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Question Number 132877 by mnjuly1970 last updated on 17/Feb/21

.... nice calculus... prove that : 𝛗=∫_0 ^( ∞) ((sin(x).log^2 (x))/x) =(π/(24))(12γ^( 2) +π^2 ) .....

$$\:\:\:\:\:\:\:\:\:\:\:\:\:....\:{nice}\:\:{calculus}... \\ $$$$\:\:\:{prove}\:{that}\:: \\ $$$$\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({x}\right).{log}^{\mathrm{2}} \left({x}\right)}{{x}} \\ $$$$\:\:\:\:\:\:\:=\frac{\pi}{\mathrm{24}}\left(\mathrm{12}\gamma^{\:\mathrm{2}} +\pi^{\mathrm{2}} \right)\:..... \\ $$

Answered by Dwaipayan Shikari last updated on 17/Feb/21

φ(α)=∫_0 ^∞ ((sinx)/x^α )dx φ(α)=(1/(Γ(α)))∫_0 ^∞ ∫_0 ^∞ t^(α−1) e^(−xt) sin(x)dtdx =(1/(2iΓ(α)))∫_0 ^∞ ∫_0 ^∞ t^(α−1) e^(−(t−i)x) −t^(α−1) e^(−(t+i)x) dx =(1/(2iΓ(α)))∫_0 ^∞ (t^(α−1) /(t−i))−(t^(α−1) /(t+i))dt =(1/(Γ(α)))∫_0 ^∞ (t^(α−1) /(t^2 +1))dt=(1/(2Γ(α)))∫_0 ^∞ (u^((α/2)−1) /((u+1)^(1−(α/2)+(α/2)) ))du =(1/(2Γ(α))).((Γ((α/2))Γ(1−(α/2)))/(Γ(1)))=(π/(2Γ(α)sin(((πα)/2)))) φ′′(α)∣_(α=1) =∫_0 ^∞ ((sin(x)log^2 (x))/x)dx= (π/(24))(12γ^2 +π^2 )

$$\phi\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}^{\alpha} }{dx} \\ $$$$\phi\left(\alpha\right)=\frac{\mathrm{1}}{\Gamma\left(\alpha\right)}\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} {t}^{\alpha−\mathrm{1}} {e}^{−{xt}} {sin}\left({x}\right){dtdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}\Gamma\left(\alpha\right)}\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} {t}^{\alpha−\mathrm{1}} {e}^{−\left({t}−{i}\right){x}} −{t}^{\alpha−\mathrm{1}} {e}^{−\left({t}+{i}\right){x}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}\Gamma\left(\alpha\right)}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\alpha−\mathrm{1}} }{{t}−{i}}−\frac{{t}^{\alpha−\mathrm{1}} }{{t}+{i}}{dt} \\ $$$$=\frac{\mathrm{1}}{\Gamma\left(\alpha\right)}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\alpha−\mathrm{1}} }{{t}^{\mathrm{2}} +\mathrm{1}}{dt}=\frac{\mathrm{1}}{\mathrm{2}\Gamma\left(\alpha\right)}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\frac{\alpha}{\mathrm{2}}−\mathrm{1}} }{\left({u}+\mathrm{1}\right)^{\mathrm{1}−\frac{\alpha}{\mathrm{2}}+\frac{\alpha}{\mathrm{2}}} }{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\Gamma\left(\alpha\right)}.\frac{\Gamma\left(\frac{\alpha}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−\frac{\alpha}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{1}\right)}=\frac{\pi}{\mathrm{2}\Gamma\left(\alpha\right){sin}\left(\frac{\pi\alpha}{\mathrm{2}}\right)} \\ $$$$\phi''\left(\alpha\right)\mid_{\alpha=\mathrm{1}} =\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}\right){log}^{\mathrm{2}} \left({x}\right)}{{x}}{dx}=\:\:\frac{\pi}{\mathrm{24}}\left(\mathrm{12}\gamma^{\mathrm{2}} +\pi^{\mathrm{2}} \right) \\ $$

Commented by Dwaipayan Shikari last updated on 17/Feb/21

φ′(α)=−((π^2 /(4Γ(α))).((cos((π/2)α))/(sin^2 ((π/2)α)))+(((π^2 /2)Γ(α).((cos((π/2)α))/(sin^2 ((π/2)α))))/(2Γ^2 (α)))+((πΓ′(α)cosec((π/2)2α))/(2Γ^2 (α)))) φ′(1)=((πγ)/2)

$$\phi'\left(\alpha\right)=−\left(\frac{\pi^{\mathrm{2}} }{\mathrm{4}\Gamma\left(\alpha\right)}.\frac{{cos}\left(\frac{\pi}{\mathrm{2}}\alpha\right)}{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}\alpha\right)}+\frac{\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\Gamma\left(\alpha\right).\frac{{cos}\left(\frac{\pi}{\mathrm{2}}\alpha\right)}{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}\alpha\right)}}{\mathrm{2}\Gamma^{\mathrm{2}} \left(\alpha\right)}+\frac{\pi\Gamma'\left(\alpha\right){cosec}\left(\frac{\pi}{\mathrm{2}}\mathrm{2}\alpha\right)}{\mathrm{2}\Gamma^{\mathrm{2}} \left(\alpha\right)}\right) \\ $$$$\phi'\left(\mathrm{1}\right)=\frac{\pi\gamma}{\mathrm{2}} \\ $$

Commented by mnjuly1970 last updated on 17/Feb/21

with thanking sir payan...

$${with}\:{thanking}\:{sir}\:{payan}... \\ $$$$ \\ $$

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