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Possible Duplicate:
Computing $ \int_a^b \frac{x^p}{\sqrt{(x-a)(b-x)}} \mathrm dx$

How would you compute $$\int_{\alpha}^{\beta} \frac{x}{\sqrt{(x-\alpha)(\beta-x)}} \ dx\space?$$

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    $\begingroup$ a more general case is discussed here: math.stackexchange.com/questions/153018/… $\endgroup$
    – Valentin
    Nov 28 '12 at 19:50
  • $\begingroup$ @Valentin: thanks for the link. (+1) $\endgroup$ Nov 28 '12 at 19:52
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    $\begingroup$ @Norbert: it can happen that some particular cases allow particular solutions that do not apply to the general case. I'm not convinced that it's a good idea to close such problems based on the fact that there is a general case already discussed. (just saying my opinion) $\endgroup$ Nov 28 '12 at 21:47
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Make a change of variables $y= \alpha+\beta - x$. $$ \mathcal{I}_{\alpha,\beta} := \int_\alpha^\beta \frac{x}{\sqrt{(x-\alpha)(\beta-x)}} \mathrm{d}x = \int_{\alpha}^\beta \frac{\alpha+\beta - y}{\sqrt{(\beta-y)(y-\alpha)}} \mathrm{d}y $$ Implying $$ \frac{2}{\alpha+\beta} \mathcal{I}_{\alpha,\beta} = \int_\alpha^\beta \frac{\mathrm{d}x}{\sqrt{(x-\alpha)(\beta-x)}} $$ Now standardize: $x = \alpha + (\beta-\alpha) u$: $$ \frac{2}{\alpha+\beta} \mathcal{I}_{\alpha,\beta} = \int_0^1 \frac{\mathrm{d} u}{\sqrt{u(1-u)}} $$ Giving: $$ \mathcal{I}_{\alpha,\beta} = \frac{\alpha+\beta}{2} \int_0^1 \frac{\mathrm{d} u}{\sqrt{u(1-u)}} = \frac{\alpha+\beta}{2} \left. 2 \arcsin(\sqrt{u}) \right|_{u=0}^{u=1} = \pi \frac{\alpha+\beta}{2} $$

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  • $\begingroup$ Beautiful! (+1) $\endgroup$ Nov 28 '12 at 19:40
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Let $$I(a,b) = \int_a^b \dfrac{x}{\sqrt{(x-a)(b-x)}} dx$$ $$I(-b,b) = \int_{-b}^b \dfrac{x}{\sqrt{(x+b)(b-x)}} dx = 0 \,\,\,\,\,\,\,(\text{Odd function})$$ Hence, $$I(a,b) = f(a+b)$$ Hence, $$f(z) = I(0,z) = \int_0^z \dfrac{\sqrt{x}}{\sqrt{z-x}} dx$$ Now $x = z-t^2 \implies dx = -2t dt$. Hence, $$f(z) = I(0,z) = \int_0^z \dfrac{\sqrt{x}}{\sqrt{z-x}} dx = \int_{\sqrt{z}}^{0} \dfrac{\sqrt{z-t^2}}{t}(-2t dt) = 2\int_0^{\sqrt{z}} \sqrt{z-t^2} dt = 2 \dfrac{\pi z}4 = \dfrac{\pi z}2$$ Hence, $$I(a,b) = f(a+b) = \dfrac{\pi(a+b)}2$$

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  • $\begingroup$ where I'd add that the last integral is evaluated by geometric interpretation. (+1) $\endgroup$ Nov 28 '12 at 20:23
  • $\begingroup$ @anon Yes. What you have written is true. I will update it accordingly. $\endgroup$
    – user17762
    Nov 29 '12 at 20:00

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