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.
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?
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.
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: