a very cool integral comparison problem

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[isthissarcasm]

The n-dimensional integral does limit to 1:

Because we are integrating over a mass of one, we can easily think of this as an expectation. In particular, we can define Y = x_1 * x_2 * ... * x_n, and think of x_1, x_2, ..., x_n and Y as random variables, and the integral is then E(Y^Y). The x_i are iid distributed uniformly on [0, 1]. As n goes to infinity, Y converges in probability to 0. Because f(x) = x^x is continuous on [0, 1] (taking f(0) = 1), this means E(Y^Y) limits to E(f(0)) = 1.


To show Y converges in probability to 0, I think it's sufficient to see that it is positive, and E(Y) converges to 0. And E(Y) = E(x_1) * E(x_2) * ... * E(x_n) = (1/2)^n, which limits to 0.

[charlesbrowne]

After staring at the thumbnail, I decided the single integral was larger. The solution was a pleasant surprise.

[txikitofandango]

Surprising! I assumed the xy one was smaller, since you're multiplying a number in (0, 1) by another number in (0, 1) which shrinks the product

[bartekltg]

Another proof the limit is 1.
It seems to be easier to work with the logarithm of that function. And since we expect the values are close 1 most of the time, that crude bound is enough.
f_n(...) = x1 x2 ...xn (log(x1)+...+log(xn))
Since
exp(x)>=1+x
\int exp(f_n(...)) >= \int_n 1 + \int f_n(...) = 1 - \int -f_n(...)
\int goes over n-th dimensional hypercbe
On the left side we have our integral from the video. On the right side we already have 1, and we subtract an integral of a positive function. Let's show it is small.
\int -f_n(...) = \int -x1 x2 ...xn (log(x1)+...+log(xn)) = \sum_i=1^n \int -log(x_i) x1...xn

Each integral is the same, so let's change names of variables and we can write it as one term
...= n* \int -log(x_1) x1 x2...xn
-x1 log(x1) is bounded by a constant (1/e to be precise)and positive, everything else is positive, so even without namedroping Hoelder inequality:
...<= n* 1/e \int x2...xn = n/e *2^-(n-1)
Now, we apply this bound the the initial inequality, remembering ghat the function under the original integral is <=1, so the internal is also <=1
1>= \int exp(f_n(...)) >= \int_n 1 + \int f_n(...) = 1 - \int - f_n(...) >= 1- n/e *2^-(n-1) ->1

[richardheiville]

No need double change of variable Perform the single change of variable u(x)=xy to obtain integral from x=0 to 1 of 1/x integrate from u=0 to x of u^u then, perform the integration by parts dv=1/x

[janihabetler]

The t = u^u substitution is invalid as u^u has a derivative = 0 somewhere (1/e, I guess) inside the interval.

[alexeycanopus]

In general, I guess the integral is equal to \frac{1}{(n-1)!} \int_0^1 u^u (1 / \ln{u})^{n-1} du, which goes to 1 if n goes to infinity

[egoreremeev]

I can't quite show directly that it's indeed would be greater than original integral for n>2, but here my steps:

Change variables to
x_1 = u_1/(u_2 * ... * u_n)
x_2 = u_2 ...
the jacobian then would be 1/(u_2 ... u_n)

The integral changes in this way:
J_n = \int _ [0, 1]^n d^n x (x_1 * ... * x_n)^(x_1 * ... * x_n) = \int_S u_1^u_1 * 1/(u_2 * ... * u_n) d^n u

A bit of thinking about parametrization so that x_1 is the "last" variable to integrate leads to these limits:

\int_S (...) = \int_0^1 du_1 u_1^(u_1) \int_(u_1)^1 du^2/u^2 \int_(u_1/u_2)^1 du_3/u_3 ... \int_(u_1/(u_2 ... u_n-1)^1 du_n/u_n

Changing variables once more (but in u_1 it's just relabeling) to ln(u_i) = y_i (pretty straightforward change)

\int_0^1 du_1 u_1^(u_1) \int_(y_1)^0 dy_2 \int_(y_1 - y_2)^0 dy_3 ... \int_(y_1 - y_2 - ... - y_(n-1)) dy_n

One can see that this is a volume of something (something like a symplex but not quite, mostly a pyramid?). Let's change the sign of variables and define:
V_n(r) = \int_0^r dx_1 * \int_0^(r-x_1) dx_2 * ... * \int_0^(r-x_1-...-x_n-1) dx_n
And we see that the original integral is:

J_n = \int _ [0, 1]^n d^n x (x_1 * ... * x_n)^(x_1 * ... * x_n) = \int_0^1 x^x dx * V_{n-1}(-ln(x))

It's easy to spot recursive relation of V's:

V_n(r) = \int_0^r du V_{n-1}(r-u) = \int_0^r du V_n-1(u); V_1(r) = r
Pretty easy to say that V_n(r) = r^k/k!

Then I dont really know how to proceed, these V_n(r)'s are not increasing if x is fixed, but they seem (in case of V_n(-ln x) like an approximating function of delta function (cuz they are normalized to 1 and localize to a spike at x=0). So in a sense J_n -|n->inf|-> lim_{x->0} x^x = 1

Why there's a feeling that they are increasing - that's because they "localize" the rise at the graph of x^x near x=0. It's pretty obvious if we get a rectangle bumps limit for delta function, but it may not be true for V_n(-ln x).

And I guess to actually calculate the limit explicitly one can try using method of steepest descent (after changing variables -ln x = u) for
J_n = \int_0^+inf e^(-u(1+e^-u))u^k / k! du

[johnpaterson]

Since they are in different dimensions, the question is poorly posed - eg is 18 feet larger or smaller than 15 square feet?

[evankalis]

My guess is that the x^x will be larger

[Teja]

So this result, that the double integral equals the single integral should be called the Junior's Dream?!

[Neo_Bones]

According to desmos they’re both equal

[Alan-zftt]

Have not watched it yet but .... thinking a quarter to a quarter is greater than a (quarter to a quarter) all squared but will it survive Michael's testabilities, analysis and computational skills?
Start the video!

[sshkbf]

What about this question dear?
The indefinte integral of x^x
Someone gave me to solve and I could not solve.
If you solve me it would be your ...