Timeline for $\displaystyle\dfrac{a(x)}{x^{2}}\int_{0}^{x}b(\tau)d\tau \sim \dfrac{2}{1-\alpha}\dfrac{\sqrt{a(x)}}{x^{3/2}}, \ \text{when} \ x \to 0.$

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Sep 19, 2022 at 19:17	comment	added	user253963		@ChristianRemling Thank you!
Sep 19, 2022 at 17:16	review	Close votes
Sep 19, 2022 at 20:15
Sep 19, 2022 at 16:56	comment	added	Christian Remling		I don't think there's anything involved here. You just multiplied both sides by $a/x^2$, which is certainly justified if $\sim$ has the standard meaning $A\sim B \iff A/B\to 1$.
Sep 19, 2022 at 16:40	comment	added	LSpice		If you're going to `\dfrac` your fractions, please also `\displaystyle` your integrals: $\displaystyle\frac{a(x)}{x^2}\int_0^x b(\tau)\mathrm d\tau$ `\displaystyle\frac{a(x)}{x^2}\int_0^x b(\tau)\mathrm d\tau` looks better than $\dfrac{a(x)}{x^2}\int_0^x b(\tau)d\tau$ `\dfrac{a(x)}{x^2}\int_0^x b(\tau)d\tau` (though it's better not to `\displaystyle` titles at all). I have edited accordingly.
Sep 19, 2022 at 16:39	history	edited	LSpice	CC BY-SA 4.0	Sizing integrals like fractions; name of paper
S Sep 19, 2022 at 16:06	review	First questions
Sep 19, 2022 at 20:22
S Sep 19, 2022 at 16:06	history	asked	user253963	CC BY-SA 4.0