Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 134147 by Ñï= last updated on 28/Feb/21

lim_(n→∞) ∫_0 ^1 xsin (2πnx)dx=?

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} {xsin}\:\left(\mathrm{2}\pi{nx}\right){dx}=? \\ $$

Answered by mathmax by abdo last updated on 28/Feb/21

let u_n =∫_0 ^1  xsin(2πnx)dx  by parts for n>0  u_n =[−(x/(2πn))cos(2πnx)]_0 ^1 +∫_0 ^1 (1/(2πn))cos(2πnx)dx  =(1/(4π^2 n^2 ))[sin(2πnx)]_0 ^1  =0 ⇒lim_(n→+∞) u_n =0

$$\mathrm{let}\:\mathrm{u}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{xsin}\left(\mathrm{2}\pi\mathrm{nx}\right)\mathrm{dx}\:\:\mathrm{by}\:\mathrm{parts}\:\mathrm{for}\:\mathrm{n}>\mathrm{0} \\ $$$$\mathrm{u}_{\mathrm{n}} =\left[−\frac{\mathrm{x}}{\mathrm{2}\pi\mathrm{n}}\mathrm{cos}\left(\mathrm{2}\pi\mathrm{nx}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}\pi\mathrm{n}}\mathrm{cos}\left(\mathrm{2}\pi\mathrm{nx}\right)\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\pi^{\mathrm{2}} \mathrm{n}^{\mathrm{2}} }\left[\mathrm{sin}\left(\mathrm{2}\pi\mathrm{nx}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\mathrm{0}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{u}_{\mathrm{n}} =\mathrm{0} \\ $$

Answered by mnjuly1970 last updated on 28/Feb/21

  method 1     2πnx=t      Ω=lim_(n→∞) (1/(4π^2 n^n ))∫_0 ^( 2πn) tsin(t)dt  =lim_(n→∞) (1/(4π^2 n^2 ))[−tcost+sin(t)(t)]_0 ^(2πn)   =lim_(n→∞) ((−1)/(2πn))=0      method2   f:[a,b]→R is ∗Reiman integrable∗  then:     lim_(t→∞) ∫_a ^( b) f(x)sin(xt)dx=0      reiman−lebesgue theorem...

$$\:\:{method}\:\mathrm{1} \\ $$$$\:\:\:\mathrm{2}\pi{nx}={t}\: \\ $$$$\:\:\:\Omega={lim}_{{n}\rightarrow\infty} \frac{\mathrm{1}}{\mathrm{4}\pi^{\mathrm{2}} {n}^{{n}} }\int_{\mathrm{0}} ^{\:\mathrm{2}\pi{n}} {tsin}\left({t}\right){dt} \\ $$$$={lim}_{{n}\rightarrow\infty} \frac{\mathrm{1}}{\mathrm{4}\pi^{\mathrm{2}} {n}^{\mathrm{2}} }\left[−{tcost}+{sin}\left({t}\right)\left({t}\right)\right]_{\mathrm{0}} ^{\mathrm{2}\pi{n}} \\ $$$$={lim}_{{n}\rightarrow\infty} \frac{−\mathrm{1}}{\mathrm{2}\pi{n}}=\mathrm{0} \\ $$$$\:\:\:\:{method}\mathrm{2} \\ $$$$\:{f}:\left[{a},{b}\right]\rightarrow\mathbb{R}\:{is}\:\ast{Reiman}\:{integrable}\ast\:\:{then}: \\ $$$$\:\:\:{lim}_{{t}\rightarrow\infty} \int_{{a}} ^{\:{b}} {f}\left({x}\right){sin}\left({xt}\right){dx}=\mathrm{0} \\ $$$$\:\:\:\:{reiman}−{lebesgue}\:{theorem}... \\ $$$$\:\:\:\:\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com