# Tricky (impossible?) Integral

1. Jan 23, 2008

### KingOrdo

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

$$\int x^{(x-1)}dx$$

2. Jan 24, 2008

### Defennder

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

3. Jan 24, 2008

### HallsofIvy

Staff Emeritus
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.

4. Jan 24, 2008

### KingOrdo

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

5. Jan 24, 2008

### HallsofIvy

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

6. Jan 24, 2008

### 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).

7. Jan 24, 2008

### ice109

do you the antiderivative or the definite integral?

8. Jan 24, 2008

### arildno

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

9. Jan 24, 2008

### 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.

10. Jan 24, 2008

### arildno

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

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

11. Jan 24, 2008

### KingOrdo

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

I am asking how to integrate it.

12. Jan 24, 2008

### morphism

The answer probably is "no one knows".

13. Jan 24, 2008

### 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-

14. Jan 24, 2008

### ice109

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

15. Jan 24, 2008

### 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.

16. Jan 24, 2008

### KingOrdo

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

Antiderivative.

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

17. Jan 24, 2008

### Gib Z

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

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 $$\int^{ab}_1 \frac{1}{t} dt = \int^b_1 \frac{1}{t} dt + \int^a_1 \frac{1}{t} dt$$. =]

18. Jan 24, 2008

### Defennder

Here's what Wikipedia says:

In other words, you don't need to know the anti-derivate to approximate the definite integral, which is pretty much what the others have said. That function doesn't have an elementary function for an anti-derivative.

Last edited by a moderator: May 3, 2017
19. Jan 24, 2008

### KingOrdo

Gib Z, I do not require a "nice analytical solution". But I do require a non-trivial one.

And I do not want to know the definite integral. Had I wanted the definite integral, I would have asked for it. When I say, 'What is A?', I want to know what A is--not what B is.

20. Jan 24, 2008

### Ben Niehoff

KO, you must understand that a great many people come to this board asking the wrong questions, or using the wrong terminology, because they don't understand what they are asking about. It is only natural for us to ask for clarifications and to clear up potential misunderstandings. That we have done so here should give you no cause to be an ass.

Anyway, there is a very easy way for you to answer your own question. First, assume that your integral has an antiderivative, and call it f:

$$f(x) = \int_A^x t^{t-1} dt$$

for some constant A. Now, suppose that f(x) has a power series representation:

$$f(x) = \sum_{k=0}^{\infty} C_k (x-x_0)^k$$

Now, you need to find some x_0 about which to expand the power series. The function x^(x-1) is undefined at x=0, so the integral might not exist there.

At any rate, once you choose an appropriate x_0, you can begin by taking derivatives of f(x), and evaluating them at x=x_0. You should then find an easy way to get all of the coefficients C_k.

Have fun.