How can we evaluate the following integral? $$\int_0^4y^3\sqrt{64-y^3}\,\mathrm dy$$ I can't find anything to substitute because all of the trigonometric identities are in square form...
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$$\begin{align}\int_0^4 dy \, y^3 (64-y^3)^{1/2} &= \underbrace{2048 \int_0^1 dx \, x^3 (1-x^3)^{1/2}}_{y=4 x} \\ &= \frac{2048}{3} \underbrace{\int_0^1 du \, u^{1/3} \, (1-u)^{1/2}}_{x=u^{1/3}}\\ &= \frac{2048}{3} \frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(\frac{3}{2}\right)}{\Gamma\left(\frac{17}{6}\right)} \end{align}$$
Ron Gordon
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how 2048 came..?? – Tamim Addari Jun 21 '13 at 12:43
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$4 \cdot 4^3 \cdot \sqrt{64} = 2^8 \cdot 2^3 = 2^{11} = 2048$ – Ron Gordon Jun 21 '13 at 12:43
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You applied the beta and gamma relation, but beta(m,n) supposed to be x^(m-1)*(1-x)^(n-1) so, how did you put directly 1/3 and 1/2? shouldnt it be something like (1/3-1) type? – Tamim Addari Jun 21 '13 at 12:58
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1@TamimAdDari: good catch. Easy to get that mixed up. – Ron Gordon Jun 21 '13 at 13:02
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1I would only like to add that $\Gamma(5/6)$ is expressible in terms of $\Gamma(1/3)$: $$\Gamma(5/6)\Gamma^2(1/3)=\frac{2^{4/3}\pi^{3/2}}{\sqrt{3}}.$$ – Start wearing purple Jun 21 '13 at 13:10
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Many thanks Ron, can you help me with another one here please? http://math.stackexchange.com/questions/426152/hint-of-integration-about-gamma-function – Tamim Addari Jun 21 '13 at 13:10
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@O.L.: Math.SE is a better place with you here. – Ron Gordon Jun 21 '13 at 13:11
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I think there's a typo in the denominator: it should be $\Gamma \left( 17/6\right) $, instead of $\Gamma \left( 11/6\right) $, because $\frac{4}{3}+ \frac{3}{2}=\frac{17}{6}$. – Américo Tavares Jun 21 '13 at 13:25
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@AméricoTavares: correct. Not my day I guess. – Ron Gordon Jun 21 '13 at 13:36
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Not your day but a nice answer: (+1). – Américo Tavares Jun 21 '13 at 13:38
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how do I find the value of gamma 4/3? – Tamim Addari Jun 21 '13 at 13:40
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Well, it is $(1/3) \Gamma(1/3)$, and you can get $\Gamma(1/3)$ here: http://mathworld.wolfram.com/GammaFunction.html . It is not pretty. – Ron Gordon Jun 21 '13 at 13:46
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:D and there is no calculation there also :( so there is no practical way to calculate gamma(1/3) by integrating? – Tamim Addari Jun 21 '13 at 14:03