Why is it difficult to integrate x^x

[latyph]

I tried doing this but could not,why is it so?

 

[dextercioby]

It's not difficult.Nobody thought of defining a special function to account for its antiderivative.

Daniel.

 

[Zurtex]

Because it can't be integrated in terms of elementary functions. Most functions are 'not easy' to integrate in this way.

 

[HallsofIvy]

Or, to say the same thing that Dextercioby and Zurtex said, in different words, because there is no elementary function whose derivative is x^x!

 

[arildno]

A better question would be:
Why is it simple to integrate f(x)=1?

 

[abia ubong]

i tried doing it ,but always get to a place i can't continue.who can help integrate [(x^x)(1+LOG[X])]^2.All help will be appreciated

 

[arildno]

i tried doing it ,but always get to a place i can't continue.who can help integrate [(x^x)(1+LOG[X])]^2.All help will be appreciated

You won't get any, since an anti-derivative of x^x is inexpressible in terms of elementary functions.

 

[goldi]

when i plugged it into the integrator of mathematica it gave it back as same...i don't know why it did not do computation.

 

[Zurtex]

when i plugged it into the integrator of mathematica it gave it back as same...i don't know why it did not do computation.

Because as said many times previously in this thread, it can not be integrated in terms of elementary functions.

 

[goldi]

i understand that yeah...but even Mathematica couldn't post the solution in terms of complex functions or whatever high level function it ocntains...
there must be a solution to it...What it is?

 

[Zurtex]

i understand that yeah...but even Mathematica couldn't post the solution in terms of complex functions or whatever high level function it ocntains...
there must be a solution to it...What it is?

There are no special functions defined in general mathematics for the integral.

If you want a function that is the anti-derivative of x^x then just define one and then you can study its properties.

 

[matt grime]

There is a solution it is the function F such that dF/dx is x^x. But we can't write it anymore nicely than that, and there is nothing surprising about it. Almost no functions have integrals that we can write out nicely and explicitly in some closed form. How many times must that be said in this thread? Shall we lock it now to stop yet another person having to say it?

 

[goldi]

Sorry but i am such a big fool that yours terminology is not clear to me...Last time-Has its integral ever calculated ...or as ppl are saying that it has such a function as integral that has never been defined./so is research going on over this

i have a question...how can we integrate x*Sec(x)
i have tried this question than any other question ever...
the point is that it was asked in my 12th class and when i plug it into Integrator i could not even understand the solution...

 

[arildno]

Any particular definite integral of x^x can, of course be calculated to an arbitrary degree of accuracy by numerical techniques.

 

[Jameson]

To integrate x*Sec(x) I would use integration by parts, but in this case the tabular method will work nicely.

\int x\sec{x}dx

\int udv = uv - \int vdu

u = x
dv = \sec{x}dx

That should get you started.

 

[dextercioby]

I think you meant

dv=\sec x \ dx

I'm not sure though...

Daniel.

 

[Icebreaker]

If you want a function that is the anti-derivative of x^x then just define one and then you can study its properties.

Has it been done before? Any interesting properties?

 

[Pyrrhus]

Look for Liouville's Principle about integration in finite terms.

 

[goldi]

To integrate x*Sec(x) I would use integration by parts, but in this case the tabular method will work nicely.

\int x\sec{x}dx

\int udv = uv - \int vdu

u = x
dv = \sec{x}dx

That should get you started.

that i would have had tried 100 times...after 1 step i am stuck and there is no way out...

 

[calculus1967]

i understand that yeah...but even Mathematica couldn't post the solution in terms of complex functions or whatever high level function it ocntains...
there must be a solution to it...What it is?

Well, there's F where
F(x) = \int_a^x t^t dt
and a can be any number greater than or equal to 0. Mathematica isn't able to find that.

 

[matt grime]

For those who think functions are *nice* (goldi etc) and aren't happy with our answers, can I ask what sin(x) is? I mean given x=2 radians say what is sin(x)? How do you define it? How do you find it? To me the answer is sin(2). There is nothing wrong with calculus's answer in the last post using the fundamental theory of calc. It is a very good function.

 

[Zurtex]

Has it been done before? Any interesting properties?

Attached below is a picture between x = 0 and x = 3 of the function:

f(x) = \lim_{a \rightarrow 0} \int_a^x t^t dt

 

[heman]

that i would have had tried 100 times...after 1 step i am stuck and there is no way out...

i checked into mathematica,,,it contains some functions of the form polylog but i think this is calculated with the knowledge of Complex Analysis.

 

[krab]

I tried doing this but could not,why is it so?

You could do it by Taylor expansion:

\int x^x dx=x + \frac{\left( -1 + <br /> 2\,\log (x) \right) \,<br /> x^2}{4} + <br /> \frac{\left( 2 - <br /> 6\,\log (x) + <br /> 9\,{\log (x)}^2 \right)<br /> \,x^3}{54} + <br /> \frac{\left( -3 + <br /> 12\,\log (x) - <br /> 24\,{\log (x)}^2 + <br /> 32\,{\log (x)}^3<br /> \right) \,x^4}{768} + <br /> \frac{\left( 24 - <br /> 120\,\log (x) + <br /> 300\,{\log (x)}^2 - <br /> 500\,{\log (x)}^3 + <br /> 625\,{\log (x)}^4<br /> \right) \,x^5}{75000} +<br /> \frac{\left( -5 + <br /> 30\,\log (x) - <br /> 90\,{\log (x)}^2 + <br /> 180\,{\log (x)}^3 - <br /> 270\,{\log (x)}^4 + <br /> 324\,{\log (x)}^5<br /> \right) \,x^6}{233280}<br /> + {O(x^7)

 

[shmoe]

I wonder if students aren't done a disservice in first year calculus classes with their sterilized examples and problems. They'll be asked to do hundreds of integration problems, all rigged to work out nicely with the techniques they've just learned. Perhaps it will be mentioned that there are functions whose antiderivatives cannot be written in a "nice" form, but examples will be scarce- e^{-x^2} being the stock one. After seeing such an unnatural ratio of nice examples to possibly one or two 'not-nice' ones, it's no surprise that many walk away believing themselves invincible and any function that itself looks 'nice' will have a 'nice' antiderivative waiting around the corner so they flap their arms around and bash their heads in frustration trying to find it. Makes me wonder if they bother to even consider why numerical techniques are taught at all?

 

[Zurtex]

I wonder if students aren't done a disservice in first year calculus classes with their sterilized examples and problems. They'll be asked to do hundreds of integration problems, all rigged to work out nicely with the techniques they've just learned. Perhaps it will be mentioned that there are functions whose antiderivatives cannot be written in a "nice" form, but examples will be scarce- e^{-x^2} being the stock one. After seeing such an unnatural ratio of nice examples to possibly one or two 'not-nice' ones, it's no surprise that many walk away believing themselves invincible and any function that itself looks 'nice' will have a 'nice' antiderivative waiting around the corner so they flap their arms around and bash their heads in frustration trying to find it. Makes me wonder if they bother to even consider why numerical techniques are taught at all?

You are very much right, most people don't understand why we did a course in numerical analysis on my degree program. I think a lot will still have some very naive views on mathematics.

 

[heman]

e^{-x^2}
This is a nice question ,Our Professor explained it yesterday only and very much amused to find the way it was done.

 

[uart]

e^{-x^2}
This is a nice question ,Our Professor explained it yesterday only and very much amused to find the way it was done.

What do you mean "the way it was done", it has no antiderivative other than as definied by a special function. Or are you referring to the integration from -infinity to infinity, that's a different thing than the general antiderivative and can be done in closed form.

I wonder if students aren't done a disservice in first year calculus classes with their sterilized examples and problems. They'll be asked to do hundreds of integration problems, all rigged to work out nicely with the techniques they've just learned. Perhaps it will be mentioned that there are functions whose antiderivatives cannot be written in a "nice" form, but examples will be scarce- LaTeX graphic is being generated. Reload this page in a moment. being the stock one.

Yep I really know exactly what you mean. I can remember when I was first learning this stuff and when exp(-x^2) was introduced it was almost like it was this bizarre pathological function just because it didn't have a nice antiderivative.

 

[saltydog]

I tried doing this but could not,why is it so?

Can I introduce Philosophy here or will you guys beat-up on me? And I don't wish to make light of all the nice responses above also. Here goes:

Why does there exist some functions not integrable in closed form? It reduces to I think, to the question of why are some problems harder than others to solve? I mean, can there be a Universe with just easy nice problems: A Universe with the absence of any functions which can't be integrated in closed form?

Perhaps, but I don't think such a world would give rise to us. So Latyph, my efforts to answer your question is this: It's difficulty to integrate it is a reflection of the type of Universe that gives rise to an intellect that can ponder the question.

Edit: I should say "some functions whose antiderivative cannot be expressed in terms of simple functions or operations" .

 

[arildno]

I don't agree, saltydog:

Integration is "basically" a task of summing up an infinite amount of individual contributions; i.e, at the very outset a Herculean task no one in their right minds would assume could actually ever be completed in an exact manner.
Thus, the basic question is rather:
Why are we on occasion able to fully complete this task?
As long as we happen to know about an anti-derivative of the integrand, FOTC guarantees us that our impossible summation effort can be completed in a trivial manner.

There exists, however, no fool-proof technique of constructing anti-derivatives other than by calculating zillions of definite integrals!

Thus, it should come as no surprise that it is only in special cases we may find a nice expression for an anti-derivative, or be able to compute some particular definite integral exactly.

 

[saltydog]

Arildno, let me first say, I yield to you sir. With that said, I hold there is a fool-proof technique of constructing the anti-derivative F(x), of any continuous function f(x) no matter how complex and this is guaranteed by the fundamental theorem of Calculus. It is:

F(x)=\int_0^x f(t) dt

How wonderful it is our world is so complex for only such a world would give rise to us.

 

[arildno]

As I said:

There exists, however, no fool-proof technique of constructing anti-derivatives other than by calculating zillions of definite integrals!

The set of function values to your anti-derivative cannot, in general, be computed in any other way than through calculating zillions of definite integrals (barring the special case where you recognize f to be the derivative of some known function F).

 

[HallsofIvy]

Warning for people who might think they are serious:
saltydog and Arildno are using different meaning for "construct the anti-derivative"!

 

[Izzhov]

Hmm... if I wanted to see a graph of the integral of x^x, would I just have to make it myself (by finding "zillions of definite integrals"), or is there any software that can graph it for me?

 

[mathwonk]

let F(x) = the area under the graph of y = x^x, between 1 and x, say for x>0. then F'(x) = x^x.

where the area is defined of course by the limit of riemann sums.

if you waNT A FORMULA, you coulod write out the powers eries for e^[ ], and subtitute to get the powers eries for x^x = e^[xln(x)], and then antidifferentiate term by term to get F(x).

as pointed out before, no one to my knowledge haS YET GIVEN A NAME TO THIS FUNCTION, SO WE CANNOT SAY ITS NAME, if that is what you mean by tell what it is, BUT WE CAN DEFINE IT AS ABOVE BY A LIMIT, AND BY A SERIES, AND that should do.

If you insist it have a name I suggest calling it Howard, or perhaps Latiph.

 

[whatta]

let's put it under different angle:

what functions should we add to the list of "elementary" ones, so that largest possible set of funcs could be closed-form-integrated?

suppose you take away exp. this will probably drag sin and cos down, too. what are we left with? X^n?

no imagine there's even more fundamental func than exp, "waiting around the corner".

p.s.: I watched "Pi" movie yesterday. You should, too.

 

[Gib Z]

You could try adding Elliptical Functions >.<

Or of course...The function F where F'(x)=f(x) and f(x) is what you want integrated :P Its definition is pretty elementary!

 

[Gib Z]

O btw since this thread is of so much interesting for such a long period of time, let's give it a name.

How about Lamb Bread? Sounds funny

And has anyone got any idea how to tell if a functions anti derivative is not elementary? That would help

 

[Amoogle]

Hmm... if I wanted to see a graph of the integral of x^x, would I just have to make it myself (by finding "zillions of definite integrals"), or is there any software that can graph it for me?

sure you can:

http://img110.imageshack.us/img110/8519/integratexxnu0.jpg

http://img110.imageshack.us/img110/7536/integrate2xxlj9.jpg

(mathematica 6)

 

[HallsofIvy]

O btw since this thread is of so much interesting for such a long period of time, let's give it a name.

How about Lamb Bread? Sounds funny

And has anyone got any idea how to tell if a functions anti derivative is not elementary? That would help

There is no way, except for the fact that, in a very specific sense, "almost all" integrable functions have non-elementary anti-derivatives.

 

[Diffy]

O btw since this thread is of so much interesting for such a long period of time, let's give it a name.

How about Lamb Bread? Sounds funny

And has anyone got any idea how to tell if a functions anti derivative is not elementary? That would help

This thread should be called, why do random people bring back to life threads that are over 3 years old.

 

[maze]

Asymptotically, how does int(x^x) behave?

 

[Marin]

Hi! I have a program to plot functions numerically. It also plots their derivatives and antiderivatives. :)

*I hope I've attached them to this post - Well it obviously didn't work

 

[HallsofIvy]

Asymptotically, how does int(x^x) behave?

There is a graph of the function in post 22 of this thread.

 

[maze]

There is a graph of the function in post 22 of this thread.

It explodes to infinity, of course, but the question is how fast? Is it faster than e^x? (I would think so). Faster than x^x (maybe I would guess so)? Slower than e^(x^2) (again that would be my guess)? Perhaps slower than e^(x^(1+epsilon))?

--
On second thought, I would think O(e^x) < O(int(x^x)) < O(x^x) as x-> infinity.

Now why? I haven't thought through this rigorously, but here's the idea.

For strictly positive monotonic increasing functions that are "slowly growing" like polynomials x, x^2, x^3, and so on, their integral is asymptotically larger than the original function.

O(p(x)) < O(int(p(x))

(eg, O(x) < O(x^2/2))

On the other hand, when when you start to consider functions that grow faster and faster, the growth of the function starts to match the accumulated area under the curve. As you go past the polynomials and get to functions asymptotically equivalent to e^x, this exactly balances and the integral is asymptotically equal to the original function.

O(e^x) = O(int(e^x))

Is this a turning point for how an integral acts on functions asymptotically? After you get to functions that grow faster than e^x, is the growth of the function so great that it outpaces the rate at which area accumulates under the curve? In other words for (monotonic positive increasing) functions BIG(x) that are asymptotically larger than e^x (as x-> infinity), is
O(BIG(x)) > O(int(BIG(x))?

One can imagine a number line of monotonic increasing functions, organized by how fast they grow:

<...ln(x)...x...x^2...x^n...e^x...x^x...e^(x^(1+epsilon))...e^(x^2)...>

The integral can be thought of as a function from this line to itself. For all the stuff less than e^x, the integral maps it larger. Functions asymptotically equivalent to e^x are a fixed point. What of functions greater than e^x? The integral is such a nice operator offhand I would think they would be mapped smaller. Of course to really figure this out a more rigorous thought must be given, considering this as an ordering of equivalence classes and trying to show properties of the integral on it, or things of that nature.

 

[mrsn321]

To derive x^x, you have to use the technique of logarithmic differentiation. Start with y = x^x, take the natural log of both sides, bring the power in front of the log on the right, then derive using the basic derivative rules. Isolate dy and you will find the derivative of x^x. The derivative is not difficult.

 

[confinement]

There is no way

T here are many statements of necessary or sufficient conditions for the anti-derivative of a function to be expressible in terms of a particular class of functions. As has already been mentioned in this thread, Liouiville gave the first result of this kind.

except for the fact that, in a very specific sense, "almost all" integrable functions have non-elementary anti-derivatives.

What specific sense is that Halls, since we know that there is no analogue of Lebesgue measure for infinite-dimensional spaces? Do you mean that the set of functions without elementary antiderivatives is dense in L2 (almost certainly, since even the smooth functions with compact support are dense) or do you mean they are prevalent (a stronger claim that has not been shown to my knowledge) ?

If a intergral of a function such as x^x cannot be expressed in terms of elementary term, would it be correct to say that NO function has an area that is the integral of x^x + C?

No, it just means that this area cannot be computed in terms of a finite number of additions, subtractions, multiplications, divisions, exponentiations, root extractions, or trigonometric or logarithmic evaluations. That's all it means.

what functions should we add to the list of "elementary" ones, so that largest possible set of funcs could be closed-form-integrated?

suppose you take away exp. this will probably drag sin and cos down, too. what are we left with? X^n?

no imagine there's even more fundamental func than exp, "waiting around the corner".

There are large classes of functions for this purpose, and they are built into Mathematica. The largest currently known family of functions that is convenient for expressing anti-derivatives is called MeijerG:

http://en.wikipedia.org/wiki/Meijer_G-function

and even this family is not large enough to contain the anti-derivative of x^x.

I also want to comment on the snobby mathematical tone in this thread, as if many are saying 'mathematicians do not dirty our hands with such matters as finding anti-derivatives, we simply give an abstract set-theoretic definition of functions and then say that we know everything about them. Now let me return to my rigorously derived trivialities in point-set topology.' The culuture of mathematics is not owned by Hardy, Bourbaki et al but so many folks these days act like it is. A healthy contrasting viewpoint comes from V.I. Arnold, who solved Hilbert's 13th problem, and who states that mathematics is a subfield of physics and that the endless abstractification has bogged down mathematical education. The point is that there is room for both points of view, the modern one that anti-derivatives and elementary functions are just accidental arbitrary questions with no deep mathematical structure, and the classic view, held by Euler, that the functions of interest should be expressible in terms of formulae.

 

[confinement]

As a thought experiment, let's see what would happen if rigor mortis had paralyzed mathematics before the discovery of the logarithmic function. This could have happened by historical accident, e.g. if high-speed computers had been availible in the 15th century then there would have been no need for large tables of logarithms to aid in arithmetic, and since necessity is the mother of invention this is a plausible scenario in which logarithms are never invented.

Then a student asks the physics forum, what is the antiderivative of 1/x ? He gets a response like this:

There is a solution it is the function F such that dF/dx is 1/x. But we can't write it anymore nicely than that, and there is nothing surprising about it. Almost no functions have integrals that we can write out nicely and explicitly in some closed form. How many times must that be said in this thread? Shall we lock it now to stop yet another person having to say it?

Is a response like this healthy for mathematics as a human activity?

Modern mathematics would be set back tremendously without the exponential function (Lie theory) and physics as we know it would hardly exist at all! For this reason it is difficult to suspend belief for this thought experiment: there are too many independent ways that the exponential and logarithmic functions would have been discovered. The point is made, however, that it is important to study specific cases because sometimes the solutions have properties which open up entire new fields of study (just as historically occurred with elliptic functions).

 

[HallsofIvy]

I'm not sure what you mean by this. It is, in fact, common to define ln(x) as
ln(x)= \int_1^x \frac{1}{t} dt
That certainly could have been done in the scenario you envision. It is not necessary to worry about the "calculating" aspect of the common logarithm.

And, of course, we define the "error function", erf(x), as
\frac{1}{\sqrt{2\pi}}\int_0^x e^{-t^2}dt[/itex].<br /> so that e^{-x^2} <b>can</b> be integrated in terms of that function.<br /> <br /> I see nothing wrong with saying that x^{-1} cannot be integrated of powers of x (as x to any other power of x can) nor with saying that e^{-x^2} cannot be integrated in terms of &quot;elementary&quot; functions. And, further, with saying that this is not because there is anything special about either x^{-1} and e^{-x^2} but rather that functions which <b>can</b> be integrated in simple terms are the &quot;special&quot; ones.

 

[confinement]

I apologize if my thought experiment was not clear --- I was trying to imagine an alternate history in which the mathematical importance of the logarithmic function was never discovered. In this alternate history when a student asks about the anti-derivative of 1/x he would receive the same type of response that Matt Grime gave to the original poster of this thread (as a form of parody the only thing I changed from his post was the function being given by 1/x instead of x^x). Notice the discouraging suggestion to "lock the thread". The point is that we should not discourage the business of finding new generalized classes of functions which contain the anti-derivative of x^x.