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It's true, according to WolframAlpha.

http://www.wolframalpha.com/input/?i=integral+0+to+1+%281+-+u%5E4%29%5E%28-1%2F2%29+du

Potato
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    Are you looking for a step-by-step solution to get this result? Or is it a "why" in the sense of how can an integral that doesn't look terribly complicated involve the gamma function? – graydad Feb 02 '15 at 04:56
  • While this is true.... this isn't probably what you want. I assume that you are a calculus 2 student? In any case, WolframAlpha LOVES to give its answers in terms of special functions and especially gamma functions (it has something to do with the symbolic logic that was put into the programming of W.Alpha). In any case, you should try to solve it by other means (in other words, if you want to know what the left hand side equals... just ignore the right hand side, and try solving it on your own) – Squirtle Feb 02 '15 at 04:57
  • @Squirtle: the antiderivative is not elementary. You might get a result in terms of an elliptic integral function... – Robert Israel Feb 02 '15 at 05:07
  • ... but there's no way you're going to avoid special functions, unless you can come up with an elementary expression for $\Gamma(3/4)$, something nobody else has managed to do. – Robert Israel Feb 02 '15 at 05:16
  • @Squirtle It also possibly has a little to do with Wolfram's own love of the special functions [see http://www.stephenwolfram.com/publications/history-future-special-functions/ ] – colormegone Feb 02 '15 at 05:24
  • Yeah, I rushed to assume that this was an easier problem than it really is, because we get many kinds like that here. But indeed it's not elementary. – Squirtle Feb 02 '15 at 05:29
  • http://math.stackexchange.com/questions/1122911/integral-int-0-infty-frac1-sqrt-alpha1x-betadx/1122990#1122990 – jimjim Feb 02 '15 at 12:41

2 Answers2

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Let $u = x^4$. Then $x = u^{1/4}$ $\implies$ $dx = (1/4)u^{-3/4}\, du$. When $x = 0$, $u = 0$ and when $x = 1$, $u = 1$. Thus \begin{align}\int_0^1 \frac{dx}{\sqrt{1 - x^4}}\, dx &= \frac{1}{4}\int_0^1 (1 - u)^{-1/2} u^{-3/4}\, du \\ &= \frac{1}{4}B(1/2, 1/4) \\ &= \frac{1}{4}\frac{\Gamma(1/2)\Gamma(1/4)}{\Gamma(3/4)} \\ &= \frac{\sqrt{\pi}\Gamma(5/4)}{\Gamma(3/4)}\end{align}

Here, the formulas $\Gamma(1/2) = \sqrt{\pi}$ and $x\Gamma(x) = \Gamma(x + 1)$ (for $x > 0$) were used.

kobe
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Using the beta and gamma functions and the relation between them: $$u=x^4\implies du=4x^3dx\implies dx=\frac{du}{4u^{3/4}}\implies$$

$$\int_0^1\frac{dx}{\sqrt{1-x^4}}=\frac14\int_0^1\frac{du}{u^{3/4}}\frac1{\sqrt{1-u}}=\frac14\int_0^1u^{1/4-1}(1-u)^{1/2-1}du=\frac14B\left(\frac14\,,\,\frac12\right)=$$

$$=\frac14\frac{\Gamma\left(\frac14\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac14+\frac12\right)}$$ and etc.

Timbuc
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