Question Number 212107 by MrGaster last updated on 01/Oct/24
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{I}=\int_{\mathrm{0}} ^{\infty} \frac{{x}\:\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} }{dx}. \\ $$$$ \\ $$
Answered by Frix last updated on 01/Oct/24
![Take a closer look: ((u/v))′=((u′v+v′u)/v^2 ) ⇔ ∫ ((u′v+v′u)/v^2 )=(u/v) With u=sin x and v=x we have the given integral ⇒ ∫_0 ^∞ ((xcos x −sin x)/x^2 )dx=[((sin x)/x)]_0 ^∞ =−1 lim_(x→0) ((sin x)/x) =1 lim_(x→∞) ((sin x)/x) =0](https://www.tinkutara.com/question/Q212109.png)
$$\mathrm{Take}\:\mathrm{a}\:\mathrm{closer}\:\mathrm{look}: \\ $$$$\left(\frac{{u}}{{v}}\right)'=\frac{{u}'{v}+{v}'{u}}{{v}^{\mathrm{2}} }\:\Leftrightarrow\:\int\:\frac{{u}'{v}+{v}'{u}}{{v}^{\mathrm{2}} }=\frac{{u}}{{v}} \\ $$$$\mathrm{With}\:{u}=\mathrm{sin}\:{x}\:\mathrm{and}\:{v}={x}\:\mathrm{we}\:\mathrm{have}\:\mathrm{the}\:\mathrm{given} \\ $$$$\mathrm{integral} \\ $$$$\Rightarrow \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{{x}\mathrm{cos}\:{x}\:−\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} }{dx}=\left[\frac{\mathrm{sin}\:{x}}{{x}}\right]_{\mathrm{0}} ^{\infty} =−\mathrm{1} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}}{{x}}\:=\mathrm{1} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}}{{x}}\:=\mathrm{0} \\ $$