Question

The function that can satisfied area under the curve and why area is continue to positive infinity ?

Answer 1

`\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.