Impossible Integral?

Factual Questions
1

New math question on behalf of my fiancee:

Does:



  /\x+t     (sin^2 s)
  |         e               ds
 \/x-t

have a solution? I see that the error function doesn’t, and this seems similar, but we’re hoping the sine makes it different.

We tried several things, and the most promising was using the Taylor expansion of e^x, but after some manipulation that gave us something like:
(1-u^2)^(n-0.5)/n! du
which is nasty to integrate.

Since the board was down, we’ve been searching all over for an answer, but we’re stumped.

Anyone know?
Thanks in advance.
Thanks,
Ron

2

Do you have Mathematica or some other such program? Try it out there. If you really need to know this, I’ll check tonight on my desktop computer.

3

This integral stumps “Maple”. I think you’re right: it’s not possible to integrate it symbolically.

4

Forgive me for butting in, but IIRC, there are actually a large number of common integrals which cannot be symbolically evaluated, true? :confused:

Isn’t one of them:

(Integral) x[sup]x[/sup] dx? I don’t know…

5

Most integrals can’t be evaluated symbolically, including many common ones. The one you listed is one of them.

6

Oh, and the integral in the OP is one of them.

7

In my travels trying to solve this, I came across www.integrals.com which uses a Mathematica backend, and it was stumped.

I also came across the un(symbolically)integratable x^x.

Thanks, ultafilter, for the confirmation that this is also one of those functions. Is there a method for determining if a given function can be integrated smbolically?

8

oops, sorry for the misspelled name ultrafilter

9

Stick it into Mathematica and see what happens?

That’s pretty much the best I can give you.

10

Oh good, I’m glad I wasn’t completely crazy.

I’ve always liked that one because it’s so simple-seeming.

11

This brings to mind an interesting question. Strictly speaking, values like sin and cos can’t be ‘simply integrated’ except with respect to each other. Their polynomial expansions by Taylor or MacLaurin are reasonably easy to integrate, but all you end up with is a sum of an infinite series again.

I think my point is that the trigonometric functions encapsulate certain very useful functions the calculus properties of which are well known. They belong to a family of functions that seem to be related under the calculus. Furthermore, we find an astonishing number of itegrals that are suddenly possible because we have these trigonometric functions.

Same thing goes with complex and hypercomplex numbers, too.

While, say, x[sup]x[/sup] may not be integrable, could we call this function, say, q(x), and see if it belongs to a family of functions that are similarly related? If so, then while we couldn’t easily intregrate x[sup]x[/sup] with existing algebraic and trigonometric tools, perhaps we can define a new set of tools that can expand the number of integrals we can do.

I’m not proposing simply calling the integral of x[sup]x[/sup] t(x) so we have the integral of q being t… that’s just a trivial renaming. But if q(x) belongs to a family of related functions like the trigonometric ones, maybe we do have something.

12

About the OP, I agree that you can’t solve it analytically. However, there is one form of the solution as an infinite series which may be useful, which nobody has mentioned yet. If you set the integral equal to f(t), then you can do a Maclaurin-series expansion. f(0) = 0, and:

f’(t) = exp(sin[sup]2[/sup](x + t)) + exp(sin[sup]2[/sup](x - t))

According to my calculator, to fourth order in t:

f(t) = A(x)t + B(x)t[sup]3[/sup] + …

A(x) = 2exp(sin[sup]2[/sup]x)
B(x) = 2(2(sin[sup]2[/sup]x + 1)cos[sup]2[/sup]x - 1) × exp(sin[sup]2[/sup]x) / 3

f(t) is, of course, an odd function, and I believe this series converges for all values of t. If you know t is small, this may be helpful.

13

William: Stuff like that is done. You might want to check out gamma function theory, which is probably the simplest example.