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

$$\frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} e^{su}\;\frac{\text{d}s}{s} = \begin{cases}0 &\text{ if } u<0\\ \frac 12 &\text{ if } u=0\\ 1 &\text{ if } u>0\end{cases}$$

for $c>0$.

We will apply this to calculate the area of the triangle

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

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

\begin{align*}&\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\infty}^{c+i\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\infty}^{c+i\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\infty}^{c+i\infty} \frac{e^{sN}}{s^3}\;\text{d}s\end{align*}

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.