Tricky (impossible?) Integral

KingOrdo

Anyone know how to do this integral? I can't figure it out analytically, nor do I find it in my tables. Thanks!

[tex]
\int x^{(x-1)}dx
[/tex]

Defennder

I used the Integrator http://integrals.wolfram.com/index.jsp and got the following, which implies no elementary solutions exist:

Mathematica could not find a formula for your integral. Most likely this means that no formula exists.

HallsofIvy

Do you have any reason to believe that such a formula does exist? "Almost all" such functions do not have an anti-derivative in any simple form.

KingOrdo

Quote by HallsofIvy

Do you have any reason to believe that such a formula does exist? "Almost all" such functions do not have an anti-derivative in any simple form.

Nope. I never claimed such, nor did I specify that I would only accept a solution in a "simple form".

HallsofIvy

Ah, well, if you don't require it in "simple form", I can definitely say that
[tex]\int x^{x-1} dx= Ivy(x)+ C[/itex]
where I have defined "Ivy(x)" to be the anti-derivative of
[tex]\int x^{x-1}dx[/tex]
such that Ivy(0)= 0!

KingOrdo

Quote by HallsofIvy

Ah, well, if you don't require it in "simple form", I can definitely say that
[tex]\int x^{x-1} dx= Ivy(x)+ C[/itex]
where I have defined "Ivy(x)" to be the anti-derivative of
[tex]\int x^{x-1}dx[/tex]
such that Ivy(0)= 0!

Jokes aside, again: It need not be in "simple form"--however such a term is defined in detail--but it must be a solution (in the sense that it must help me calculate, in principle at least, the integral).

*ice109*

do you the antiderivative or the definite integral?

arildno

Quote by KingOrdo

Jokes aside, again: It need not be in "simple form"--however such a term is defined in detail--but it must be a solution (in the sense that it must help me calculate, in principle at least, the integral).

Why do you need an anti-derivative in order to get good value estimates of definite integrals?

KingOrdo

Quote by ice109

do you the antiderivative or the definite integral?

I can't answer this until I know what verb you left out of your question.

Quote by arildno

Why do you need an anti-derivative in order to get good value estimates of definite integrals?

You don't. No one claimed such a thing.

arildno

Quote by KingOrdo

Jokes aside, again: It need not be in "simple form"--however such a term is defined in detail--but it must be a solution (in the sense that it must help me calculate, in principle at least, the integral).

What do you mean by "must be a solution"?

The integrand can be shown to be integrable, that's all you need to check.

KingOrdo

Quote by arildno

What do you mean by "must be a solution"?

I mean it must not be trivial (cf. HallsofIvy's "solution").

Quote by arildno

The integrand can be shown to be integrable, that's all you need to check.

I am asking how to integrate it.

morphism

Quote by KingOrdo

I am asking how to integrate it.

The answer probably is "no one knows".

Gib Z

HallsofIvy's solution is the best you will get, and theres actually nothing wrong with it either.

Here I quote Courant from page 242, Volume 1;

After stating that in the 19th century it was proven certain elementary integrands did not have elementary antiderivatives-

If the object of the integral Calculus were to integrate functions in terms of elementary functions, we should have come to a definite halt. But such a restricted object has no intrinsic justification; indeed, it is of a somewhat artificial nature. We know that the integral of every continuous function exists and is itself a continuous function of the upper limit, and this fact has nothing to do with the question whether the integral can be expressed in terms of elementary functions or not. The distinguishing features of the elementary functions are based on the fact that their properties are easily recognized, that their application to numerical problems is often facilitated by convenient tables or, as in the case of the rational functions, that they can easily be calculated with as great a degree of accuracy as we please.

Where the integral of a function cannot be expressed by means of functions with which we are already acquainted, there is nothing to hinder us from introducing this integral as a new "higher" function in analysis, which really means no more than giving it a name. Whether the introduction of such a function is convenient or not depends on the properties which it possesses, the frequency with which it occurs, and the ease with which it can be manipulated in theory and in practice.

*ice109*

Quote by KingOrdo

I can't answer this until I know what verb you left out of your question.

You don't. No one claimed such a thing.

do you need the antiderivative or the definite integral over some interval

Gib Z

O ice's post reminds me; if your a desperate little one, less than a year ago I remember a paper of arVix or however you spell that site, its well known, about using hyper geometric series in a method of "approximate solutions for antiderivatives". I am not sure about the credibility of the paper, and when I skimmed through it, it seemed very elaborated. Obviously it didn't get much attention from the mathematical community, as it didn't really have any real use. Which once again brings us back to the fact that Hall's solution really actually is the best you can do in this case.

If its a definite integral, nothings stopping you from getting any degree of accuracy you want using numerical methods.

KingOrdo

Quote by Gib Z

HallsofIvy's solution is the best you will get, and theres actually nothing wrong with it either.

HallsofIvy's "solution" is in fact worse than incorrect, because it is useless.

Quote by ice109

do you need the antiderivative or the definite integral over some interval

Antiderivative.

Quote by Gib Z

O ice's post reminds me; if your a desperate little one, less than a year ago I remember a paper of arVix or however you spell that site, its well known, about using hyper geometric series in a method of "approximate solutions for antiderivatives". I am not sure about the credibility of the paper, and when I skimmed through it, it seemed very elaborated. Obviously it didn't get much attention from the mathematical community, as it didn't really have any real use. Which once again brings us back to the fact that Hall's solution really actually is the best you can do in this case.

Thanks; I'll take a look at the arXiv. Any other suggestions along these lines would be appreciated.

Gib Z

Ok. Lets go back to the days when the natural logarithm was not yet defined, and some poor souls ran into [tex]\int^x_1 \frac{1}{t} dt[/tex].

Yes, most just complained that it didn't have any nice anti derivative at all! How could they cope?

But some of the bright ones said, "well nothings going to stop me from defining a new function, and showing it has these nice properties, like f(ab) = f(a) + f(b), etc etc. Ooh, hold on a second, these properties are the same ones the inverse of the exponential function must have! Hooray!"

Point is, you don't need a nice analytical solution to everything to work out somethings properties and actually still achieve a "solution".

It may be a nice exercise for you to prove [tex]\int^{ab}_1 \frac{1}{t} dt = \int^b_1 \frac{1}{t} dt + \int^a_1 \frac{1}{t} dt[/tex]. =]

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