\(\displaystyle\int_0^\infty f(x)\,\mathrm dx=\lim_{t\to\infty}\int_0^tf(x)\,\mathrm dx\)
Consider a partition of the interval \([0,t]\) such that
\(\displaystyle[0,t]=[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-1},x_n]=\bigcup_{k=0}^{n-1}[x_k,x_{k+1}]\)
Within each subinterval, select \(\hat{x_k}\in[x_k,x_{k+1}]\) so that the value of \(f(\hat{x_k})=\max\{[x_k,x_{k+1}]\}\). In other words, \(\hat{x_k}\) is specially chosen to guarantee that \(f(\hat{x_k})\) is the maximum value of \(f(x)\) within each corresponding subinterval.
So now we have an upper bound for the integral in terms of a Riemann sum,
\(\displaystyle\int_0^tf(x)\,\mathrm dx\le\sum_{k=0}^{n-1}f(\hat{x_k})\Delta x_k\)
with \(\Delta x_k=x_{k+1}-x_k\). In the limit, these two will be exactly equal:
\(\displaystyle\int_0^\infty f(x)\,\mathrm dx=\lim_{t\to\infty}\int_0^tf(x)\,\mathrm dx=\lim_{n\to\infty}\sum_{k=0}^{n-1}\underbrace{f(\hat{x_k})\Delta x_k}_{a_k}\)
The main point here is that the right hand side is the sum of the terms of a convergent sequence \(a_k\to2\) as \(k\to\infty\).
But \(\sum_ka_k\) diverges if \(a_k
ot\to0\), so the integral must also diverge.
