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 $x$.

  • $\begingroup$ what is $a$ here? $\endgroup$ – Dr. Sonnhard Graubner Jan 4 '17 at 17:44
  • $\begingroup$ I would think that an analytical expression for it doesn't exist, if Wolfram Alpha couldn't solve it quickly enough. $\endgroup$ – Nathan H. Jan 4 '17 at 17:45
  • $\begingroup$ Wolfram worked for me: wolframalpha.com/input/?i=int_c%5E1+(1-x)%5Ea%2Fx%5Ea+dx $\endgroup$ – David Peterson Jan 4 '17 at 17:46
  • $\begingroup$ @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. $\endgroup$ – Arthur Jan 4 '17 at 17:48
  • $\begingroup$ @DavidP The incomplete beta function makes that result basically a rewriting, not a solution. $\endgroup$ – Arthur Jan 4 '17 at 17:50

\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.

  • $\begingroup$ Thanks @poweierstrass, it also looks like the hypergeometric function 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. $\endgroup$ – mikeck Jan 4 '17 at 20:39
  • $\begingroup$ 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. $\endgroup$ – poweierstrass Jan 4 '17 at 21:04

What about that?

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

  • $\begingroup$ Thanks @nikolay, unfortunately I can't really tell how you got to that solution. Can you elaborate? $\endgroup$ – mikeck Jan 4 '17 at 20:42

Define the incomplete beta function as

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


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

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

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

  • $\begingroup$ 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? $\endgroup$ – mikeck Jan 4 '17 at 20:41
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    $\begingroup$ @mikeck if you make the substitution $x \to 1-x$ the answer will look simpler. Both answers are correct. $\endgroup$ – Zaid Alyafeai Jan 4 '17 at 22:07
  • $\begingroup$ You could further simplify since $\Gamma(1+a)\Gamma(1-a)=\pi a \csc (\pi a)$ $\endgroup$ – Claude Leibovici Jan 5 '17 at 5:55
  • $\begingroup$ @ClaudeLeibovici, yes sure using the reflection formula. $\endgroup$ – Zaid Alyafeai Jan 5 '17 at 6:19
  • $\begingroup$ No idea why would someone downvote this . $\endgroup$ – Zaid Alyafeai Jan 9 '17 at 12:17

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