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Question Number 212001 by mnjuly1970 last updated on 26/Sep/24

prove that: Σ_(k∈Z) (( (−1)^k )/( x + kπ)) = (1/(sin(x))) −−−−−−−−−

provethat:∑k∈Z(−1)kx+kπ=1sin(x)−−−−−−−−−

Answered by Berbere last updated on 27/Sep/24

Σ_(k∈Z) ∫_0 ^1 (−1)^k t^(x−1+kπ) dt=Σ_(k∈Z) (((−1)^k )/(x+kπ))=S(x) S(x)=S_1 (x)−S_1 (−x)+(1/x) S_1 (x)=Σ_(k≥0) (((−1)^k )/(x+kπ))=Σ_(k≥0) ∫_0 ^1 (−t^π )^k t^(x−1) dt=∫_0 ^1 (t^(x−1) /(1+t^π ))dt =∫_0 ^1 ((t^(x−1) (1−t^π ))/(1−t^(2π) ))dt=∫_0 ^1 ((z^((x−1)/(2π)) (1−z^(1/2) ))/(1−z)).z^((1/(2π))−1) (dz/((2π))) =(1/((2π)))∫_0 ^1 (z^((x/(2π))−1) −z^((x/(2π))−(1/2)) )(1−z)^(−1) dz Ψ(1+z)=−γ+∫_0 ^1 ((1−t^(z−1) )/(1−t))dt =(1/((2π)))[Ψ((x/(2π))+(1/2))−Ψ((x/(2π)))] S=(1/((2π) ))[Ψ((x/(2π))+(1/2))−Ψ((x/(2π)))−Ψ((1/2)−(x/(2π)))+Ψ(−(x/(2π)))] =(1/(2π))π[cot(π((x/(2π))+(1/2))−[Ψ(1+(x/(2π)))−((2π)/x)]+Ψ(−(x/(2π)))] =(1/(2π))[−((2π)/x)+πcot((x/2)+(π/2))+πcot(π+(x/2))] =−(1/x)+(1/2)[((sin((x/(2())))/(cos((x/2))))+((cos((x/2)))/(sin((x/2))))]=−(1/x)+(1/(2cos((x/2))sin((x/2)))) =−(1/x)+(1/(sin(x))) S=S1−S_1 (−x)+(1/x)+(1/(sin(x)))=(1/(sin(x)))

∑k∈Z∫01(−1)ktx−1+kπdt=∑k∈Z(−1)kx+kπ=S(x)S(x)=S1(x)−S1(−x)+1xS1(x)=∑k⩾0(−1)kx+kπ=∑k⩾0∫01(−tπ)ktx−1dt=∫01tx−11+tπdt=∫01tx−1(1−tπ)1−t2πdt=∫01zx−12π(1−z12)1−z.z12π−1dz(2π)=1(2π)∫01(zx2π−1−zx2π−12)(1−z)−1dzΨ(1+z)=−γ+∫011−tz−11−tdt=1(2π)[Ψ(x2π+12)−Ψ(x2π)]S=1(2π)[Ψ(x2π+12)−Ψ(x2π)−Ψ(12−x2π)+Ψ(−x2π)]=12ππ[cot(π(x2π+12)−[Ψ(1+x2π)−2πx]+Ψ(−x2π)]=12π[−2πx+πcot(x2+π2)+πcot(π+x2)]=−1x+12[sin(x2()cos(x2)+cos(x2)sin(x2)]=−1x+12cos(x2)sin(x2)=−1x+1sin(x)S=S1−S1(−x)+1x+1sin(x)=1sin(x)

Commented by mnjuly1970 last updated on 27/Sep/24

thanks alot Sir...

thanksalotSir...

Commented by Berbere last updated on 27/Sep/24

Withe Pleasur

WithePleasur

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