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Question Number 219911 by Nicholas666 last updated on 03/May/25

∫_( 0) ^( ∞) (Σ_(n≥1) ((sin(2πnx))/n))(dx/x^(s+1) )

$$ \\ $$$$\:\:\:\:\int_{\:\mathrm{0}} ^{\:\infty} \left(\underset{{n}\geqslant\mathrm{1}} {\sum}\:\frac{{sin}\left(\mathrm{2}\pi{nx}\right)}{{n}}\right)\frac{{dx}}{{x}^{{s}+\mathrm{1}} } \\ $$$$ \\ $$

Commented by MrGaster last updated on 03/May/25

Σ_(n≥1) ^((?)) Only the lower limit is givenb ut the upper limit is missing?

$$\underset{{n}\geqslant\mathrm{1}} {\overset{\left(?\right)} {\sum}}\:\mathrm{Only}\:\mathrm{the}\:\mathrm{lower}\:\mathrm{limit}\:\mathrm{is}\:\mathrm{givenb} \\ $$$$\mathrm{ut}\:\mathrm{the}\:\mathrm{upper}\:\mathrm{limit}\:\mathrm{is}\:\mathrm{missing}? \\ $$

Answered by MrGaster last updated on 03/May/25

if ∃(Σ_(n≥1) ^∞ ((sin(2πnx))/n))(dx/x^(s+1) )⇒=Σ_(n=1) ^∞ (1/n)∫_0 ^∞ ((sin(2πnx))/x^(s+1) )dx =^(t=2πnx) Σ_(n=1) ^∞ (((2πn)^s )/n)∫_0 ^∞ ((sin t)/t^(s+1) )dt =(2π)^s ζ(1−s)∫_0 ^∞ ((sin t)/t^(s+1) )dt =(2π)^s ζ(1−s)∙(π/(2Γ(s+1)sin(((πs)/2)))) =(((2π)^s πζ(1−s))/(2Γ(s+1)sin(((πs)/2)))) =(((2π)^s πζ(1−s))/(2^1 Γ(s+1)))∙(1/(sin(((πs)/2)))) =(((2π)^s Γ(1−s)ζ(1−s))/2)∙(π/(Γ(s+1)sin(((πs)/2))Γ(1−s))) =(((2π)^s ζ(1−s))/2)∙(π/(Γ(s+1)sin(((πs)/2))Γ(1−s))) =(((2π)^s πζ(1−s))/2)∙(π/(Γ(s+1)sin(((πs)/2)))) =(((2π)^s ζ(1−s))/2)∙((Γ((s/2)))/(Γ(((1−s)/2)))) =(((2π)^s ζ(1−s)Γ((s/2)))/(2Γ(((1−s)/2)))) =((π^(1−s) ζ(1−s))/(2 sin(((πs)/2))))

$$\mathrm{if}\:\exists\left(\underset{{n}\geqslant\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left(\mathrm{2}\pi{nx}\right)}{{n}}\right)\frac{{dx}}{{x}^{{s}+\mathrm{1}} }\Rightarrow=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left(\mathrm{2}\pi{nx}\right)}{{x}^{{s}+\mathrm{1}} }{dx} \\ $$$$\overset{{t}=\mathrm{2}\pi{nx}} {=}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2}\pi{n}\right)^{{s}} }{{n}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\:{t}}{{t}^{{s}+\mathrm{1}} }{dt} \\ $$$$=\left(\mathrm{2}\pi\right)^{{s}} \zeta\left(\mathrm{1}−{s}\right)\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\:{t}}{{t}^{{s}+\mathrm{1}} }{dt} \\ $$$$=\left(\mathrm{2}\pi\right)^{{s}} \zeta\left(\mathrm{1}−{s}\right)\centerdot\frac{\pi}{\mathrm{2}\Gamma\left({s}+\mathrm{1}\right)\mathrm{sin}\left(\frac{\pi{s}}{\mathrm{2}}\right)} \\ $$$$=\frac{\left(\mathrm{2}\pi\right)^{{s}} \pi\zeta\left(\mathrm{1}−{s}\right)}{\mathrm{2}\Gamma\left({s}+\mathrm{1}\right)\mathrm{sin}\left(\frac{\pi{s}}{\mathrm{2}}\right)} \\ $$$$=\frac{\left(\mathrm{2}\pi\right)^{{s}} \pi\zeta\left(\mathrm{1}−{s}\right)}{\mathrm{2}^{\mathrm{1}} \Gamma\left({s}+\mathrm{1}\right)}\centerdot\frac{\mathrm{1}}{\mathrm{sin}\left(\frac{\pi{s}}{\mathrm{2}}\right)} \\ $$$$=\frac{\left(\mathrm{2}\pi\right)^{{s}} \Gamma\left(\mathrm{1}−{s}\right)\zeta\left(\mathrm{1}−{s}\right)}{\mathrm{2}}\centerdot\frac{\pi}{\Gamma\left({s}+\mathrm{1}\right)\mathrm{sin}\left(\frac{\pi{s}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−{s}\right)} \\ $$$$=\frac{\left(\mathrm{2}\pi\right)^{{s}} \zeta\left(\mathrm{1}−{s}\right)}{\mathrm{2}}\centerdot\frac{\pi}{\Gamma\left({s}+\mathrm{1}\right)\mathrm{sin}\left(\frac{\pi{s}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−{s}\right)} \\ $$$$=\frac{\left(\mathrm{2}\pi\right)^{{s}} \pi\zeta\left(\mathrm{1}−{s}\right)}{\mathrm{2}}\centerdot\frac{\pi}{\Gamma\left({s}+\mathrm{1}\right)\mathrm{sin}\left(\frac{\pi{s}}{\mathrm{2}}\right)} \\ $$$$=\frac{\left(\mathrm{2}\pi\right)^{{s}} \zeta\left(\mathrm{1}−{s}\right)}{\mathrm{2}}\centerdot\frac{\Gamma\left(\frac{{s}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{1}−{s}}{\mathrm{2}}\right)} \\ $$$$=\frac{\left(\mathrm{2}\pi\right)^{{s}} \zeta\left(\mathrm{1}−{s}\right)\Gamma\left(\frac{{s}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{\mathrm{1}−{s}}{\mathrm{2}}\right)} \\ $$$$=\frac{\pi^{\mathrm{1}−{s}} \zeta\left(\mathrm{1}−{s}\right)}{\mathrm{2}\:\mathrm{sin}\left(\frac{\pi{s}}{\mathrm{2}}\right)} \\ $$$$ \\ $$

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