I have to evaluate following integral.

$$ \int_{-c}^c \sqrt{b+\frac{c^6}{x^6-c^6}}dx$$

I know that final answer includes gamma function entries like ($\frac{\Gamma(1/6)\Gamma(1/2)}{\Gamma(2/3)})$.

But, I'm having difficulty figuring out how to integrate this. Any help will be highly appreciated.

Thank You,

  • $\begingroup$ Notice that $\dfrac 1 6 + \dfrac 1 2 = \dfrac 2 3$ and $$ \frac{\Gamma(1/6) \Gamma(1/2)}{\Gamma((1/6) + (1/2))} = \int_0^1 x^{(1/6)-1} (1-x)^{(1/2)-1} \, dx. $$ $\endgroup$ – Michael Hardy Jul 19 '17 at 5:53
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    $\begingroup$ Thank you @MichaelHardy . But, I can't figure out which transformations should be applied to reduce the problem to that form. $\endgroup$ – Coderzz Jul 19 '17 at 5:57
  • $\begingroup$ You have $-c<x<c.$ If $u = (x+c)/(2c)$ then $0<u<1$ and $dx = 2c\,du.$ Just how this is related to the integral in my comment above may be something to look at. $\endgroup$ – Michael Hardy Jul 19 '17 at 6:17
  • $\begingroup$ This can't be correct $\frac{\Gamma(1/6)\Gamma(1/2)}{\Gamma(2/3)}=7.2859...$ and now plug for example $b=1$ and $c=1$ to obtain a value of $0.86...i$ for your integral. $\endgroup$ – Math-fun Jul 19 '17 at 12:27
  • $\begingroup$ @ Coderzz : Are you aware that your integral is not real ? $\endgroup$ – JJacquelin Jul 19 '17 at 18:05

The result (below) as expected includes $ \frac{\Gamma(1/6)\Gamma(1/2)}{\Gamma(2/3)} = \sqrt{\pi}\frac{\Gamma(1/6)}{\Gamma(2/3)}$ and an hypergeometric function.

See Eq.(9) in : http://mathworld.wolfram.com/AppellHypergeometricFunction.html

For some particular values of $b$, the hypergeometric function reduces to functions of lower level.

enter image description here

  • $\begingroup$ Thank you @JJacquelin.. I tried to evaluate this in Mathematica using your steps. $$\frac{1}{3} c \int_0^1 \frac{\sqrt{b t-1}}{(1-t)^{5/6} \sqrt{t}} \, dt$$ But got a different answer(Given Below). Can you please explain why these are different. I also eager to know how to get an answer even without a gamma function. $$\frac{\sqrt{\pi } \sqrt[6]{1-b} \sqrt{b-1} c \Gamma \left(\frac{1}{6}\right) \, Hypergeometric2F1\left(\frac{1}{6},\frac{7}{6};\frac{2}{3};b\right)}{3 \Gamma \left(\frac{2}{3}\right)}$$ thanks, $\endgroup$ – Coderzz Jul 20 '17 at 3:42
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    $\begingroup$ The two answers are not different, they are equal. Hypergeometric2F1$(-1/2,1/2,2/3,x)=$ $(1-x)^{1/6}\sqrt{1-x}$Hypergeometric2F1$(1/6,7/6,2/3,x)$. Do not expect simpler answer if you don't specify the value of $b$. $\endgroup$ – JJacquelin Jul 20 '17 at 13:25
  • $\begingroup$ They are a lot of relationships between Hypergeometric functions : functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/… So, the same formula can be written on a lot of apparently different forms, but equal in fact. $\endgroup$ – JJacquelin Jul 20 '17 at 13:38

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