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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,

Coderzz
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  • Notice that $\dfrac 1 6 + \dfrac 1 2 = \dfrac 2 3$ and $$ \frac{\Gamma(1/6) \Gamma(1/2)}{\Gamma((1/6) + (1/2))} = \int_0^1 x^{(1/6)-1} (1-x)^{(1/2)-1} , dx. $$ – Michael Hardy Jul 19 '17 at 05:53
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    Thank you @MichaelHardy . But, I can't figure out which transformations should be applied to reduce the problem to that form. – Coderzz Jul 19 '17 at 05:57
  • 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. – Michael Hardy Jul 19 '17 at 06:17
  • 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. – Math-fun Jul 19 '17 at 12:27
  • @ Coderzz : Are you aware that your integral is not real ? – JJacquelin Jul 19 '17 at 18:05

1 Answers1

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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

JJacquelin
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  • 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,

     – Coderzz Jul 20 '17 at 03:42
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    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 $b$. – JJacquelin Jul 20 '17 at 13:25
  • They are a lot of relationships between Hypergeometric functions : http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/17/02/07/ So, the same formula can be written on a lot of apparently different forms, but equal in fact. – JJacquelin Jul 20 '17 at 13:38