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Question Number 157003 by MathSh last updated on 18/Oct/21

let n∈Z^+ shov that ∫_( 0) ^( ∞) ((sin(x^(-n) )ln(x))/x) dx = ((π𝛄)/(2n^2 )) where 𝛄 is the Euler-Mascheroni constan

$$\mathrm{let}\:\:\boldsymbol{\mathrm{n}}\in\mathbb{Z}^{+} \\ $$$$\mathrm{shov}\:\mathrm{that}\:\:\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{sin}\left(\mathrm{x}^{-\boldsymbol{\mathrm{n}}} \right)\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{x}}\:\mathrm{dx}\:=\:\frac{\pi\boldsymbol{\gamma}}{\mathrm{2n}^{\mathrm{2}} }\: \\ $$$$\mathrm{where}\:\:\boldsymbol{\gamma}\:\:\mathrm{is}\:\mathrm{the}\:\mathrm{Euler}-\mathrm{Mascheroni}\:\mathrm{constan}\: \\ $$

Answered by mindispower last updated on 18/Oct/21

∫_0 ^∞ ((sin(x^n )ln(x))/x)dx=Δ =∫_0 ^∞ ((sin(x^n )ln(x^n ))/(nx^n )).x^(n−1) dx=(1/n^2 )∫_0 ^∞ ((sin(x^n )ln(x^n ))/x^n )dx^n =(1/n^2 )∫_0 ^∞ ((sin(t))/t)ln(t) f(a)=∫_0 ^∞ sin(t)t^a dt=Im∫_0 ^∞ e^(it) t^a dt =Im.∫_0 ^(i∞) e^(−t) (it)^a .idt =Imi^(a+1) ∫_0 ^∞ t^a e^(−t) dt=sin(((a+1)/2)π)Γ(1+a) Δ=(1/n^2 )f′(−1)=(π/(2n^2 ))lim_(a→0) (cos((a/2)π)Γ(a)+Γ′(a)a) Γ′(a)=Ψ(a)Γ(a) =lim_(a→0) (π/(2n^2 ))(Γ(a)+Ψ(a)Γ(a)a) Ψ(1+a)=Ψ(a)+(1/a)⇒aΨ(a)=aΨ(1+a)−1 =lim_(a→0) (π/(2n^2 ))(Γ(a)+aΨ(1+a)Γ(a)−Γ(a)) =lim_(a→0) .(π/(2n^2 ))(Γ(1+a)Ψ(1+a))=(π/(2n^2 ))Ψ(1)=((πγ)/(2n^2 ))

$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}^{{n}} \right){ln}\left({x}\right)}{{x}}{dx}=\Delta \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}^{{n}} \right){ln}\left({x}^{{n}} \right)}{{nx}^{{n}} }.{x}^{\boldsymbol{{n}}−\mathrm{1}} \boldsymbol{{dx}}=\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}^{{n}} \right){ln}\left({x}^{{n}} \right)}{{x}^{{n}} }{dx}^{{n}} \\ $$$$=\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({t}\right)}{{t}}{ln}\left({t}\right) \\ $$$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} {sin}\left({t}\right){t}^{{a}} {dt}={Im}\int_{\mathrm{0}} ^{\infty} {e}^{{it}} {t}^{{a}} {dt} \\ $$$$={Im}.\int_{\mathrm{0}} ^{{i}\infty} {e}^{−{t}} \left({it}\right)^{{a}} .{idt} \\ $$$$={Imi}^{{a}+\mathrm{1}} \int_{\mathrm{0}} ^{\infty} {t}^{{a}} {e}^{−{t}} {dt}={sin}\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\pi\right)\Gamma\left(\mathrm{1}+{a}\right) \\ $$$$\Delta=\frac{\mathrm{1}}{{n}^{\mathrm{2}} }{f}'\left(−\mathrm{1}\right)=\frac{\pi}{\mathrm{2}{n}^{\mathrm{2}} }\underset{{a}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({cos}\left(\frac{{a}}{\mathrm{2}}\pi\right)\Gamma\left({a}\right)+\Gamma'\left({a}\right){a}\right) \\ $$$$\Gamma'\left({a}\right)=\Psi\left({a}\right)\Gamma\left({a}\right) \\ $$$$=\underset{{a}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\pi}{\mathrm{2}{n}^{\mathrm{2}} }\left(\Gamma\left({a}\right)+\Psi\left({a}\right)\Gamma\left({a}\right){a}\right) \\ $$$$\Psi\left(\mathrm{1}+{a}\right)=\Psi\left({a}\right)+\frac{\mathrm{1}}{{a}}\Rightarrow{a}\Psi\left({a}\right)={a}\Psi\left(\mathrm{1}+{a}\right)−\mathrm{1} \\ $$$$=\underset{{a}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\pi}{\mathrm{2}{n}^{\mathrm{2}} }\left(\Gamma\left({a}\right)+{a}\Psi\left(\mathrm{1}+{a}\right)\Gamma\left({a}\right)−\Gamma\left({a}\right)\right) \\ $$$$=\underset{{a}\rightarrow\mathrm{0}} {\mathrm{lim}}.\frac{\pi}{\mathrm{2}{n}^{\mathrm{2}} }\left(\Gamma\left(\mathrm{1}+{a}\right)\Psi\left(\mathrm{1}+{a}\right)\right)=\frac{\pi}{\mathrm{2}{n}^{\mathrm{2}} }\Psi\left(\mathrm{1}\right)=\frac{\pi\gamma}{\mathrm{2}{n}^{\mathrm{2}} } \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by MathSh last updated on 18/Oct/21

Perfect dear Ser, thank you

$$\mathrm{Perfect}\:\mathrm{dear}\:\mathrm{Ser},\:\mathrm{thank}\:\mathrm{you} \\ $$

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