Consider the following sum:
$\displaystyle \int_1^{n/x} \sum_{ \gamma < y \leq n/x} e(\alpha x y)\mathrm{d}\gamma/\gamma$, where $e(r):=\exp(2i\pi r)$. I don't understand the notation at all. It seems that I can think of this as $\int_1^{n/x} A(\gamma)\mathrm{d}\gamma/\gamma$ where $A(\gamma):=\sum_{ \gamma < y \leq n/x} e(\alpha x y)$. But how am I supposed to even compute the integral if I replace the function $e(\alpha x y)$ by something as simple as $\alpha x y$?
I apologise if this is a stupid question but I cannot seem to understand this notation and get how in Vaughan's book on the circle method, he writes $\displaystyle \sum_{y\leq n/x} e(\alpha x y) \log y=\int_1^{n/x} \sum_{ \gamma < y \leq n/x} e(\alpha x y)\mathrm{d}\gamma/\gamma$. Thanks!
Let $e(t)= \exp(2i\pi t)$ and $b_y = e(\alpha xy)$ if $y\le n/x$ and $b_y=0$ otherwise then
$$\sum_{y\ge 1} b_y \log y=\sum_{y\ge 1} b_y \int_1^y \frac1{\gamma} d\gamma =\int_1^\infty (\sum_{y\ge \gamma} b_y) \frac1{\gamma} d\gamma=\int_1^{n/x}(\sum_{y\ge \gamma,y\le n/x} e(\alpha x y)) \frac1{\gamma} d\gamma$$