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Question Number 166872 by mnjuly1970 last updated on 01/Mar/22
∫_0 ^( ∞) (( e^( −x) .ln(x).sin(x))/x) dx = −(π/8) (2γ + ln(2))
$$ \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{e}^{\:−{x}} .{ln}\left({x}\right).{sin}\left({x}\right)}{{x}}\:{dx}\:=\:−\frac{\pi}{\mathrm{8}}\:\left(\mathrm{2}\gamma\:+\:{ln}\left(\mathrm{2}\right)\right) \\ $$$$ \\ $$
Answered by qaz last updated on 01/Mar/22
∫_0 ^∞ ((e^(−x) lnx∙sin x)/x)dx =∫_0 ^∞ L{e^(−x) sin x}∙L^(−1) {((lnx)/x)}dx =−∫_0 ^∞ ((lnx+γ)/((x+1)^2 +1))dx =−γarctan (x+1)∣_0 ^∞ −∫_1 ^∞ ((ln(x−1))/(x^2 +1))dx =−(π/4)γ−∫_0 ^1 ((ln((1/x)−1))/(1+x^2 ))dx =−(π/4)γ−∫_0 ^(π/4) ln(cot x−1)dx =−(π/4)γ−(1/8)πln2
$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{e}^{−\mathrm{x}} \mathrm{lnx}\centerdot\mathrm{sin}\:\mathrm{x}}{\mathrm{x}}\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \mathscr{L}\left\{\mathrm{e}^{−\mathrm{x}} \mathrm{sin}\:\mathrm{x}\right\}\centerdot\mathscr{L}^{−\mathrm{1}} \left\{\frac{\mathrm{lnx}}{\mathrm{x}}\right\}\mathrm{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{lnx}+\gamma}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}\mathrm{dx} \\ $$$$=−\gamma\mathrm{arctan}\:\left(\mathrm{x}+\mathrm{1}\right)\mid_{\mathrm{0}} ^{\infty} −\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{x}−\mathrm{1}\right)}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx} \\ $$$$=−\frac{\pi}{\mathrm{4}}\gamma−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{x}}−\mathrm{1}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=−\frac{\pi}{\mathrm{4}}\gamma−\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \mathrm{ln}\left(\mathrm{cot}\:\mathrm{x}−\mathrm{1}\right)\mathrm{dx} \\ $$$$=−\frac{\pi}{\mathrm{4}}\gamma−\frac{\mathrm{1}}{\mathrm{8}}\pi\mathrm{ln2} \\ $$
Answered by mindispower last updated on 02/Mar/22
((sin(x))/x)=Σ_(n≥0) (((−1)^n )/((2n+1)!))x^(2n) ∫_0 ^∞ x^(2n) e^(−x) ln(x)dx=∂_(a=2n+1) ∫_0 ^∞ x^a e^(−x) dx =∂_(a=2n+1) Γ(a+1)=Γ′(2n+1)=(2n)!Ψ(2n+1) =Σ_(n≥0) (((−1)^n )/((2n+1)))Ψ(2n+1)etcka Ψ(2n+1)=H_(2n+1) −γ =Σ_(n≥0) (−1)^n ((H_(2n+1) −γ)/(2n+1))=−γtan^(−1) (1)+Σ_(n≥0) (((−1)^n H_(2n+1) )/(2n+1)) =Σ_(n≥1) x^n H_n =−((ln(1−x))/(1−x)) ⇒Σ_(n≥1) (x^(n+1) /(n+1))H_n =(1/2)ln^2 (1−x)....E E=i⇒Σ_(n≥1) (i^(n+1) /(n+1))H_n =(1/2)(ln(1−i))^2 We tack ImaginrayPart⇒ImΣ_(n≥0) (i^(n+1) /(n+1))H_n =Σ_(n≥0) (((−1)^n H_(2n+1) )/(2n+1)) =Im(1/2)(ln(1−i))^2 =Im(1/2)(ln((√2))+i.(−(π/4)))^2 =−(π/4)ln((√2))=−(π/8)ln(2) ⇔=−(π/4)γ−(π/8)ln(2)=−(π/8)(2γ+ln(2))
$$\frac{{sin}\left({x}\right)}{{x}}=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{x}^{\mathrm{2}{n}} \\ $$$$\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{2}{n}} {e}^{−{x}} {ln}\left({x}\right){dx}=\partial_{{a}=\mathrm{2}{n}+\mathrm{1}} \int_{\mathrm{0}} ^{\infty} {x}^{{a}} {e}^{−{x}} {dx} \\ $$$$=\partial_{{a}=\mathrm{2}{n}+\mathrm{1}} \Gamma\left({a}+\mathrm{1}\right)=\Gamma'\left(\mathrm{2}{n}+\mathrm{1}\right)=\left(\mathrm{2}{n}\right)!\Psi\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)}\Psi\left(\mathrm{2}{n}+\mathrm{1}\right){etcka} \\ $$$$\Psi\left(\mathrm{2}{n}+\mathrm{1}\right)={H}_{\mathrm{2}{n}+\mathrm{1}} −\gamma \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{n}} \frac{{H}_{\mathrm{2}{n}+\mathrm{1}} −\gamma}{\mathrm{2}{n}+\mathrm{1}}=−\gamma\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}\right)+\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} {H}_{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\underset{{n}\geqslant\mathrm{1}} {\sum}{x}^{{n}} {H}_{{n}} =−\frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}} \\ $$$$\Rightarrow\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}{H}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)….{E} \\ $$$${E}={i}\Rightarrow\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{i}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}{H}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\mathrm{1}−{i}\right)\right)^{\mathrm{2}} \\ $$$${We}\:{tack}\:{ImaginrayPart}\Rightarrow{Im}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{{i}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}{H}_{{n}} =\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} {H}_{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$={Im}\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\mathrm{1}−{i}\right)\right)^{\mathrm{2}} ={Im}\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\sqrt{\mathrm{2}}\right)+{i}.\left(−\frac{\pi}{\mathrm{4}}\right)\right)^{\mathrm{2}} \\ $$$$=−\frac{\pi}{\mathrm{4}}{ln}\left(\sqrt{\mathrm{2}}\right)=−\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right) \\ $$$$\Leftrightarrow=−\frac{\pi}{\mathrm{4}}\gamma−\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right)=−\frac{\pi}{\mathrm{8}}\left(\mathrm{2}\gamma+{ln}\left(\mathrm{2}\right)\right) \\ $$$$ \\ $$

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