Topic: Is there a solution to this infinite integration problem?

[Sleestak]

Homework Statement

Evaluate the limit
1 1 1
lim ∫ ∫ ... ∫ cos^2((pi/2n)(x1 + x2 +... xn))dx1 dx2 ... dxn
0 0 0
n→∞

Homework Equations

Well, I know that we can change this using a double angle rule, so that the integrals become 1/2 + 1/2 cos (2*pi/2n)(x1 + x2 +... xn))dx1 dx2 ... dxn

and the integral over the 1/2 just becomes 1/2, but the other side baffles me.

The Attempt at a Solution

My professor tried to do this, but I don't agree with his methodology. When he integrated it, he got pi/n out front, and if you keep integrating, this would go to pi^n / n^n . However, the integral should produce n^n / pi^n if I'm not mistaken, meaning this would diverge to infinity, and not go to zero like he said. Any ideas?

 

[Simon Bridge]

Show the working for the definite integral over x1 first.

 

[Sleestak]

Okay, so the first integral gives n/pi (sin(pi)), and the sin(pi) = 0, so you'd be integrating 0 from there on out. Is that correct?

 

[Mark44]

Homework Statement

Evaluate the limit
1 1 1
lim ∫ ∫ ... ∫ cos^2((pi/2n)(x1 + x2 +... xn))dx1 dx2 ... dxn
0 0 0
n→∞

Homework Equations

Well, I know that we can change this using a double angle rule

Why? $cos^2(\pi/(2n))$ is a constant as far as any of the integrations are concerned. You can bring it out of all of the integrals.

Edit: Never mind. I didn't count enough parentheses.

, so that the integrals become 1/2 + 1/2 cos (2*pi/2n)(x1 + x2 +... xn))dx1 dx2 ... dxn

and the integral over the 1/2 just becomes 1/2, but the other side baffles me.

The Attempt at a Solution

My professor tried to do this, but I don't agree with his methodology. When he integrated it, he got pi/n out front, and if you keep integrating, this would go to pi^n / n^n . However, the integral should produce n^n / pi^n if I'm not mistaken, meaning this would diverge to infinity, and not go to zero like he said. Any ideas?

 

[Sleestak]

Whoops, that's my bad. All of the x terms are inside the cosine as well, and all of them are multiplied by the pi/2n, so

cos^2( (pi/2n)*(x1 + x2 +... xn) )

 

[Simon Bridge]

...putting $X_m=\sum_{i=m}^n x_i$ (saves typing)
$$\int_0^1 \cos \frac{\pi}{n}(x_1+X_2)\;dx_1 = \frac{n}{\pi}\big[\sin \frac{\pi}{n}(1+X_2)-\sin \frac{\pi}{n}X_2\big]$$ ... but I'm a little distracted.

@Mark44
The original problem is:
$$\lim_{n\to\infty}\int_0^1\cdots\int_0^1 \cos^2\left(\frac{\pi}{2n}X_1\right)\;dx_1\cdots dx_n$$

 

[Mark44]

Whoops, that's my bad. All of the x terms are inside the cosine as well, and all of them are multiplied by the pi/2n, so

cos^2( (pi/2n)*(x1 + x2 +... xn) )

No, my bad. What you wrote was clear, but I misread it.

...putting $X_m=\sum_{i=m}^n x_i$ (saves typing)
$$\int_0^1 \cos \frac{\pi}{n}(x_1+X_2)\;dx_1 = \frac{n}{\pi}\big[\sin \frac{\pi}{n}(1+X_2)-\sin \frac{\pi}{n}X_2\big]$$ ... but I'm a little distracted.

@Mark44
The original problem is:
$$\lim_{n\to\infty}\int_0^1\cdots\int_0^1 \cos^2\left(\frac{\pi}{n}X_1\right)\;dx_1\cdots dx_n$$

Gotcha.

 

[Ray Vickson]

Homework Statement

Evaluate the limit
1 1 1
lim ∫ ∫ ... ∫ cos^2((pi/2n)(x1 + x2 +... xn))dx1 dx2 ... dxn
0 0 0
n→∞

Homework Equations

Well, I know that we can change this using a double angle rule, so that the integrals become 1/2 + 1/2 cos (2*pi/2n)(x1 + x2 +... xn))dx1 dx2 ... dxn

and the integral over the 1/2 just becomes 1/2, but the other side baffles me.

The Attempt at a Solution

My professor tried to do this, but I don't agree with his methodology. When he integrated it, he got pi/n out front, and if you keep integrating, this would go to pi^n / n^n . However, the integral should produce n^n / pi^n if I'm not mistaken, meaning this would diverge to infinity, and not go to zero like he said. Any ideas?

It might help to use the fact that $\cos(w)$ is the real part of $e^{iw}$, so you need to integrate
\int_0^1 \int_0^1 \cdots \int_0^1 e^{i c (x_1 + x_2 + \cdots + x_n)} \, dx_1 \, dx_2 \cdots \, dx_n,
where $c = \pi/n$. The exponential factors into separate factors for each variable, and so the n-fold integration is just the product of single-variable integrals. You can pull the "real part" operator outside all the integrals (why?).