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While reading a paper on the circle method, I came accross the following exercise:
Let $e(x) = e^{2 \pi i x}$, prove that for $n,m \in \mathbb {Z}$, $\int_0^1 e(nx)e(-mx)=\delta _{mn}$ (Kronecker Delta).
I have tried to combine the exponentials and integrating through u-substitution, I have tried to leave them uncombined and integrate by parts, I have used Euler's Identity and integrated the sine and cosine, and no matter what I do, I get an $n-m$ term in the denominator, so whenever $m=n$ division by zero will occur. The limit is $1$, but the problem specifically says that the function will equal $1$, what am I missing?

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  • $\begingroup$ Did you solve the case when $m\neq n$? That's the difficult part. For the other, just pay attention to the fact that, when $m=n$, $e(mx)e(-mx) = 1$ for every $x \in [0,1]$. $\endgroup$ – Hugo Feb 27 '16 at 20:09
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The primitive is $$ \int dx e^{2\pi i (n-m)x}=\frac{i e^{-2 i \pi x (m-n)}}{2 \pi (m-n)}\ , $$ for $m\neq n$, and $=x$ for $m=n$. So for $m\neq n$, computing the primitive between the bounds $0$ and $1$ yields $$ \frac{i e^{-2 i \pi (m-n)}}{2 \pi (m-n)}-\frac{i }{2 \pi (m-n)}=0 $$ using $e^{-2\pi i k}=1$ for $k\in\mathbb{Z}$, while for $m=n$ the primitive between $0$ and $1$ yields $1$.

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