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Question Number 103198 by Dwaipayan Shikari last updated on 13/Jul/20

∫_0 ^1 sin(logx)dx

$$\int_{\mathrm{0}} ^{\mathrm{1}} {sin}\left({logx}\right){dx} \\ $$

Commented by Dwaipayan Shikari last updated on 13/Jul/20

sinx=((e^(ix) −e^(−ix) )/(2i)) sin(logx)=((x^i −x^(−i) )/(2i)) ∫_0 ^1 ((x^i −x^(−i) )/(2i))dx=(1/(2i))∫x^i −x^(−i) dx=(1/(2i)).[((x^(i+1) /(i+1))−(x^(−i+1) /(1−i)))]_0 ^1 =(1/(2i))((1/(i+1))−(1/(1−i)))=−(1/2) Is it true???????

$${sinx}=\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}} \\ $$$${sin}\left({logx}\right)=\frac{{x}^{{i}} −{x}^{−{i}} }{\mathrm{2}{i}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{i}} −{x}^{−{i}} }{\mathrm{2}{i}}{dx}=\frac{\mathrm{1}}{\mathrm{2}{i}}\int{x}^{{i}} −{x}^{−{i}} {dx}=\frac{\mathrm{1}}{\mathrm{2}{i}}.\left[\left(\frac{{x}^{{i}+\mathrm{1}} }{{i}+\mathrm{1}}−\frac{{x}^{−{i}+\mathrm{1}} }{\mathrm{1}−{i}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left(\frac{\mathrm{1}}{{i}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{1}−{i}}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{Is}}\:\boldsymbol{\mathrm{it}}\:\boldsymbol{\mathrm{true}}??????? \\ $$

Answered by OlafThorendsen last updated on 13/Jul/20

I = ∫_0 ^1 sin(lnx)dx x = e^u ⇒ dx = e^u du I = ∫_(−∞) ^0 sinu.e^u du I = ∫_(−∞) ^0 ((e^(iu) −e^(−iu) )/(2i))e^u du I = (1/(2i))∫_(−∞) ^0 (e^((1+i)u) −e^((1−i)u) )du I = (1/(2i))[(e^((1+i)u) /(1+i))−(e^((1−i)u) /(1−i))]_(−∞) ^0 I = (1/(2i))[(1/(1+i))−(1/(1−i))] I = (1/(2i))[((1−i)/(1−i^2 ))−((1+i)/(1−i^2 ))] I = (1/(2i))(((−2i)/2)) = −(1/2)

$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}\left(\mathrm{ln}{x}\right){dx} \\ $$$${x}\:=\:{e}^{{u}} \:\Rightarrow\:{dx}\:=\:{e}^{{u}} {du} \\ $$$$\mathrm{I}\:=\:\int_{−\infty} ^{\mathrm{0}} \mathrm{sin}{u}.{e}^{{u}} {du} \\ $$$$\mathrm{I}\:=\:\int_{−\infty} ^{\mathrm{0}} \frac{{e}^{{iu}} −{e}^{−{iu}} }{\mathrm{2}{i}}{e}^{{u}} {du} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{−\infty} ^{\mathrm{0}} \left({e}^{\left(\mathrm{1}+{i}\right){u}} −{e}^{\left(\mathrm{1}−{i}\right){u}} \right){du} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{2}{i}}\left[\frac{{e}^{\left(\mathrm{1}+{i}\right){u}} }{\mathrm{1}+{i}}−\frac{{e}^{\left(\mathrm{1}−{i}\right){u}} }{\mathrm{1}−{i}}\right]_{−\infty} ^{\mathrm{0}} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{2}{i}}\left[\frac{\mathrm{1}}{\mathrm{1}+{i}}−\frac{\mathrm{1}}{\mathrm{1}−{i}}\right] \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{2}{i}}\left[\frac{\mathrm{1}−{i}}{\mathrm{1}−{i}^{\mathrm{2}} }−\frac{\mathrm{1}+{i}}{\mathrm{1}−{i}^{\mathrm{2}} }\right] \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{2}{i}}\left(\frac{−\mathrm{2}{i}}{\mathrm{2}}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by Smail last updated on 13/Jul/20

By parts u=sin(lnx)⇒u′=(1/x)cos(lnx) v′=1⇒v=x ∫_0 ^1 sin(lnx)dx=[xsin(lnx)]_0 ^1 −∫_0 ^1 cos(lnx)dx u=cos(lnx)⇒u′=−(1/x)sin(lnx) v′=1⇒v=x ∫_0 ^1 sin(lnx)dx=−[xcos(lnx)]_0 ^1 −∫_0 ^1 sin(lnx)dx 2∫_0 ^1 sin(lnx)dx=−cos(ln1) ∫_0 ^1 sin(lnx)dx=((−1)/2)

$${By}\:{parts} \\ $$$${u}={sin}\left({lnx}\right)\Rightarrow{u}'=\frac{\mathrm{1}}{{x}}{cos}\left({lnx}\right) \\ $$$${v}'=\mathrm{1}\Rightarrow{v}={x} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {sin}\left({lnx}\right){dx}=\left[{xsin}\left({lnx}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} {cos}\left({lnx}\right){dx} \\ $$$${u}={cos}\left({lnx}\right)\Rightarrow{u}'=−\frac{\mathrm{1}}{{x}}{sin}\left({lnx}\right) \\ $$$${v}'=\mathrm{1}\Rightarrow{v}={x} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {sin}\left({lnx}\right){dx}=−\left[{xcos}\left({lnx}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} {sin}\left({lnx}\right){dx} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {sin}\left({lnx}\right){dx}=−{cos}\left({ln}\mathrm{1}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {sin}\left({lnx}\right){dx}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$

Answered by mathmax by abdo last updated on 13/Jul/20

A =∫_0 ^1 sin(lnx)dx changement lnx =−t give A =−∫_0 ^∞ sin(−t)(−e^(−t) )dt =−∫_0 ^∞ e^(−t) sint dt =−Im(∫_0 ^∞ e^(−t+it) dt) and ∫_0 ^∞ e^((−1+i)t ) dt =[(1/(−1+i))e^((−1+i)t) ]_0 ^∞ =−(1/(−1+i)) =(1/(1−i)) =((1+i)/2) ⇒ A =−(1/2)

$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{sin}\left(\mathrm{lnx}\right)\mathrm{dx}\:\:\mathrm{changement}\:\mathrm{lnx}\:=−\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{A}\:=−\int_{\mathrm{0}} ^{\infty} \:\mathrm{sin}\left(−\mathrm{t}\right)\left(−\mathrm{e}^{−\mathrm{t}} \right)\mathrm{dt}\:=−\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{t}} \:\mathrm{sint}\:\mathrm{dt} \\ $$$$=−\mathrm{Im}\left(\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{t}+\mathrm{it}} \mathrm{dt}\right)\:\mathrm{and}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{\left(−\mathrm{1}+\mathrm{i}\right)\mathrm{t}\:} \:\mathrm{dt}\:=\left[\frac{\mathrm{1}}{−\mathrm{1}+\mathrm{i}}\mathrm{e}^{\left(−\mathrm{1}+\mathrm{i}\right)\mathrm{t}} \right]_{\mathrm{0}} ^{\infty} \\ $$$$=−\frac{\mathrm{1}}{−\mathrm{1}+\mathrm{i}}\:=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{i}}\:=\frac{\mathrm{1}+\mathrm{i}}{\mathrm{2}}\:\Rightarrow\:\mathrm{A}\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by Aziztisffola last updated on 13/Jul/20

let x=e^t ⇒dx=e^t dt ∫_0 ^1 sin(logx)dx=∫_(−∞) ^( 0) e^t sin(t)dt=∫_0 ^(+∞) −e^t sin(t)dt With Laplace transform: L{−e^t sin(t)}= L{−sin(t)}(s−1)=((−1)/((s−1)^2 +1)) then ∫_0 ^( +∞) −e^(−st) e^t sin(t)dt=((−1)/((s−1)^2 +1)) let s=0 ⇒ ∫_0 ^( +∞) −e^t sin(t)dt=((−1)/((−1)^2 +1))=((−1)/2) Hence ∫_(−∞) ^( 0) e^t sin(t)dt=((−1)/2) ∴ ∫_0 ^1 sin(logx)dx= ((−1)/2)

$$\:\mathrm{let}\:\mathrm{x}=\mathrm{e}^{\mathrm{t}} \:\Rightarrow\mathrm{dx}=\mathrm{e}^{\mathrm{t}} \mathrm{dt} \\ $$$$\:\int_{\mathrm{0}} ^{\mathrm{1}} {sin}\left({logx}\right){dx}=\int_{−\infty} ^{\:\mathrm{0}} \mathrm{e}^{\mathrm{t}} \mathrm{sin}\left(\mathrm{t}\right)\mathrm{dt}=\int_{\mathrm{0}} ^{+\infty} −\mathrm{e}^{\mathrm{t}} \mathrm{sin}\left(\mathrm{t}\right)\mathrm{dt} \\ $$$$\:\mathrm{With}\:\mathrm{Laplace}\:\mathrm{transform}: \\ $$$$\:\mathcal{L}\left\{−\mathrm{e}^{\mathrm{t}} \mathrm{sin}\left(\mathrm{t}\right)\right\}=\:\mathcal{L}\left\{−\mathrm{sin}\left(\mathrm{t}\right)\right\}\left(\mathrm{s}−\mathrm{1}\right)=\frac{−\mathrm{1}}{\left(\mathrm{s}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$$\:\mathrm{then}\:\int_{\mathrm{0}} ^{\:+\infty} −\mathrm{e}^{−\mathrm{st}} \mathrm{e}^{\mathrm{t}} \mathrm{sin}\left(\mathrm{t}\right)\mathrm{dt}=\frac{−\mathrm{1}}{\left(\mathrm{s}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$$\:\:\mathrm{let}\:\mathrm{s}=\mathrm{0}\:\Rightarrow\:\int_{\mathrm{0}} ^{\:+\infty} −\mathrm{e}^{\mathrm{t}} \mathrm{sin}\left(\mathrm{t}\right)\mathrm{dt}=\frac{−\mathrm{1}}{\left(−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$$\:\mathrm{Hence}\:\:\int_{−\infty} ^{\:\mathrm{0}} \mathrm{e}^{\mathrm{t}} \mathrm{sin}\left(\mathrm{t}\right)\mathrm{dt}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$$\therefore\:\int_{\mathrm{0}} ^{\mathrm{1}} {sin}\left({logx}\right){dx}=\:\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$$\: \\ $$

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