Worst Way to Calculate Area of a Triangle

A useful technique from Complex Analysis helps us to identify whether a real number $u$ is positive or negative by checking a discontinuous integral. We have

$$\frac{1}{2\pi i}{\int }_{c-i\mathrm{\infty }}^{c+i\mathrm{\infty }}{e}^{su}\frac{\text{d}s}{s}=\{\begin{array}{ll}0& \text{ if }u<0\\ \frac{1}{2}& \text{ if }u=0\\ 1& \text{ if }u>0\end{array}$$

for $c>0$.

We will apply this to calculate the area of the triangle

$$T(N):=\{(x,y)\in {\mathbb{R}}_{\ge 0}:ax+by\le N\}$$

with $(a,b)=1$. This area is given by

$$\begin{array}{rl}& {\int }_{x,y\ge 0}{\mathbf{1}}_{N-ax-by\ge 0}\text{d}x\text{d}y\\ =& {\int }_{x,y\ge 0}\frac{1}{2\pi i}{\int }_{s=c-i\mathrm{\infty }}^{c+i\mathrm{\infty }}{e}^{s(N-ax-by)}\frac{\text{d}s}{s}\text{d}x\text{d}y\\ =& \frac{1}{2\pi i}{\int }_{s=c-i\mathrm{\infty }}^{c+i\mathrm{\infty }}\left({\int }_{x\ge 0}{e}^{-asx}\text{d}x\right)\left({\int }_{y\ge 0}{e}^{-bsy}\text{d}y\right)\frac{e^{sN}}{s}\text{d}s\\ =& \frac{1}{ab}\cdot \frac{1}{2\pi i}{\int }_{s=c-i\mathrm{\infty }}^{c+i\mathrm{\infty }}\frac{e^{sN}}{s^3}\text{d}s\end{array}$$

by the above mentioned trick. Notice that the integral with $N-ax-by=0$ is zero, and hence was ignored.

Now, moving the contour far to the left we pick up the residue from the pole of order $3$ at $s=0$ and hence the area comes out to be

$$\frac{1}{ab}\cdot \frac{N^2}{2}=\frac{N^2}{2ab}$$

which matches with the area calculated by other known methods!

Acknowledgement : I learnt this from lectures by Prof. Andrew Granville for the course Distribution of Prime Numbers.