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I am trying to understand this proof (problem 2 pages 1-2) which shows that if a rectangle can be tiled by smaller rectangles each of which has at least one integer side, then the tiled rectangle has at least one integer side.

I can't seem to understand why "the integral of $f$ over any horizontal or vertical line segment with integer length is zero.". How can this be proven mathematically? Could someone help me with the proof linked?

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It means that $f(x,y)=\exp(2\pi i(x+y))$ has the property that $$\int_{a}^{a+m}f(x,y)\,dx=0$$ for any $a$, $y$ and integer $m$, and $$\int_{b}^{b+n}f(x,y)\,dy=0$$ for any $b$, $x$ and integer $n$.

The key to the proof is that $$\int_a^b\int_c^d f(x,y)\,dx\,dy=0$$ if and only if at least one of $b-a$ and $d-c$ is an integer.

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From $$\iint_{[a,b]\times[c,d]}f\,\mathrm dA=\frac{e^{2\pi i b}-e^{2\pi i a}}{2\pi i}\cdot \frac{e^{2\pi i d}-e^{2\pi i c}}{2\pi i} $$ we conclude that the integral over a rectangle is zero iff at least one of the factors is zero, i.e., iff at least one side lenth s an integer.

From the additivity of the integral, we conclude that the integral over a rectangle with one-integer-side rectangle can be written as sum of integrals each of which is zero, hence is zero itself, hence the rectangle had an integer side.

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Hint: that integral is $0$ because of the periodicity of $e^{it}$:

$$ \int_a^b e^{2\pi i (x+y)}dx = e^{2\pi i y}\int_a^b e^{2\pi i x}dx = \frac{e^{2\pi i y}}{2 \pi i}\left( e^{2\pi i b} - e^{2\pi i a}\right) = \frac{e^{2\pi i y}} {2 \pi i}e^{2 \pi i a} \left( e^{2\pi i (b-a)} - 1 \right). $$

The expression in the last set of parentheses is $0$ when $b-a$ is an integer. (That follows from the periodicity of $\sin$ and $\cos$.)

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  • $\begingroup$ Could you expand, I'm really struggling sorry? $\endgroup$
    – Tom Finet
    Feb 7 '18 at 22:17
  • $\begingroup$ See my edit ... $\endgroup$ Feb 7 '18 at 23:55

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