Question Number 197380 by mokys last updated on 15/Sep/23
Commented by witcher3 last updated on 19/Sep/23
$$\mathrm{verry}\:\mathrm{Nice}\:\mathrm{one} \\ $$$$\mathrm{nice}\:\mathrm{Result} \\ $$
Commented by mokys last updated on 19/Sep/23
$${thank}\:{you}\:{sir}\:{i}\:{think}\:{by}\:{resideo}\:{is}\:{easy}\: \\ $$
Commented by witcher3 last updated on 20/Sep/23
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcom}\:\mathrm{what}\:\mathrm{is}\:\mathrm{this}\:\mathrm{files} \\ $$$$\mathrm{somme}\:\mathrm{nice}\:\mathrm{quations} \\ $$
Commented by mokys last updated on 20/Sep/23
$${this}\:{is}\:{files}\:{is}\:{very}\:{old}\: \\ $$
Commented by witcher3 last updated on 20/Sep/23
$$\mathrm{you}\:\mathrm{have}\:\mathrm{The}\:\mathrm{nam}\:\mathrm{Please} \\ $$
Answered by witcher3 last updated on 19/Sep/23
![(1/(ln^2 (x)+π^2 ))=(1/π)Re(1/(π+iln(x))) =(1/π)Re∫_0 ^∞ e^(−(π+iln(x))t) dt f(a)=∫_0 ^∞ (dx/((x^2 +a^2 )(ln^2 (x)+π^2 ))) =(1/π)Re∫_0 ^∞ ∫_0 ^∞ (e^(−(π+iln(x))t) /(x^2 +a^2 ))dxdt=((Re)/π)∫_0 ^∞ e^(−πt) ∫_0 ^∞ (x^(−it) /(x^2 +a^2 ))dxdt ∫_0 ^∞ (x^(−it) /(x^2 +a^2 ))dx=∫_0 ^∞ ((a^(−it) y^(−((it+1)/2)) )/(a^2 (1+y))).(dy/2)“x=a(√y)” =(a^(−it) /(2a))∫_0 ^∞ (y^(−((it+1)/2)) /(1+y))=(a^(−it) /(2a)).(π/(sin((π/2)(1−it)))) =((a^(−it) 2iπ)/(2a(ie^(π/2) +ie^(−(π/2)) )))=(π/a).(a^(−it) /(e^((π/2)t) +e^(−(π/2)t) )) f(a)=((Re)/π)∫_0 ^∞ e^(−πt) .((πa^(−it) )/(2a(e^((π/2)t) +e^(−(π/2)t) )))dt =(1/a)Re∫_0 ^∞ (e^(−t(π+(π/2)+iln(a))) /(1+e^(−πt) ))dt=(1/a)Re{g(a)} g(a)=∫_0 ^∞ Σ_(n≥0) (−1)^n e^(−t(((3π)/2)+nπ+iln(a))) dt =Σ_(n≥0) (((−1)^n )/(((3π)/2)+nπ+iln(a)))=Σ_(n≥0) ((1/(2nπ+((3π)/2)+iln(a)))−(1/(2nπ+((5π)/2)+iln(a)))) =(1/(2π))Σ_(n≥0) (1/(n+(3/4)+((iln(a))/(2π))))−(1/(n+(5/4)+i((ln(a))/(2π)))) Ψ(z)=−γ+Σ_(m≥0) (1/(m+1))−(1/(m+z)) g(a)=(1/(2π))(Ψ((5/4)+((iln(a))/(2π)))−Ψ((3/4)+((iln(a))/(2π)))) f(a)=(1/a).(1/(2π))Re(Ψ((5/4)+((iln(a))/(2π)))−Ψ((3/4)+i((ln(a))/(2π)))) =(1/(4πa))(Ψ((5/4)+((iln(a))/(2π)))+Ψ((5/4)−((iln(a))/(2π)))−Ψ((3/4)+((iln(a))/(2π)))−Ψ((3/4)−((iln(a))/(2π)))) =(1/(4πa))[(1/((1/4)+i((ln(a))/(2π))))+(1/((1/4)−((iln(a))/(2π))))+Ψ((1/4)+i((ln(a))/(2π)))+Ψ((1/4)−((iln(a))/(2π))) −Ψ(1−((1/4)−((iln(a))/(2π)))−Ψ(1−((1/4)+i((ln(a))/(2π))))] Ψ(1−z)−Ψ(z)=πcot(πz) f(a)=(π/(2a)).(1/(ln^2 (a)+(π^2 /4)))+(1/(4πa)){−cot(π((1/4)+((iln(a))/(2π)))−πcot(π((1/4)−((iln(a))/(2π))))} cot(πz)=i((e^(2πiz) +1)/(e^(2πiz) −1)) =(π/(2a)).(1/(ln^2 (a)+(π^2 /4)))−(i/(4a))(((e^((π/2)i−ln(a)) +1)/(e^(((πi)/2)−ln(a)) −1))+((e^(((iπ)/2)+ln(a)) +1)/(e^(((iπ)/2)+ln(a)) −1))) =(π/(2a)).(1/(ln^2 (a)+(π^2 /4)))−(i/(4a))((((i/a)+1)/((i/a)−1))+((ia+1)/(ia−1))) =(π/(2a)).(1/(ln^2 (a)+(π^2 /4)))−(i/(4a))(((1−ia)/(1+ia))+((1+ia)/(ia−1))) =(π/(2a))(1/(ln^2 (a)+(π^2 /4)))−(i/(4a))(((4ia)/(−1−a^2 ))) =(π/(2a)) (1/(ln^2 (a)+(π^2 /4)))−(1/(1+a^2 )) ∫_0 ^∞ (dx/((x^2 +a^2 )(ln^2 (x)+π^2 )))=(π/(2a)) (1/(ln^2 (a)+(π^2 /4)))−(1/(1+a^2 )) ∀a>0](https://www.tinkutara.com/question/Q197504.png)
$$\frac{\mathrm{1}}{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)+\pi^{\mathrm{2}} }=\frac{\mathrm{1}}{\pi}\mathrm{Re}\frac{\mathrm{1}}{\pi+\mathrm{iln}\left(\mathrm{x}\right)} \\ $$$$=\frac{\mathrm{1}}{\pi}\mathrm{Re}\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\left(\pi+\mathrm{iln}\left(\mathrm{x}\right)\right)\mathrm{t}} \mathrm{dt} \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \right)\left(\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)+\pi^{\mathrm{2}} \right)}\: \\ $$$$=\frac{\mathrm{1}}{\pi}\mathrm{Re}\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{e}^{−\left(\pi+\mathrm{iln}\left(\mathrm{x}\right)\right)\mathrm{t}} }{\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }\mathrm{dxdt}=\frac{\mathrm{Re}}{\pi}\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\pi\mathrm{t}} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{−\mathrm{it}} }{\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }\mathrm{dxdt} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{−\mathrm{it}} }{\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }\mathrm{dx}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{a}^{−\mathrm{it}} \mathrm{y}^{−\frac{\mathrm{it}+\mathrm{1}}{\mathrm{2}}} }{\mathrm{a}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{y}\right)}.\frac{\mathrm{dy}}{\mathrm{2}}“\mathrm{x}=\mathrm{a}\sqrt{\mathrm{y}}'' \\ $$$$=\frac{\mathrm{a}^{−\mathrm{it}} }{\mathrm{2a}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{y}^{−\frac{\mathrm{it}+\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}+\mathrm{y}}=\frac{\mathrm{a}^{−\mathrm{it}} }{\mathrm{2a}}.\frac{\pi}{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\mathrm{it}\right)\right)} \\ $$$$=\frac{\mathrm{a}^{−\mathrm{it}} \mathrm{2i}\pi}{\mathrm{2a}\left(\mathrm{ie}^{\frac{\pi}{\mathrm{2}}} +\mathrm{ie}^{−\frac{\pi}{\mathrm{2}}} \right)}=\frac{\pi}{\mathrm{a}}.\frac{\mathrm{a}^{−\mathrm{it}} }{\mathrm{e}^{\frac{\pi}{\mathrm{2}}\mathrm{t}} +\mathrm{e}^{−\frac{\pi}{\mathrm{2}}\mathrm{t}} } \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{\mathrm{Re}}{\pi}\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\pi\mathrm{t}} .\frac{\pi\mathrm{a}^{−\mathrm{it}} }{\mathrm{2a}\left(\mathrm{e}^{\frac{\pi}{\mathrm{2}}\mathrm{t}} +\mathrm{e}^{−\frac{\pi}{\mathrm{2}}\mathrm{t}} \right)}\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{a}}\mathrm{Re}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{e}^{−\mathrm{t}\left(\pi+\frac{\pi}{\mathrm{2}}+\mathrm{iln}\left(\mathrm{a}\right)\right)} }{\mathrm{1}+\mathrm{e}^{−\pi\mathrm{t}} }\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{a}}\mathrm{Re}\left\{\mathrm{g}\left(\mathrm{a}\right)\right\} \\ $$$$\mathrm{g}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{e}^{−\mathrm{t}\left(\frac{\mathrm{3}\pi}{\mathrm{2}}+\mathrm{n}\pi+\mathrm{iln}\left(\mathrm{a}\right)\right)} \mathrm{dt} \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\frac{\mathrm{3}\pi}{\mathrm{2}}+\mathrm{n}\pi+\mathrm{iln}\left(\mathrm{a}\right)}=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{\mathrm{2n}\pi+\frac{\mathrm{3}\pi}{\mathrm{2}}+\mathrm{iln}\left(\mathrm{a}\right)}−\frac{\mathrm{1}}{\mathrm{2n}\pi+\frac{\mathrm{5}\pi}{\mathrm{2}}+\mathrm{iln}\left(\mathrm{a}\right)}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\mathrm{n}+\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{iln}\left(\mathrm{a}\right)}{\mathrm{2}\pi}}−\frac{\mathrm{1}}{\mathrm{n}+\frac{\mathrm{5}}{\mathrm{4}}+\mathrm{i}\frac{\mathrm{ln}\left(\mathrm{a}\right)}{\mathrm{2}\pi}} \\ $$$$\Psi\left(\mathrm{z}\right)=−\gamma+\underset{\mathrm{m}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\mathrm{m}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{m}+\mathrm{z}} \\ $$$$\mathrm{g}\left(\mathrm{a}\right)=\frac{\mathrm{1}}{\mathrm{2}\pi}\left(\Psi\left(\frac{\mathrm{5}}{\mathrm{4}}+\frac{\mathrm{iln}\left(\mathrm{a}\right)}{\mathrm{2}\pi}\right)−\Psi\left(\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{iln}\left(\mathrm{a}\right)}{\mathrm{2}\pi}\right)\right) \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{\mathrm{1}}{\mathrm{a}}.\frac{\mathrm{1}}{\mathrm{2}\pi}\mathrm{Re}\left(\Psi\left(\frac{\mathrm{5}}{\mathrm{4}}+\frac{\mathrm{iln}\left(\mathrm{a}\right)}{\mathrm{2}\pi}\right)−\Psi\left(\frac{\mathrm{3}}{\mathrm{4}}+\mathrm{i}\frac{\mathrm{ln}\left(\mathrm{a}\right)}{\mathrm{2}\pi}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\pi\mathrm{a}}\left(\Psi\left(\frac{\mathrm{5}}{\mathrm{4}}+\frac{\mathrm{iln}\left(\mathrm{a}\right)}{\mathrm{2}\pi}\right)+\Psi\left(\frac{\mathrm{5}}{\mathrm{4}}−\frac{\mathrm{iln}\left(\mathrm{a}\right)}{\mathrm{2}\pi}\right)−\Psi\left(\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{iln}\left(\mathrm{a}\right)}{\mathrm{2}\pi}\right)−\Psi\left(\frac{\mathrm{3}}{\mathrm{4}}−\frac{\mathrm{iln}\left(\mathrm{a}\right)}{\mathrm{2}\pi}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\pi\mathrm{a}}\left[\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{i}\frac{\mathrm{ln}\left(\mathrm{a}\right)}{\mathrm{2}\pi}}+\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{iln}\left(\mathrm{a}\right)}{\mathrm{2}\pi}}+\Psi\left(\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{i}\frac{\mathrm{ln}\left(\mathrm{a}\right)}{\mathrm{2}\pi}\right)+\Psi\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{iln}\left(\mathrm{a}\right)}{\mathrm{2}\pi}\right)\right. \\ $$$$−\Psi\left(\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{iln}\left(\mathrm{a}\right)}{\mathrm{2}\pi}\right)−\Psi\left(\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{i}\frac{\mathrm{ln}\left(\mathrm{a}\right)}{\mathrm{2}\pi}\right)\right)\right] \\ $$$$\Psi\left(\mathrm{1}−\mathrm{z}\right)−\Psi\left(\mathrm{z}\right)=\pi\mathrm{cot}\left(\pi\mathrm{z}\right) \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi}{\mathrm{2a}}.\frac{\mathrm{1}}{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{a}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{4}}}+\frac{\mathrm{1}}{\mathrm{4}\pi\mathrm{a}}\left\{−\mathrm{cot}\left(\pi\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{iln}\left(\mathrm{a}\right)}{\mathrm{2}\pi}\right)−\pi\mathrm{cot}\left(\pi\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{iln}\left(\mathrm{a}\right)}{\mathrm{2}\pi}\right)\right)\right\}\right. \\ $$$$\mathrm{cot}\left(\pi\mathrm{z}\right)=\mathrm{i}\frac{\mathrm{e}^{\mathrm{2}\pi\mathrm{iz}} +\mathrm{1}}{\mathrm{e}^{\mathrm{2}\pi\mathrm{iz}} −\mathrm{1}} \\ $$$$=\frac{\pi}{\mathrm{2a}}.\frac{\mathrm{1}}{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{a}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{4}}}−\frac{\mathrm{i}}{\mathrm{4a}}\left(\frac{\mathrm{e}^{\frac{\pi}{\mathrm{2}}\mathrm{i}−\mathrm{ln}\left(\mathrm{a}\right)} +\mathrm{1}}{\mathrm{e}^{\frac{\pi\mathrm{i}}{\mathrm{2}}−\mathrm{ln}\left(\mathrm{a}\right)} −\mathrm{1}}+\frac{\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{2}}+\mathrm{ln}\left(\mathrm{a}\right)} +\mathrm{1}}{\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{2}}+\mathrm{ln}\left(\mathrm{a}\right)} −\mathrm{1}}\right) \\ $$$$=\frac{\pi}{\mathrm{2a}}.\frac{\mathrm{1}}{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{a}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{4}}}−\frac{\mathrm{i}}{\mathrm{4a}}\left(\frac{\frac{\mathrm{i}}{\mathrm{a}}+\mathrm{1}}{\frac{\mathrm{i}}{\mathrm{a}}−\mathrm{1}}+\frac{\mathrm{ia}+\mathrm{1}}{\mathrm{ia}−\mathrm{1}}\right) \\ $$$$=\frac{\pi}{\mathrm{2a}}.\frac{\mathrm{1}}{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{a}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{4}}}−\frac{\mathrm{i}}{\mathrm{4a}}\left(\frac{\mathrm{1}−\mathrm{ia}}{\mathrm{1}+\mathrm{ia}}+\frac{\mathrm{1}+\mathrm{ia}}{\mathrm{ia}−\mathrm{1}}\right) \\ $$$$=\frac{\pi}{\mathrm{2a}}\frac{\mathrm{1}}{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{a}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{4}}}−\frac{\mathrm{i}}{\mathrm{4a}}\left(\frac{\mathrm{4ia}}{−\mathrm{1}−\mathrm{a}^{\mathrm{2}} }\right) \\ $$$$=\frac{\pi}{\mathrm{2a}}\:\:\frac{\mathrm{1}}{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{a}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{4}}}−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{a}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \right)\left(\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)+\pi^{\mathrm{2}} \right)}=\frac{\pi}{\mathrm{2a}}\:\frac{\mathrm{1}}{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{a}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{4}}}−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{a}^{\mathrm{2}} } \\ $$$$\forall\mathrm{a}>\mathrm{0} \\ $$$$ \\ $$
Commented by mokys last updated on 19/Sep/23
$${thank}\:{you}\:{sir} \\ $$
Commented by witcher3 last updated on 19/Sep/23
$$\mathrm{withe}\:\mathrm{Pleasur} \\ $$