Menu Close

Question-201544

Tinku Tara 0 Comments
Pin




Question Number 201544 by sonukgindia last updated on 08/Dec/23
Answered by aleks041103 last updated on 09/Dec/23
J=∫_0 ^( 1) ((4sin^2 (ln(x)))/(ln(1/x)))dx=−∫_0 ^( 1) ((4sin^2 (1.ln(x)))/(ln(x)))dx I(s)=−∫_0 ^( 1) ((4sin^2 (slnx))/(ln(x)))dx, I(0)=0, J=I(1) I ′(s)=−∫_0 ^( 1) ((8sin(s lnx)cos(s lnx) ln(x))/(ln(x)))dx= =−∫_0 ^( 1) 4sin(2slnx)dx=−4Im(∫_0 ^( 1) e^(2isln(x)) dx)= =−4Im(∫_0 ^( 1) x^(2is) dx)=−4Im((1/(1+2is)))= =−4Im(((1−2is)/(1+4s^2 )))=((8s)/(1+4s^2 ))=((4(2s))/(1+4s^2 ))= =(((1+4s^2 )′)/(1+4s^2 ))=(ln(1+4s^2 ))′ ⇒I(s)=ln(1+4s^2 )+c I(0)=0=ln(1)+c⇒c=0 ⇒I(s)=ln(1+4s^2 ) ⇒J=∫_0 ^( 1) ((4sin^2 (ln(x)))/(ln(1/x)))dx=I(1)=ln(5)
$${J}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{4}{sin}^{\mathrm{2}} \left({ln}\left({x}\right)\right)}{{ln}\left(\mathrm{1}/{x}\right)}{dx}=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{4}{sin}^{\mathrm{2}} \left(\mathrm{1}.{ln}\left({x}\right)\right)}{{ln}\left({x}\right)}{dx} \\ $$$${I}\left({s}\right)=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{4}{sin}^{\mathrm{2}} \left({slnx}\right)}{{ln}\left({x}\right)}{dx},\:{I}\left(\mathrm{0}\right)=\mathrm{0},\:{J}={I}\left(\mathrm{1}\right) \\ $$$${I}\:'\left({s}\right)=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{8}{sin}\left({s}\:{lnx}\right){cos}\left({s}\:{lnx}\right)\:{ln}\left({x}\right)}{{ln}\left({x}\right)}{dx}= \\ $$$$=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{4}{sin}\left(\mathrm{2}{slnx}\right){dx}=−\mathrm{4}{Im}\left(\int_{\mathrm{0}} ^{\:\mathrm{1}} {e}^{\mathrm{2}{isln}\left({x}\right)} {dx}\right)= \\ $$$$=−\mathrm{4}{Im}\left(\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\mathrm{2}{is}} {dx}\right)=−\mathrm{4}{Im}\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}{is}}\right)= \\ $$$$=−\mathrm{4}{Im}\left(\frac{\mathrm{1}−\mathrm{2}{is}}{\mathrm{1}+\mathrm{4}{s}^{\mathrm{2}} }\right)=\frac{\mathrm{8}{s}}{\mathrm{1}+\mathrm{4}{s}^{\mathrm{2}} }=\frac{\mathrm{4}\left(\mathrm{2}{s}\right)}{\mathrm{1}+\mathrm{4}{s}^{\mathrm{2}} }= \\ $$$$=\frac{\left(\mathrm{1}+\mathrm{4}{s}^{\mathrm{2}} \right)'}{\mathrm{1}+\mathrm{4}{s}^{\mathrm{2}} }=\left({ln}\left(\mathrm{1}+\mathrm{4}{s}^{\mathrm{2}} \right)\right)' \\ $$$$\Rightarrow{I}\left({s}\right)={ln}\left(\mathrm{1}+\mathrm{4}{s}^{\mathrm{2}} \right)+{c} \\ $$$${I}\left(\mathrm{0}\right)=\mathrm{0}={ln}\left(\mathrm{1}\right)+{c}\Rightarrow{c}=\mathrm{0} \\ $$$$\Rightarrow{I}\left({s}\right)={ln}\left(\mathrm{1}+\mathrm{4}{s}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{J}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{4}{sin}^{\mathrm{2}} \left({ln}\left({x}\right)\right)}{{ln}\left(\mathrm{1}/{x}\right)}{dx}={I}\left(\mathrm{1}\right)={ln}\left(\mathrm{5}\right) \\ $$
Answered by Mathspace last updated on 09/Dec/23
I=−4∫_0 ^1 ((sin^2 (lnx))/(lnx))dx (lnx=−t) =−4∫_∞ ^0 ((sin^2 (t))/(−t))×(−e^(−t) )dt =∫_0 ^∞ ((e^(−t) sin^2 t)/t)dt f(λ)=∫_0 ^∞ ((e^(−λt) sin^2 t)/t)dt (λ>0) f^′ (λ)=−∫_0 ^∞ e^(−λt ) sin^2 t dt =−∫_0 ^∞ e^(−λt) ((1−cos(2t))/2)dt =(1/2)∫_0 ^∞ e^(−λt) cos(2t)−(1/2)∫_0 ^∞ e^(−λt) dt but ∫_0 ^∞ e^(−λt) dt=[−(1/λ)e^(−λt) ]_0 ^∞ =(1/λ) ∫_0 ^∞ e^(−λt) cos(2t)dt=Re(∫_0 ^∞ e^((−λ+2i)t) dt) and ∫_0 ^∞ e^((−λ+2i)t) dt =(1/(−λ+2i)) e^((−λ+2i)t) ]_0 ^∞ =((−1)/(−λ+2i))=(1/(λ−2i))=((λ+2i)/(λ^2 +4)) ⇒∫_0 ^∞ e^(−λt) cos(2t)dt=(λ/(λ^2 +4)) ⇒ f^′ (λ)=(λ/(2(λ^2 +4)))−(1/(2λ)) ⇒ f(λ)=(1/4)ln(λ^2 +4)−(1/2)lnλ +σ =(1/2)ln((√(λ^2 +4)))−(1/2)lnλ +σ =(1/2)ln(((√(λ^2 +4))/λ))+σ lim_(λ→∞) f(λ)=0=σ ∫_0 ^∞ e^(−t) ((sin^2 t)/t)dt=f(1)=(1/2)ln((√5)) =((ln5)/4)
$${I}=−\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{sin}^{\mathrm{2}} \left({lnx}\right)}{{lnx}}{dx}\:\:\:\left({lnx}=−{t}\right) \\ $$$$=−\mathrm{4}\int_{\infty} ^{\mathrm{0}} \frac{{sin}^{\mathrm{2}} \left({t}\right)}{−{t}}×\left(−{e}^{−{t}} \right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{t}} \:{sin}^{\mathrm{2}} {t}}{{t}}{dt} \\ $$$${f}\left(\lambda\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−\lambda{t}} \:{sin}^{\mathrm{2}} {t}}{{t}}{dt}\:\:\:\:\left(\lambda>\mathrm{0}\right) \\ $$$${f}^{'} \left(\lambda\right)=−\int_{\mathrm{0}} ^{\infty} {e}^{−\lambda{t}\:} {sin}^{\mathrm{2}} {t}\:{dt} \\ $$$$=−\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\lambda{t}} \:\frac{\mathrm{1}−{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:{e}^{−\lambda{t}} {cos}\left(\mathrm{2}{t}\right)−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {e}^{−\lambda{t}} {dt} \\ $$$${but} \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−\lambda{t}} {dt}=\left[−\frac{\mathrm{1}}{\lambda}{e}^{−\lambda{t}} \right]_{\mathrm{0}} ^{\infty} =\frac{\mathrm{1}}{\lambda} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−\lambda{t}} {cos}\left(\mathrm{2}{t}\right){dt}={Re}\left(\int_{\mathrm{0}} ^{\infty} {e}^{\left(−\lambda+\mathrm{2}{i}\right){t}} {dt}\right) \\ $$$${and}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{\left(−\lambda+\mathrm{2}{i}\right){t}} {dt} \\ $$$$\left.=\frac{\mathrm{1}}{−\lambda+\mathrm{2}{i}}\:{e}^{\left(−\lambda+\mathrm{2}{i}\right){t}} \right]_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{−\mathrm{1}}{−\lambda+\mathrm{2}{i}}=\frac{\mathrm{1}}{\lambda−\mathrm{2}{i}}=\frac{\lambda+\mathrm{2}{i}}{\lambda^{\mathrm{2}} +\mathrm{4}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} {e}^{−\lambda{t}} {cos}\left(\mathrm{2}{t}\right){dt}=\frac{\lambda}{\lambda^{\mathrm{2}} +\mathrm{4}}\:\Rightarrow \\ $$$${f}^{'} \left(\lambda\right)=\frac{\lambda}{\mathrm{2}\left(\lambda^{\mathrm{2}} +\mathrm{4}\right)}−\frac{\mathrm{1}}{\mathrm{2}\lambda}\:\Rightarrow \\ $$$${f}\left(\lambda\right)=\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\lambda^{\mathrm{2}} +\mathrm{4}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\lambda\:+\sigma \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\sqrt{\lambda^{\mathrm{2}} +\mathrm{4}}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\lambda\:+\sigma \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\sqrt{\lambda^{\mathrm{2}} +\mathrm{4}}}{\lambda}\right)+\sigma \\ $$$${lim}_{\lambda\rightarrow\infty} {f}\left(\lambda\right)=\mathrm{0}=\sigma \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} \frac{{sin}^{\mathrm{2}} {t}}{{t}}{dt}={f}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\sqrt{\mathrm{5}}\right) \\ $$$$=\frac{{ln}\mathrm{5}}{\mathrm{4}} \\ $$$$ \\ $$
Commented by Mathspace last updated on 09/Dec/23
sorry I=4∫_0 ^∞ ((e^(−t) sin^2 t)/t)dt ⇒ ★I=(√5)★
$${sorry}\:{I}=\mathrm{4}\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{t}} {sin}^{\mathrm{2}} {t}}{{t}}{dt}\:\Rightarrow \\ $$$$\bigstar{I}=\sqrt{\mathrm{5}}\bigstar \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *