# Sign of a function inside of an integral obtained from the asymptotic behavior of the integral

Let $$\alpha>0$$ and $$A>0$$. Suppose that $$R(X)=\int_{1}^{X}L(t)t^{-\alpha}\,dt,\,\,X>1$$ satisfies $$\lim_{X\to\infty}R(X)=-A$$. Then $$L(t)<0$$ for all $$t$$ sufficiently large.

Is it true?

• Hint: By writing $L(t)=t^{\alpha}f(t)$ (where $f(t):=t^{-\alpha}L(t)$) and noting that the sign of $L(t)$ is the same as that of $f(t)$, you can simply ask this question for $f(t)$ considering $\int_1^X f(t)\, dt$. Now, do you know any function $f(t)$ that is not negative for all sufficiently large $t$ but will still give a negative integral as $X\to \infty$? – Minus One-Twelfth Jun 4 at 18:47
• Yes...something like $f(t)=-1$, for $1\leq t<2$ and $f(t)=1/t^2$, for $t\geq 2$. – GoldSoundz Jun 4 at 19:05