Question and Answers Forum

All Questions      Topic List

Differential Equation Questions

Previous in All Question      Next in All Question      

Previous in Differential Equation      Next in Differential Equation      

Question Number 212107 by MrGaster last updated on 01/Oct/24

$$$$$$I={\int }_0^{\mathrm{\infty }}\frac{x\cos x-\sin x}{x^2}dx.$$$$$$

Answered by Frix last updated on 01/Oct/24

$$\mathrm{Take}\mathrm{a}\mathrm{closer}\mathrm{look}:$$$${\left(\frac{u}{v}\right)}^{\prime }=\frac{u^{\prime }v+v^{\prime }u}{v^2}\iff \int \frac{u^{\prime }v+v^{\prime }u}{v^2}=\frac{u}{v}$$$$\mathrm{With}u=\sin x\mathrm{and}v=x\mathrm{we}\mathrm{have}\mathrm{the}\mathrm{given}$$$$\mathrm{integral}$$$$\Rightarrow$$$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{x\cos x-\sin x}{x^2}dx={\left[\frac{\sin x}{x}\right]}_0^{\mathrm{\infty }}=-1$$$$\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\sin x}{x}=0$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com