integral of $(\frac{1 - x}{x})^a$

Question

I'm looking for the result of the integration $$ \int_{x_0}^1 \left(\frac{1 - x}{x}\right)^a~dx $$ where $a$ is a constant and $0 < x_0 \leq 1$. I tried using Wolfram Alpha but it couldn't achieve it within the computation time available to free users.Is there an analytical expression for this integral, or is numerical integration needed?

EDIT $a \geq 0$ is a constant not dependent on $x$.

I would think that an analytical expression for it doesn't exist, if Wolfram Alpha couldn't solve it quickly enough. — 2017-01-04
Wolfram worked for me: http://www.wolframalpha.com/input/?i=int_c%5E1+(1-x)%5Ea%2Fx%5Ea+dx — 2017-01-04
@NathanH. While in general a good rule of thumb, the are tricks and simplifications that WA doesn't do. I have seen integrals that are solvable by hand that WA couldn't do. — 2017-01-04
@DavidP The incomplete beta function makes that result basically a rewriting, not a solution. — 2017-01-04

user id:358056 2k · Accepted Answer

\begin{align} \int\limits_{x_{0}}^{1} (1-x)^{a} x^{-a} dx &= \int\limits_{0}^{1} (1-x)^{a} x^{-a} dx - \int\limits_{0}^{x_{0}} (1-x)^{a} x^{-a} dx \\ &= \mathrm{B}(1-a,1+a) - \mathrm{B}_{x_{0}}(1-a,1+a) \\ &= \Gamma(1-a)\Gamma(1+a) - \frac{x_{0}^{1-a}}{1-a} \, {}_{2}\mathrm{F}_{1}(1-a,-1;2-a;x_{0}) \end{align}

We have used the incomplete beta function and Gauss's hypergeometric function.

Thanks @poweierstrass, it also looks like the [hypergeometric function](https://en.wikipedia.org/wiki/Hypergeometric_function#The_hypergeometric_series) simplifies for cases where $a$ or $b$ is a nonpositive integer and $|z| < 1$, which are both satisfied by this solution ($x_0 < 1$ for all practical cases). I still need to use a program to compute the incomplete beta function, doesn't look like there's any possible simplification there. — 2017-01-04
Click on the link to the wikipedia article on the incomplete beta function and go to the section "Software implementation". There, for instance, you will see how to compute the incomplete beta function in SymPy. — 2017-01-04

user id:269576 749 · Accepted Answer

What about that?

$$ \pi a \csc (\pi a)-x_0 \left(\frac{1}{x_0}-1\right){}^a \left(a \Gamma (1-a) \, _2\tilde{F}_1\left(1,1;2-a;x_0\right)+1\right) $$

Thanks @nikolay, unfortunately I can't really tell how you got to that solution. Can you elaborate? — 2017-01-04

user id:87813 9k · Accepted Answer

Define the incomplete beta function as

$$\beta(z; a,b) = \int^z_0 x^{a-1} (1-x) ^{b-1} \,dx$$

Then

$$\int_{x_0}^1 (1-x)^{a} x^{-a}~dx = \int^{1-x_0}_0 x^{a} (1-x)^{-a}~dx = \beta(1-x_0; 1+a,1-a)$$

For $x_0 = 0$ we have the complete Beta function

$$ \int^{1}_0 x^{a} (1-x)^{-a}~dx = \beta(1; 1+a,1-a) = \beta(1+a,1-a) = \Gamma(1+a)\Gamma(1-a)$$

Thanks @zaid, this looks simpler than the answer by poweierstrass but I see you upvoted that one. Can you explain why the other answer is preferable? — 2017-01-04
@mikeck if you make the substitution $x \to 1-x$ the answer will look simpler. Both answers are correct. — 2017-01-04
You could further simplify since $\Gamma(1+a)\Gamma(1-a)=\pi a \csc (\pi a)$ — 2017-01-05

integral of $(\frac{1 - x}{x})^a$

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