Mastering the Challenging Integral: Tips and Tricks for Solving x^(x-1)dx

[KingOrdo]

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

<br /> \int x^{(x-1)}dx<br />

 

[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]

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
\int x^{x-1} dx= Ivy(x)+ C[/itex] <br /> where I have defined &quot;Ivy(x)&quot; to be the anti-derivative of <br /> \int x^{x-1}dx <br /> such that Ivy(0)= 0!

 

[KingOrdo]

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] <br /> where I have defined &quot;Ivy(x)&quot; to be the anti-derivative of <br /> \int x^{x-1}dx <br /> such that Ivy(0)= 0!

<br /> <br /> Jokes aside, again: It need not be in &quot;simple form&quot;--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]

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]

do you the antiderivative or the definite integral?

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

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]

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]

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

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

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

I am asking how to integrate it.

 

[morphism]

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 there's 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]

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]

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

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

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

Antiderivative.

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. Let's 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. =]

 

[Defennder]

Here's what Wikipedia says:

if[/PLAIN] a function has no elementary antiderivative (for instance, exp(x2)), its definite integral can be approximated using numerical integration

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.

 

[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.

 

[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.

 

[KingOrdo]

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.

You are the first to use a pejorative word (viz. "ass"). I try to be rigorous in both my mathematics and my language. What other posters say, mean, or confuse is irrelevant, and it is not fair to ascribe their mistakes--or their strengths--to me.

Anyway, there is a very easy way for you to answer your own question.

This is semantically false. If I knew how to "answer [my] own question", I wouldn't be posing it here. What you mean is, 'There is a very easy way to answer your 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.

Excellent; thanks.

 

[Gib Z]

Ooh yes, I should have remembered that as well. Taylor series of the integrand, integrate term by term. But really that series is no elementary function either. Analytical functions always have an antiderivative in terms of a series, but that series may be even worse than numerical methods, it probably converges horribly slow.

 

[HallsofIvy]

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

Yes, I was being facetious. You did leave yourself open to that when you said you were not just looking for a "simple" function!

And, I had a serious point. "Almost all" integrable functions don't have an antiderivative in any form we've seen. Many important functions have to be defined as the anti-derivative of a given function (they just won't let me do it!). Some calculus texts define ln(x) as \int_1^x (/1/t)dt. The Error function is defined as \int_0^x e^{-t^2}dt.

That also happens in differential equations. Arey's function is defined as a specific solution to a differential equation. Bessel's functions are defined as solutions to differential equations.

 

[Ben Niehoff]

Ooh yes, I should have remembered that as well. Taylor series of the integrand, integrate term by term. But really that series is no elementary function either. Analytical functions always have an antiderivative in terms of a series, but that series may be even worse than numerical methods, it probably converges horribly slow.

True that, and its radius of convergence might be minuscule to boot! But KO only wanted an answer; not a good answer. ;)

 

[KingOrdo]

True that, and its radius of convergence might be minuscule to boot! But KO only wanted an answer; not a good answer. ;)

You are claiming something false. I do not understand why so many people on this board are snarky--and, worse, unhelpful.

I just need to solve the integral. My request is extremely simple to understand (high school calculus should be sufficient), even if not to answer. A solution given in terms of elementary functions would be best. If no such solution exists (as seems to be the case), explanations of solutions using series (e.g.), or via numerical methods, would be appreciated.

I do not want trivialities, obfuscations, or any other kind of non-rigorous nonsense.

 

[jostpuur]

I would very much like to know the proof for why that integral function cannot be expressed as a finite composition of elementary functions.

 

[pkleinod]

I := \int x^{x-1}dx = \int e^{x\log x}d\log x.
Now let u = log x so that x = e^u. Then
I = \int (1 +y + \frac{y^2}{2!} + \frac{y^3}{3!} + \ldots )du,
where y = u e^u. I think each term in this series can be integrated
in closed form. For 0 < x < 1, u < 0, so that you might be able to use this
series in a real calculation, but for x > 1, this formula is probably useless.
But it fulfills your wish, at least partially (unless, of course, I have made
a mistake, which is highly probable.) I have no idea whether this approach is better or worse than the suggestion to use a Taylor series.

Please note: hell is the place where your every wish is fulfilled;
heaven is where you have no wishes and
the coffee tastes the same as it smells.

 

[Gib Z]

That suggestion is no worse or better than using a Taylor series, as it is the same =]