Mathematics desk
< November 15	<< Oct \| November \| Dec >>	November 17 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

November 16

Integrals without closed form antiderivatives

I was speculating about this but do not have the knowledge unfortunately to make progress on this question. This is not homework. Roughly: can the antiderivative of any sum, product, or ratio of elementary functions always be expressed as a finite linear combination of elementary functions and named integrals (i.e., the Bessel function, the sin integral, the cos integral)? Or are there an infinite number of unnamed intractable integrals that cannot be transformed into a known named integral? Thanks. 160.39.130.28 (talk) 06:00, 16 November 2014 (UTC)[reply]

"Named integrals" is an ill-defined set. But I can already say it is definitely not possible. For example there is no elementary form for

\int e^{x^{n}}dx

for n>1 (the error function can provide the case n=2 but I know of no named function for n 3 or up, infinite series not counting).

I was going to ask a similar question: can we define a set of functions that is closed under the operation of antidifferentiation, that is not something like the set of all continuous functions, yet includes any finite composition of any of its members?--Jasper Deng (talk) 06:28, 16 November 2014 (UTC)[reply]

For positive values of n, we have

\int _{0}^{\infty }e^{-x^{n}}dx=\Gamma {\Big (}1+{\tfrac {1}{n}}{\Big )}

and

\int e^{-x^{n}}dx=-\Gamma {\Big (}1+{\tfrac {1}{n}},x^{n}{\Big )}

See gamma function and incomplete gamma function — 79.118.167.116 (talk) 03:55, 17 November 2014 (UTC)[reply]

One could start with any finite set of functions, integrate them and take all compositions. This gives another set of functions, also finite. Do it again, etc. This gives a countable family of functions, and so cannot be all continuous functions. It's doubtful that there is any family besides polynomials and all continuous functions that can easily be identified, though. The requirement that it be closed under composition makes it highly non-trivial. Sławomir Biały (talk) 11:15, 16 November 2014 (UTC)[reply]

I may get this question wrong, but my hunch is that even among the nicest families of functions (say continuous ones), most functions are totally intractable in every respect except maybe for their unifying feature (e.g. continuity). YohanN7 (talk) 11:46, 16 November 2014 (UTC)[reply]

To OP: Even if you confine attention to integrals of algebraic function, there is no finite set of functions that can express all of the others. You get generalized elliptic functions that live on the Jacobians of curves, and there is no known way to parametrize all of these. Sławomir Biały (talk) 12:14, 16 November 2014 (UTC)[reply]

You may be interested in Differential Galois theory,Liouville's theorem, and Risch algorithm; they explore related notions. The following papers give some introduction and notions of things: [1], [2], [3], and [4].Phoenixia1177 (talk) 07:02, 17 November 2014 (UTC)[reply]