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Question Number 126465 by mnjuly1970 last updated on 20/Dec/20

... nice calculus... Evaluate ... φ = ∫_0 ^( 1) ((sin(ln(x))−ln(x))/(ln^2 (x)))dx Ans :: ln((√2) )+(π/4) −1 ...

$$\:\:\:\:\:\:\:\:\:\:\:\:...\:{nice}\:\:{calculus}... \\ $$$$\:\:\:\:\:\mathscr{E}{valuate}\:... \\ $$$$\:\:\:\:\:\:\:\:\phi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{sin}\left({ln}\left({x}\right)\right)−{ln}\left({x}\right)}{{ln}^{\mathrm{2}} \left({x}\right)}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathscr{A}{ns}\:::\:{ln}\left(\sqrt{\mathrm{2}}\:\right)+\frac{\pi}{\mathrm{4}}\:−\mathrm{1}\:... \\ $$

Commented by talminator2856791 last updated on 22/Dec/20

do you have more difficult question than this

$$\:\mathrm{do}\:\mathrm{you}\:\mathrm{have}\:\mathrm{more}\:\mathrm{difficult}\:\mathrm{question}\:\mathrm{than}\:\mathrm{this} \\ $$

Answered by mindispower last updated on 23/Dec/20

−ln(x)=t =∫_0 ^∞ ((t−sin(t))/t^2 )e^(−t) sin(t)−t=Σ_(k≥1) (((−1)^k t^(2k+1) )/((2k+1)!)) =Σ_(k≥1) ∫_0 ^∞ (((−1)^k )/((2k+1)!))t^(2k−1) e^(−t) dt =Σ_(k≥1) (((−1)^k )/((2k+1)!))Γ(2k) =Σ_(k≥1) (((−1)^k )/(2k(2k+1)))=Σ_(k≥1) (((−1)^k )/(2k))−Σ_(k≥1) (((−1)^k )/(2k+1)) =(1/2)Σ_(k≥1) (−1)^k ∫_0 ^1 x^(k−1) =(1/2)∫_0 ^1 (1/x)Σ_(k≥1) (−x)^k dx−Σ_(k≥1) ∫_0 ^1 (−1)^k x^(2k) dx =−(1/2)∫_0 ^1 (1/(1+x))dx−∫_0 ^1 ((−x^2 )/(1+x^2 ))dx =−(1/2)ln(2)+∫_0 ^1 dx+∫_0 ^1 (dx/(1+x^2 )) =1+(π/4)−ln((√2))

$$−{ln}\left({x}\right)={t} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{t}−{sin}\left({t}\right)}{{t}^{\mathrm{2}} }{e}^{−{t}} \\ $$$${sin}\left({t}\right)−{t}=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} {t}^{\mathrm{2}{k}+\mathrm{1}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!} \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!}{t}^{\mathrm{2}{k}−\mathrm{1}} {e}^{−{t}} {dt} \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!}\Gamma\left(\mathrm{2}{k}\right) \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}\left(\mathrm{2}{k}+\mathrm{1}\right)}=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}}−\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}\geqslant\mathrm{1}} {\sum}\left(−\mathrm{1}\right)^{{k}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{k}−\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{x}}\underset{{k}\geqslant\mathrm{1}} {\sum}\left(−{x}\right)^{{k}} {dx}−\underset{{k}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} {x}^{\mathrm{2}{k}} {dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{x}}{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)+\int_{\mathrm{0}} ^{\mathrm{1}} {dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\mathrm{1}+\frac{\pi}{\mathrm{4}}−{ln}\left(\sqrt{\mathrm{2}}\right) \\ $$

Commented by mnjuly1970 last updated on 24/Dec/20

thanks a lot..

$${thanks}\:{a}\:{lot}.. \\ $$$$ \\ $$

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