2
$\begingroup$

I wish to prove or disprove the following statement:

$f\in L^1[0,\infty]$ if and only then $f$ has an improper Riemann integral on $[0,\infty)$.

I think $(\Leftarrow)$ is false. If we let $f(x) = \frac{\sin x}{x},~x\gt 0$, then $$\int_0^\infty \frac{\sin x}{x}~dx = \frac{\pi}{2}, $$

but $$\int_0^\infty \left|\frac{\sin x}{x}\right|~dx = \infty.$$ So $f\notin L^1[0,\infty)$.

How about $(\Rightarrow)$? I can't seem to think of any counterexample, and I don't see how to show that it is true.

1 Answers 1