Is there any problem on evaluating the indefinite integral that couldn't be solved?

Question

In another word, is there any indefinite integral that couldn't find another representation

This is one of them: $$ \int_0^{\Large\frac\pi4} \frac {\sin x} {x \cos^2 x} \mathrm d x. $$At least, how to solve it analytically is unknown yet. Wanna give a try? :) — Tunk-Fey, May 02 '14 at 20:45
One source of examples would be "cheating" by encoding unsolved problems as integral equalities. Sometimes this might be even useful/natural. — Marcin Łoś, May 02 '14 at 20:45
There are general algorithms that give the indefinite integral in terms of given primitives, or show it can't be done. This is more the realm of symbolic computation. — vonbrand, May 02 '14 at 20:45
Something else you could do is look at unanswered questions on SE : http://math.stackexchange.com/questions/tagged/indefinite-integrals?sort=unanswered&pageSize=50 — user88595, May 02 '14 at 20:56
See http://math.stackexchange.com/questions/155/how-can-you-prove-that-a-function-has-no-closed-form-integral — Martin Sleziak, May 02 '14 at 22:23

score 1 · Accepted Answer · edited Jun 04 '16 at 13:26

1

For example, $$\int \frac{1}{x-\sin x} dx$$

edited Jun 04 '16 at 13:26

Bérénice

answered May 03 '14 at 05:15

arslan ansari

score 0 · Answer 2 · edited May 02 '14 at 22:25

0

here is an example:

$$\int \frac{(x+1)e^x}{x(1+xe^x)} dx$$

There should be a lot of these I think

edited May 02 '14 at 22:25

Martin Sleziak

answered May 02 '14 at 21:30

Victor

2 Answers2