1
$\begingroup$

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 $x$.

$\endgroup$
5
  • $\begingroup$ what is $a$ here? $\endgroup$ Jan 4, 2017 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, 2017 at 17:45
  • $\begingroup$ Wolfram worked for me: wolframalpha.com/input/?i=int_c%5E1+(1-x)%5Ea%2Fx%5Ea+dx $\endgroup$
    – David P
    Jan 4, 2017 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, 2017 at 17:48
  • $\begingroup$ @DavidP The incomplete beta function makes that result basically a rewriting, not a solution. $\endgroup$
    – Arthur
    Jan 4, 2017 at 17:50

3 Answers 3

3
$\begingroup$

\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.

$\endgroup$
2
  • $\begingroup$ 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, 2017 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. $\endgroup$ Jan 4, 2017 at 21:04
0
$\begingroup$

What about that?

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

$\endgroup$
1
  • $\begingroup$ Thanks @nikolay, unfortunately I can't really tell how you got to that solution. Can you elaborate? $\endgroup$
    – mikeck
    Jan 4, 2017 at 20:42
0
$\begingroup$

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

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

$\endgroup$
5
  • $\begingroup$ 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, 2017 at 20:41
  • 1
    $\begingroup$ @mikeck if you make the substitution $x \to 1-x$ the answer will look simpler. Both answers are correct. $\endgroup$ Jan 4, 2017 at 22:07
  • $\begingroup$ You could further simplify since $\Gamma(1+a)\Gamma(1-a)=\pi a \csc (\pi a)$ $\endgroup$ Jan 5, 2017 at 5:55
  • $\begingroup$ @ClaudeLeibovici, yes sure using the reflection formula. $\endgroup$ Jan 5, 2017 at 6:19
  • $\begingroup$ No idea why would someone downvote this . $\endgroup$ Jan 9, 2017 at 12:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .