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I came across an integral of this form:

$$\int_{0}^{a}\frac{dx}{(a^{n}-x^{n})^{1/n}}$$ How do I solve this integral? I tried using this substitution: $x=asin^{2}(x)$ in order to reduce this to a beta form integral, but I a man getting the value $\frac{B(0,1)}{2}$. This happens to tending towards minus infinity as $B(0,1)=\frac{\Gamma{(0)}\Gamma{(1)}}{\Gamma{(1)}}\rightarrow-\infty$

Any help is appreciated.

Calculation:

inserting ansatz, the new integral obtained is: $$\int_{0}^{\frac{\pi}{2}}\frac{2acos(\theta)sin(\theta)d\theta}{a^{n/n}(1-sin^{2n}(\theta))^{1/n}}$$ This further reduces to:

$$\int_{0}^{\frac{\pi}{2}}{2(cos(\theta))^{-1}(sin(\theta))^{1}d\theta}$$ Comparing this to the standard Beta integral we get p to be 0 and q to be 1 and thus the integral tends towards minus infinity.

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  • $\begingroup$ Where is $n$ gone ? Double check your calculation. $\endgroup$
    – user65203
    Feb 13, 2017 at 13:22
  • $\begingroup$ I see that you ignored my comment. $\endgroup$
    – user65203
    Feb 13, 2017 at 13:37
  • $\begingroup$ Nope, I have edited in the calculations. $\endgroup$ Feb 13, 2017 at 13:39
  • $\begingroup$ Not at all, this is still wrong. $\endgroup$
    – user65203
    Feb 13, 2017 at 13:40
  • $\begingroup$ You mean the answer? $\endgroup$ Feb 13, 2017 at 13:41

1 Answer 1

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WLOG, $a=1$.

Then with $x^n=t$,

$$I=\frac1n\int_0^1t^{1/n-1}(1-t)^{-1/n}dt=\frac1n\text{B}\left(\frac1n,1-\frac1n\right)=\frac1n\Gamma\left(\frac1n\right)\Gamma\left(1-\frac1n\right)=\frac{\dfrac\pi n}{\sin\dfrac\pi n}.$$

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    $\begingroup$ A natural question at this point is whether one can obtain this answer directly rather than through special function properties. I suspect the right approach is to use a substitution to map $[0,1]\to [0,\infty)$ and then use a keyhole contour. But this kind of argument can likely be found in a proof of Euler's reflection formula. $\endgroup$ Feb 13, 2017 at 13:52
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    $\begingroup$ @Semiclassical: the Beta/Gamma relation is usually established through a convolution-like identity involving the Laplace transform. Maybe some simplification can arise for these exponents (?) $\endgroup$
    – user65203
    Feb 13, 2017 at 13:57
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    $\begingroup$ @Semiclassical you an also use a dogbone contour directly on orignial integrand $\endgroup$
    – tired
    Feb 13, 2017 at 14:10

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