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Question Number 138041 by mnjuly1970 last updated on 09/Apr/21
.......advanced ... ... ... calculus...... prove that ::: 𝛗=∫_0 ^( 1) ψ(x).sin(2πx)dx=−(π/2) ...✓
$$\:\:\:\:\:\:\:\:\:\:\:…….{advanced}\:…\:…\:…\:{calculus}…… \\ $$$$\:\:{prove}\:\:{that}\:::: \\ $$$$\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \psi\left({x}\right).{sin}\left(\mathrm{2}\pi{x}\right){dx}=−\frac{\pi}{\mathrm{2}}\:…\checkmark \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 09/Apr/21
∫_0 ^1 ψ(x)sin(2πx)dx =[sin(2πx)log(Γ(x))]_0 ^1 −2π∫_0 ^1 log(Γ(x))cos(2πx) =−π∫_0 ^1 log(π)cos(2πx)+π∫_0 ^1 log(sin(πx))cos(2πx)dx =∫_0 ^π log(sin(u))cos(2u)du =[(1/2)log(sin(u))sin(2u)]_0 ^π −(1/2)∫_0 ^π ((sin(2u)cos(u))/(sin(u)))du =−(1/2)∫_0 ^π 1+cos(2u)du=−(π/2)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \psi\left({x}\right){sin}\left(\mathrm{2}\pi{x}\right){dx} \\ $$$$=\left[{sin}\left(\mathrm{2}\pi{x}\right){log}\left(\Gamma\left({x}\right)\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\mathrm{2}\pi\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left(\Gamma\left({x}\right)\right){cos}\left(\mathrm{2}\pi{x}\right) \\ $$$$=−\pi\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left(\pi\right){cos}\left(\mathrm{2}\pi{x}\right)+\pi\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left({sin}\left(\pi{x}\right)\right){cos}\left(\mathrm{2}\pi{x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi} {log}\left({sin}\left({u}\right)\right){cos}\left(\mathrm{2}{u}\right){du} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{2}}{log}\left({sin}\left({u}\right)\right){sin}\left(\mathrm{2}{u}\right)\right]_{\mathrm{0}} ^{\pi} −\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \frac{{sin}\left(\mathrm{2}{u}\right){cos}\left({u}\right)}{{sin}\left({u}\right)}{du} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \mathrm{1}+{cos}\left(\mathrm{2}{u}\right){du}=−\frac{\pi}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 09/Apr/21
verynice mr payan.. thank you so much..
$${verynice}\:{mr}\: \\ $$$${payan}..\:{thank}\:{you}\:{so}\:{much}.. \\ $$
Answered by mnjuly1970 last updated on 09/Apr/21
𝛗=∫_0 ^( 1) ψ(x).sin(2πx)dx 𝛗=−∫_0 ^( 1) ψ(1−x).sin(2πx)dx 2𝛗=∫_0 ^( 1) {(ψ(x)−ψ(1−x)}sin(2πx)dx =−π∫_0 ^( 1) cot(πx).sin(2πx)dx =−2π∫_0 ^( 1) ((cos(πx))/(sin(πx))){sin(πx).cos(πx)}dx ∴ 𝛗=−π∫_0 ^( 1) ((1+cos(2πx))/2)dx =−(π/2) .....✓
$$\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \psi\left({x}\right).{sin}\left(\mathrm{2}\pi{x}\right){dx} \\ $$$$\:\:\:\:\boldsymbol{\phi}=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \psi\left(\mathrm{1}−{x}\right).{sin}\left(\mathrm{2}\pi{x}\right){dx} \\ $$$$\:\:\:\:\mathrm{2}\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left\{\left(\psi\left({x}\right)−\psi\left(\mathrm{1}−{x}\right)\right\}{sin}\left(\mathrm{2}\pi{x}\right){dx}\right. \\ $$$$\:\:\:\:\:\:\:\:\:=−\pi\int_{\mathrm{0}} ^{\:\mathrm{1}} {cot}\left(\pi{x}\right).{sin}\left(\mathrm{2}\pi{x}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:=−\mathrm{2}\pi\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{cos}\left(\pi{x}\right)}{{sin}\left(\pi{x}\right)}\left\{{sin}\left(\pi{x}\right).{cos}\left(\pi{x}\right)\right\}{dx} \\ $$$$\:\:\:\:\:\:\therefore\:\boldsymbol{\phi}=−\pi\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}+{cos}\left(\mathrm{2}\pi{x}\right)}{\mathrm{2}}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{\pi}{\mathrm{2}}\:…..\checkmark \\ $$

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