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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

BigM
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bartop
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2 Answers2

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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 $f(n)=n$.

André Nicolas
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  • If I may ask, what would be integral of such a function? – bartop Mar 12 '14 at 22:16
  • For the first function I gave, the integral from $0$ to $\infty$ is the sum of the areas of the triangles. The isosceles triangle centered on $n$ has area $\frac{1}{2^n}$. Sum from $n=1$ to $\infty$. This is a geometric series, sum $1$. So the definite integral is $1$. The second one I suggested (height $n$ at $n$) turns out to be also a doable sum, it is $2$. – André Nicolas Mar 12 '14 at 23:21
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How about $\int_0^1\ln(x)dx=-1$;)

BigM
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