0
$\begingroup$

I am trying to find an appropriate substitution in the following indefinite integral $\int \frac{x^n}{\sqrt{x-x^2}}$ for an arbitrary power $n$ to obtainthe intermidiate steps to the same answer as in Mathematica expressed through the hypergeometric function as following

$$\int\frac{x^ndx}{\sqrt{x(1-x)}}=-\sqrt{1-x}\;_2F_1\left(\frac{1}{2},\frac{1}{2}-n;\frac{3}{2};1-x\right)$$

Using this formula I could also compute the definite integral from $0$ to $1$, which is given as

$$\int_{0}^{1} \frac{x^{n}dx}{\sqrt{x(1-x)}}=\frac{\sqrt{\pi}\Gamma\left(n+\frac{1}{2}\right)}{\Gamma(n+1)}$$

Any ideas are very welcome!

$\endgroup$
4
$\begingroup$

If you only need the definite integral, then you can write it directly as the definition of the Beta function.

$$=\int_{0}^{1}x^{n-1/2}(1-x)^{-1/2}=B(n-1/2+1,-1/2+1)=B(n+1/2,1/2)=\frac{\Gamma(n+1/2)\Gamma(1/2)}{\Gamma(n+1/2+1/2)}$$

Finally, use that $\Gamma(1/2)=\sqrt{\pi}$.

$\endgroup$
  • $\begingroup$ Indeed, this can also be used to obtain a solution for the indefinite integral. One has to use the expression of the incomplete beta-function and the Chebychev integral $$\int x^p(1-x)^qdx=B(x;1+p,1+q)$$ and then its relation to a hypergeometric function $$B(x;1+p,1+q)=\frac{x^{1+p}}{1+p} _2F_1(1+p,-q;2+p;x)$$ Thanks a lot! $\endgroup$ – Paul Frey Jul 17 '18 at 8:13
0
$\begingroup$

Let $$x=\sin^2(t)\implies dx=2 \sin (t) \cos (t)\,dt$$ to make $$\int \frac{x^n}{\sqrt{x-x^2}}\,dx=2\int \sin^{2n}(t)\,dt$$ and use the reduction formula (have a look here).

$\endgroup$
  • 1
    $\begingroup$ This approach does not help to proceed to the generalized formula which would be also expressed through a hypergeometric function. This brings me back to my original question... $\endgroup$ – Paul Frey Jul 17 '18 at 7:10

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.