(E-U)}$ with diameter given by thw two points $(U,0)$ and $(\alpha,0)$. Then the infinitesimal arc length is easily computed as $$\mathrm{d}s=\frac{R}{y},\mathrm{d}E,$$ where $R=(\alpha-U)/2$ is the radius of the circle. This tells that your integral is equal to the circumference of the semicircle divided by its radius, which is precisely $\pi$. – Sangchul Lee May 20 '23 at 10:42
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Essentially the same question: Can't solve integral of Landau Lifschitz Classical Mechanics – Dave L. Renfro May 20 '23 at 13:50
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https://math.stackexchange.com/q/2853673/686284 – Quanto May 20 '23 at 15:21
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Assuming $U<\alpha$, we may define a new variable of integration $\theta$ through the equation $E=U+(\alpha-U)\sin^2\theta$. Then \begin{align} \int_{U}^{\alpha} \frac{dE}{\sqrt{(\alpha-E)(E-U)}}&= \int_0^{\pi/2}\frac{2(\alpha-U)\sin\theta\cos\theta\,d\theta}{\sqrt{(\alpha-U)\cos^2\theta\,(\alpha-U)\sin^2\theta}} \\ &=2\int_0^{\pi/2}d\theta = \pi. \end{align}
Gonçalo
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Some clarification:
The correct domain should be $a \ge U$. Otherwise, we result in a negative area.
How did the integral calculator arrive at $0$? The entire graph of the integrand is above the horizontal axis, = no negative area/sign.
$$(\alpha - E)(E - U) = -\alpha U + (\alpha + U)E - E^2 \\ \xrightarrow{\text{completeSquare}} \left(\dfrac{\alpha - U}{2}\right)^2 - \left(E - \dfrac{\alpha + U}{2}\right)^2$$
$$\therefore \int{\dfrac{dE}{\sqrt{\left(\dfrac{\alpha - U}{2}\right)^2 - \left(E - \dfrac{\alpha + U}{2}\right)^2}}} \\ = -\arcsin{\left( \dfrac{2E - U - \alpha}{U - \alpha} \right)} + C, \quad \alpha \ge U $$
Hence,
$$\int^{\alpha}_{U}{\dfrac{1}{\sqrt{(\alpha - E)(E - U)}}} = \left. -\arcsin{\left( \dfrac{2E - U - \alpha}{U - \alpha} \right)} \right|_{U}^{\alpha} \\ = \left(-\arcsin{\left( \dfrac{2\alpha - U -\alpha}{U - \alpha} \right)} +\arcsin{\left( \dfrac{2U - U -\alpha}{U - \alpha} \right)}\right) \\ = -\arcsin{(-1)} + \arcsin{(1)} \\ = \pi$$
Dstarred
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