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I was going over some practice questions for an upcoming calculus exam and I came across this question: "Give an example of two functions $f(x)$ and $g(x)$ such that $\int _1^{\infty }f(x)dx$ and $\int _1^{\infty }g(x)dx$ both converge respectively, but $\int _1^{\infty }f(x)g(x)dx$ diverges." I am having a very difficult time with this question and cannot find a good answer. I've tried with polynomials, trigonometric functions, and exponential functions with no avail.

BlastKast
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    Also: https://math.stackexchange.com/q/3329279/42969. – Martin R Aug 22 '21 at 21:04
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    The functions $f(x)$ and $g(x)$ can have bumps in their graphs as $x \to +\infty$ that are high but at the same time narrow enough that they don't prevent $\int f$ and $\int g$ from converging. If you pick the bumps in $f$ and in $g$ to be in the same places, then on multiplying $f$ and $g$ they may become so high that now $\int fg$ doesn'r converge. – Anonymous Aug 22 '21 at 21:16
  • The question doesn't required boundedness of the functions. Let $f(x)=g(x)=\frac{1}{\sqrt{x-1}}$ for $1\le x\le 2$ and $=0$ otherwise. – herb steinberg Aug 22 '21 at 21:30
  • For example, let $f=g=\sin(x)/\sqrt{x}$. – Dispersion Aug 22 '21 at 23:06
  • $\sin(x^2)$ definitely works by itself. $\int_0^{\infty}\sin^2(x^2)$ does not converge – Dionel Jaime Aug 22 '21 at 23:18

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Set f(x) = g(x) such that they are $n^2$ on each interval $[n,n + 1/n^{4}]$ and zero otherwise.

Salcio
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