Question Number 125133 by mnjuly1970 last updated on 08/Dec/20
$$\:\:\:\:\:\:\:\:\:\:\:\:\:...\:\blacktriangleleft{advanced}\:\:\:{calculus}\blacktriangleright... \\ $$$$\:\:\:\:\:\:{prove}\:\:{that}\:::: \\ $$$$\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left\{\frac{{cos}\left({log}\left({x}\right)\right)−\mathrm{1}}{{log}\left({x}\right)}\right\}{dx}=\frac{{log}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:...\ast{adopted}\:{from}\:{youtube}\ast...\:\:\: \\ $$$$\:\ast\:\ast\:{youtube}\:{solution}\:{is}\:{not}\:{considered}\:\ast\:\ast \\ $$$$\:\: \\ $$
Answered by mathmax by abdo last updated on 08/Dec/20
![I=∫_0 ^1 ((cos(logx)−1)/(logx)) we do the chamgement logx=−t ⇒x=e^(−t) I=∫_∞ ^0 ((cost−1)/(−t))(−e^(−t) )dt =−∫_0 ^∞ ((cost−1)/t)e^(−t) dt let f(a) =∫_0 ^∞ ((cost−1)/t)e^(−at) dt with a>0 we have f^′ (a)=−∫_0 ^∞ (cost−1)e^(−at) dt =−∫_0 ^∞ e^(−at) cost +∫_0 ^∞ e^(−at) dt but ∫_0 ^∞ e^(−at) dt =[−(1/a)e^(−at) ]_0 ^∞ =(1/a) ∫_0 ^∞ e^(−at) cost dt =Re(∫_0 ^∞ e^(−at+it) dt) snd ∫_0 ^∞ e^((−a+i)t) dt =[(1/(−a+i))e^((−a+i)t) ]_0 ^∞ =((−1)/(a−i)){−1} =(1/(a−i))=((a+i)/(a^2 +1)) ⇒∫_0 ^∞ e^(−at) cost dt =(a/(a^2 +1)) ⇒f^′ (a)=−(a/(1+a^2 ))+(1/a) ⇒ f(a)=lna−(1/2)ln(1+a^2 )+C =ln((a/( (√(a^2 +1))))) +c ∣f(a)∣≤(m/a)→0 (a→+∞) ⇒c=0 ⇒f(a)=ln((a/( (√(1+a^2 ))))) I =−f(1)=−ln((1/( (√2))))=ln((√2))=((ln(2))/2)](Q125135.png)
$$\mathrm{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{cos}\left(\mathrm{logx}\right)−\mathrm{1}}{\mathrm{logx}}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{chamgement}\:\mathrm{logx}=−\mathrm{t}\:\Rightarrow\mathrm{x}=\mathrm{e}^{−\mathrm{t}} \\ $$$$\mathrm{I}=\int_{\infty} ^{\mathrm{0}} \:\frac{\mathrm{cost}−\mathrm{1}}{−\mathrm{t}}\left(−\mathrm{e}^{−\mathrm{t}} \right)\mathrm{dt}\:=−\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cost}−\mathrm{1}}{\mathrm{t}}\mathrm{e}^{−\mathrm{t}} \mathrm{dt} \\ $$$$\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cost}−\mathrm{1}}{\mathrm{t}}\mathrm{e}^{−\mathrm{at}} \:\mathrm{dt}\:\:\mathrm{with}\:\mathrm{a}>\mathrm{0}\:\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)=−\int_{\mathrm{0}} ^{\infty} \:\left(\mathrm{cost}−\mathrm{1}\right)\mathrm{e}^{−\boldsymbol{\mathrm{a}}\mathrm{t}} \mathrm{dt}\:=−\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\boldsymbol{\mathrm{a}}\mathrm{t}} \mathrm{cost}\:+\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\boldsymbol{\mathrm{a}}\mathrm{t}} \:\mathrm{dt} \\ $$$$\mathrm{but}\:\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\boldsymbol{\mathrm{a}}\mathrm{t}} \mathrm{dt}\:=\left[−\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}}\mathrm{e}^{−\boldsymbol{\mathrm{a}}\mathrm{t}} \right]_{\mathrm{0}} ^{\infty} \:=\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\boldsymbol{\mathrm{a}}\mathrm{t}} \mathrm{cost}\:\mathrm{dt}\:=\mathrm{Re}\left(\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\boldsymbol{\mathrm{a}}\mathrm{t}+\mathrm{it}} \mathrm{dt}\right)\:\mathrm{snd} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{\left(−\boldsymbol{\mathrm{a}}+\mathrm{i}\right)\mathrm{t}} \mathrm{dt}\:=\left[\frac{\mathrm{1}}{−\boldsymbol{\mathrm{a}}+\mathrm{i}}\mathrm{e}^{\left(−\boldsymbol{\mathrm{a}}+\mathrm{i}\right)\mathrm{t}} \right]_{\mathrm{0}} ^{\infty} \:=\frac{−\mathrm{1}}{\boldsymbol{\mathrm{a}}−\mathrm{i}}\left\{−\mathrm{1}\right\}\:=\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}−\mathrm{i}}=\frac{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{i}}}{\boldsymbol{\mathrm{a}}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{at}} \mathrm{cost}\:\mathrm{dt}\:=\frac{\mathrm{a}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{a}\right)=−\frac{\mathrm{a}}{\mathrm{1}+\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{a}}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\mathrm{lna}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \right)+\mathrm{C}\:=\mathrm{ln}\left(\frac{\mathrm{a}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{1}}}\right)\:+\mathrm{c} \\ $$$$\mid\mathrm{f}\left(\mathrm{a}\right)\mid\leqslant\frac{\mathrm{m}}{\mathrm{a}}\rightarrow\mathrm{0}\:\left(\mathrm{a}\rightarrow+\infty\right)\:\Rightarrow\mathrm{c}=\mathrm{0}\:\Rightarrow\mathrm{f}\left(\mathrm{a}\right)=\mathrm{ln}\left(\frac{\mathrm{a}}{\:\sqrt{\mathrm{1}+\mathrm{a}^{\mathrm{2}} }}\right) \\ $$$$\mathrm{I}\:=−\mathrm{f}\left(\mathrm{1}\right)=−\mathrm{ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)=\mathrm{ln}\left(\sqrt{\mathrm{2}}\right)=\frac{\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 08/Dec/20
$${thank}\:{you}\:{so}\:{much} \\ $$$${mr}\:{max}\:.{grateful}.... \\ $$
Commented by mathmax by abdo last updated on 08/Dec/20
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}\:\mathrm{sir} \\ $$
Answered by Dwaipayan Shikari last updated on 08/Dec/20
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{cos}\left({logx}\right)−\mathrm{1}}{{logx}}{dx} \\ $$$$=\int_{−\infty} ^{\mathrm{0}} {e}^{{t}} \:\frac{{cos}\left({t}\right)−\mathrm{1}}{{t}}\:{dt}\:\:\:\:\:{logx}={t}\Rightarrow\frac{\mathrm{1}}{{x}}=\frac{{dt}}{{dx}}\:\:\:\:\:\:\:\:\:\:\: \\ $$$${I}\left({a}\right)=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{cost}−\mathrm{1}}{{t}}{e}^{−{at}} {dt} \\ $$$${I}'\left({a}\right)=−\int_{\mathrm{0}} ^{\infty} {coste}^{−{at}} +\int_{\mathrm{0}} ^{\infty} {e}^{−{at}} {dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}\left({a}−{i}\right)} +{e}^{−{t}\left({i}+{a}\right)} {dt}\:+\frac{\mathrm{1}}{{a}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{a}−{i}}+\frac{\mathrm{1}}{{i}+{a}}\right)+\frac{\mathrm{1}}{{a}}\:=−\frac{{a}}{{a}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{{a}} \\ $$$${I}\left({a}\right)=−\frac{\mathrm{1}}{\mathrm{2}}{log}\left({a}^{\mathrm{2}} +\mathrm{1}\right)+{log}\left({a}\right)+{C} \\ $$$${a}\rightarrow\infty\:\:{C}=\mathrm{0} \\ $$$${I}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\mathrm{2}\right) \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 08/Dec/20
$${bravo}\:{mr}\:{payan}.{thank}\:{you} \\ $$$${so}\:{much}.. \\ $$
Answered by mnjuly1970 last updated on 08/Dec/20
![solution: Ω=Re(∫_0 ^( 1) ((x^i −1)/(log(x)))dx)=Re(ln(1+i)) =Re(ln((√2) e^((iπ)/4) ))=Re[ln((√2) )+i(π/4)] ∴ Ω =((ln(2))/2) ✓](Q125143.png)
$${solution}: \\ $$$$\:\Omega={Re}\left(\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{x}^{{i}} −\mathrm{1}}{{log}\left({x}\right)}{dx}\right)={Re}\left({ln}\left(\mathrm{1}+{i}\right)\right) \\ $$$$={Re}\left({ln}\left(\sqrt{\mathrm{2}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\right)={Re}\left[{ln}\left(\sqrt{\mathrm{2}}\:\right)+{i}\frac{\pi}{\mathrm{4}}\right] \\ $$$$\:\:\therefore\:\:\Omega\:=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\:\:\:\checkmark \\ $$
Commented by mathmax by abdo last updated on 08/Dec/20
$$\mathrm{you}\:\mathrm{are}\:\mathrm{taking}\:\mathrm{route}\:\mathrm{66}\:\mathrm{in}\:\mathrm{america}..! \\ $$
Commented by mnjuly1970 last updated on 08/Dec/20
$$\:\:{sincerely}\:{yours}\: \\ $$$$\:{m}.{n}.\mathrm{1970} \\ $$
Commented by mnjuly1970 last updated on 09/Dec/20
$$\:\:\:{sir}\:{max} \\ $$$$\:{since}\::{some}\:{integrals}\:{and}\:{problems}\: \\ $$$${were}\:{solved}\:{many}\:{time}\:\:{therefore}\:\:{i}\:{tried} \\ $$$${to}\:{shorten}\:{the}\:{solution}. \\ $$$${example}\:: \\ $$$$\:\:{ln}\left({a}+\mathrm{1}\right)=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{{x}^{{a}} −\mathrm{1}}{{ln}\left({x}\right)}\right){dx}\:,.... \\ $$