[tex]\displaystyle\int_0^\infty f(x)\,\mathrm dx=\lim_{t\to\infty}\int_0^tf(x)\,\mathrm dx[/tex]
Consider a partition of the interval [tex][0,t][/tex] such that
[tex]\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}][/tex]
Within each subinterval, select [tex]\hat{x_k}\in[x_k,x_{k+1}][/tex] so that the value of [tex]f(\hat{x_k})=\max\{[x_k,x_{k+1}]\}[/tex]. In other words, [tex]\hat{x_k}[/tex] is specially chosen to guarantee that [tex]f(\hat{x_k})[/tex] is the maximum value of [tex]f(x)[/tex] within each corresponding subinterval.
So now we have an upper bound for the integral in terms of a Riemann sum,
[tex]\displaystyle\int_0^tf(x)\,\mathrm dx\le\sum_{k=0}^{n-1}f(\hat{x_k})\Delta x_k[/tex]
with [tex]\Delta x_k=x_{k+1}-x_k[/tex]. In the limit, these two will be exactly equal:
[tex]\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}[/tex]
The main point here is that the right hand side is the sum of the terms of a convergent sequence [tex]a_k\to2[/tex] as [tex]k\to\infty[/tex].
But [tex]\sum_ka_k[/tex] diverges if [tex]a_k\not\to0[/tex], so the integral must also diverge.
