Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 88288 by Chi Mes Try last updated on 09/Apr/20

Commented by abdomathmax last updated on 10/Apr/20

I =∫_0 ^1  ((sin(ln∣x∣))/(ln∣x∣))dx  changement  ln(x)=−u ⇒  I =∫_0 ^1   ((sin(lnx))/(lnx))dx =−∫_0 ^(+∞)  ((−sinu)/(−u))(−e^(−u) )du  = ∫_0 ^∞   ((sinu)/u) e^(−u)  du  let f(t) =∫_0 ^∞   ((sinu)/u)e^(−tu)  du  with t>0  f^′ (t) =−∫_0 ^∞  e^(−tu)  sinu du   =−Im(∫_0 ^∞  e^(−tu+iu) du) =−Im(∫_0 ^∞  e^((−t+i)u)  du)  ∫_0 ^∞  e^((−t+i)u) du =[(1/(−t+i))e^((−t+i)u) ]_0 ^(+∞)   =((−1)/(−t+i)) =(1/(t−i)) =((t+i)/(t^2  +1)) ⇒f^′ (t)=−(1/(1+t^2 )) ⇒  f(t) =k−arctant  lim_(t→0) f(t) =∫_0 ^∞  ((sinu)/u)du =(π/2)=k ⇒  f(t) =(π/2) −arcrant   (t>0)  I =f(1) =(π/2)−(π/4) =(π/4)

$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{sin}\left({ln}\mid{x}\mid\right)}{{ln}\mid{x}\mid}{dx}\:\:{changement}\:\:{ln}\left({x}\right)=−{u}\:\Rightarrow \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{sin}\left({lnx}\right)}{{lnx}}{dx}\:=−\int_{\mathrm{0}} ^{+\infty} \:\frac{−{sinu}}{−{u}}\left(−{e}^{−{u}} \right){du} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sinu}}{{u}}\:{e}^{−{u}} \:{du}\:\:{let}\:{f}\left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sinu}}{{u}}{e}^{−{tu}} \:{du} \\ $$$${with}\:{t}>\mathrm{0} \\ $$$${f}^{'} \left({t}\right)\:=−\int_{\mathrm{0}} ^{\infty} \:{e}^{−{tu}} \:{sinu}\:{du}\: \\ $$$$=−{Im}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{tu}+{iu}} {du}\right)\:=−{Im}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{\left(−{t}+{i}\right){u}} \:{du}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{\left(−{t}+{i}\right){u}} {du}\:=\left[\frac{\mathrm{1}}{−{t}+{i}}{e}^{\left(−{t}+{i}\right){u}} \right]_{\mathrm{0}} ^{+\infty} \\ $$$$=\frac{−\mathrm{1}}{−{t}+{i}}\:=\frac{\mathrm{1}}{{t}−{i}}\:=\frac{{t}+{i}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow{f}^{'} \left({t}\right)=−\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\Rightarrow \\ $$$${f}\left({t}\right)\:={k}−{arctant} \\ $$$${lim}_{{t}\rightarrow\mathrm{0}} {f}\left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{sinu}}{{u}}{du}\:=\frac{\pi}{\mathrm{2}}={k}\:\Rightarrow \\ $$$${f}\left({t}\right)\:=\frac{\pi}{\mathrm{2}}\:−{arcrant}\:\:\:\left({t}>\mathrm{0}\right) \\ $$$${I}\:={f}\left(\mathrm{1}\right)\:=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\:=\frac{\pi}{\mathrm{4}} \\ $$

Answered by mind is power last updated on 09/Apr/20

u=−ln(x)⇒x=e^(−u)   dx=−e^(−u) du  =∫_0 ^(+∞) ((sin(u))/u)e^(−u) du  f(t)=∫_0 ^(×∞) ((sin(u))/u)e^(−ut) du  f′(t)=∫_0 ^(+∞) −sin(u)e^(−ut) du  =−Im∫_0 ^(+∞) e^(u(i−t)) du  =Im{(1/(i−t))}=((−1)/(1+t^2 ))  f(t)=−arctan(t)+(π/2)  f(1)=(π/2)−(π/4)=(π/4)=∫_0 ^1 ((sin(ln(x)))/(ln(x)))dx=(π/4)

$${u}=−{ln}\left({x}\right)\Rightarrow{x}={e}^{−{u}} \\ $$$${dx}=−{e}^{−{u}} {du} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \frac{{sin}\left({u}\right)}{{u}}{e}^{−{u}} {du} \\ $$$${f}\left({t}\right)=\int_{\mathrm{0}} ^{×\infty} \frac{{sin}\left({u}\right)}{{u}}{e}^{−{ut}} {du} \\ $$$${f}'\left({t}\right)=\int_{\mathrm{0}} ^{+\infty} −{sin}\left({u}\right){e}^{−{ut}} {du} \\ $$$$=−{Im}\int_{\mathrm{0}} ^{+\infty} {e}^{{u}\left({i}−{t}\right)} {du} \\ $$$$={Im}\left\{\frac{\mathrm{1}}{{i}−{t}}\right\}=\frac{−\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${f}\left({t}\right)=−{arctan}\left({t}\right)+\frac{\pi}{\mathrm{2}} \\ $$$${f}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{4}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sin}\left({ln}\left({x}\right)\right)}{{ln}\left({x}\right)}{dx}=\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com