- #1
KingOrdo
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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]
[tex]
\int x^{(x-1)}dx
[/tex]
Mathematica could not find a formula for your integral. Most likely this means that no formula exists.
HallsofIvy said: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.
HallsofIvy said: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!
Why do you need an anti-derivative in order to get good value estimates of definite integrals?KingOrdo said: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).
I can't answer this until I know what verb you left out of your question.ice109 said:do you the antiderivative or the definite integral?
You don't. No one claimed such a thing.arildno said:Why do you need an anti-derivative in order to get good value estimates of definite integrals?
KingOrdo said: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).
I mean it must not be trivial (cf. HallsofIvy's "solution").arildno said:What do you mean by "must be a solution"?
I am asking how to integrate it.arildno said:The integrand can be shown to be integrable, that's all you need to check.
The answer probably is "no one knows".KingOrdo said:I am asking how to integrate it.
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.
KingOrdo said: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.
HallsofIvy's "solution" is in fact worse than incorrect, because it is useless.Gib Z said:HallsofIvy's solution is the best you will get, and there's actually nothing wrong with it either.
Antiderivative.ice109 said:do you need the antiderivative or the definite integral over some interval
Thanks; I'll take a look at the arXiv. Any other suggestions along these lines would be appreciated.Gib Z said: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.
[PLAIN]http://en.wikipedia.org/wiki/Antiderivative said:if[/PLAIN] a function has no elementary antiderivative (for instance, exp(x2)), its definite integral can be approximated using numerical integration
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.Ben Niehoff said: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.
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.'Ben Niehoff said:Anyway, there is a very easy way for you to answer your own question.
Excellent; thanks.Ben Niehoff said:First, assume that your integral has an antiderivative, and call it f:
[tex]f(x) = \int_A^x t^{t-1} dt[/tex]
for some constant A. Now, suppose that f(x) has a power series representation:
[tex]f(x) = \sum_{k=0}^{\infty} C_k (x-x_0)^k[/tex]
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.
Yes, I was being facetious. You did leave yourself open to that when you said you were not just looking for a "simple" function!KingOrdo said:HallsofIvy's "solution" is in fact worse than incorrect, because it is useless.
Gib Z said: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.
You are claiming something false. I do not understand why so many people on this board are snarky--and, worse, unhelpful.Ben Niehoff said:True that, and its radius of convergence might be minuscule to boot! But KO only wanted an answer; not a good answer. ;)
The purpose of mastering this challenging integral is to develop a deeper understanding of integral calculus and to improve problem-solving skills. This integral is particularly difficult due to the presence of both a variable base and exponent.
Yes, there are several techniques that can be used to solve this integral. One approach is to use the power rule for integration and then apply the substitution method. Another method is to use the logarithmic differentiation technique.
Some tips for solving this integral include simplifying the expression by factoring out a common factor, using the properties of logarithms, and breaking down the integral into smaller parts. It is also helpful to practice solving similar integrals and familiarize oneself with common patterns and techniques.
Yes, some common mistakes to avoid include incorrect application of the power rule, forgetting to use the chain rule when applying substitution, and making errors in simplifying the expression. It is important to carefully check each step of the solution to avoid these mistakes.
Mastering this integral can be beneficial in real-world applications as it is a fundamental concept in many fields such as physics, engineering, and economics. Being able to solve challenging integrals can help in solving real-world problems involving rates of change, optimization, and finding areas under curves.