5
$\begingroup$

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,

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.