0
$\begingroup$

I find myself needing to compute the integral $$\int_{x = 0}^X\left(1+x^2\right)^{8/3}\mathrm{d}x$$ for real, positive $X$. Gradshteyn and Ryzhik (2007 edition) paragraph 2.202 implies that the answer can't be expressed in terms of elementary functions, but I really don't mind it being expressed in terms of special functions. Does this integral look familiar to anyone as the definition of a special function, please?

(If it helps, I found substitutions that convert the integral to $\int\cosh^{19/3}\left(\theta\right)\mathrm{d}\theta$ or $\int\sec^{22/3}\left(\theta\right)\mathrm{d}\theta$; or repeated integration by parts could reduce it to $\int_{x = 0}^Xx^6\left(1+x^2\right)^{-1/3}\mathrm{d}x$.)

$\endgroup$
3
  • 1
    $\begingroup$ There is an incomplete beta function solution $\endgroup$ Commented Nov 17, 2023 at 17:07
  • $\begingroup$ @TymaGaidash I think I see it. $\mathrm{B}\left(-X^2;1/2,11/3\right)$, yes? Care to make that an answer? $\endgroup$ Commented Nov 17, 2023 at 17:24
  • $\begingroup$ @ТymaGaidash Yep, confirmed that, on differentiating that incomplete beta function in Maxima, I recover the original integrand. Thanks again. $\endgroup$ Commented Nov 17, 2023 at 17:54

1 Answer 1

2
$\begingroup$

According to Maple, it's $$X \; {}_{2}{{{F_{1}^{}}}}\! \left(-\frac{8}{3},\frac{1}{2};\frac{3}{2};-X^{2}\right)$$

And more generally,

$$ \int (1+x^2)^p \; dx = x \; {}_{2}{{{F_{1}^{}}}}\! \left(-p,\frac{1}{2};\frac{3}{2};-x^{2}\right) + c$$

EDIT:

From the definition, $$x\; {}_2F_1\left(-p,1/2; 3/2; -x^2\right) = \sum_{n=0}^\infty \frac{\Gamma(n-p) (-1)^n x^{2n+1}}{\Gamma(-p) (2n+1) n!} $$ Differentiate term-by-term and you get $$ \sum_{n=0}^\infty \frac{\Gamma(n-p) (-1)^n x^{2n}}{\Gamma(-p) n!} = \sum_{n=0}^\infty \frac{\Gamma\left(p+1\right) x^{2n}}{\Gamma\left(p-n+1\right) n!} = (1+x^2)^p $$ by the binomial series.

$\endgroup$
5
  • $\begingroup$ Thanks. $_2F_1$ means the Gauss hypergeometric function, right? $\endgroup$ Commented Nov 17, 2023 at 16:04
  • 1
    $\begingroup$ Yes, that's right. $\endgroup$ Commented Nov 17, 2023 at 17:15
  • $\begingroup$ Thanks again. Tried to confirm by working backwards, differentiating that answer in Maxima, but so far I can't get the derivative of that answer to match the original integrand. $\endgroup$ Commented Nov 17, 2023 at 17:26
  • $\begingroup$ ... although the NIST Digital Library of Mathematical Functions has an equivalence theorem between incomplete beta functions and hypergeometric functions, with which a form somewhat (but not, I think, exactly) like this could be derived from the solution @TymaGaidash and I have been discussing in the comments to the question. $\endgroup$ Commented Nov 18, 2023 at 19:30
  • 1
    $\begingroup$ I've added a confirmation to the answer. $\endgroup$ Commented Nov 18, 2023 at 23:23

You must log in to answer this question.

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