[Math] Convergent integral of divergent function

calculusconvergence-divergencefunctionsimproper-integrals

On one of my calculus lectures I was told that there exist convergent improper integrals (in infinity) of divergent function. I was searching for an example in the internet, but I didn't find any. Has any of you idea how would this kind of function look like? Thanks in advance

Best Answer

For any positive integer $n$, let $f(x)=0$ at $x=n-\frac{1}{2^{n}}$. Let it climb linearly to $1$ at $x=n$, and then fall linearly to $0$ at $x=n+\frac{1}{2^{n}}$. Let $f(x)=0$ everywhere else.

So $f$ has fast-narrowing "spikes" of height $1$ around every positive integer. The combined area of these spikes is small.

We can modify the construction to make the maximum height of the spikes become arbitrarily large as $n\to\infty$, for example by replacing $f(n)=1$ by $f(n)=n$.

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