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Question-190700




Question Number 190700 by Rupesh123 last updated on 09/Apr/23
Answered by 07049753053 last updated on 09/Apr/23
∫_0 ^∞ ((sin(x)ln(x))/x)dx  (d/da)∣_(a=1) ∫_0 ^∞ ((sin(x)x^a )/x)dx=(d/da)∣_(a=1) ∫_0 ^∞ sin(x)x^(a−1) dx  from euler formula  e^(−ix) =cos(x)+sin(x)  sin(x)=Im(e^(−ix) )  (d/da)∣_(a=1) Im∫_0 ^∞ e^(−ix) x^(a−1) dx  let ix=u x=(u/i) dx=(du/i)  (d/da)∣_(a=1 ) Im∫_0 ^∞ e^(−u) ((u/i))^(a−1) (du/i)  (d/da)∣_(a=1)  Im((1/i^a ))∫_0 ^∞ e^(−u) u^(a−1) du  (d/da)∣_(a=1)  Im((1/i^a ))∫_0 ^∞ e^(−u) u^((a−1+1)−1) du  (d/da)∣_(a=1)  Im(((𝚪(a))/i^a ))=−(d/da)∣_(a=1) (−𝚪(a)sin(((𝛑a)/2)))=−(1/2)𝚪(a)(𝛑cos(((𝛑a)/2))+2sin(((𝛑a)/2))𝛙(a))∣_(a=1)   −(1/2)𝚪(1)(𝛑cos((𝛑/2))+2𝛑sin((𝛑/2))𝛙(1))=−((𝛄𝛑)/2)
$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}\right){ln}\left({x}\right)}{{x}}{dx} \\ $$$$\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\mid_{\boldsymbol{\mathrm{a}}=\mathrm{1}} \int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{a}}} }{\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{dx}}=\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\mid_{\boldsymbol{\mathrm{a}}=\mathrm{1}} \int_{\mathrm{0}} ^{\infty} \boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{a}}−\mathrm{1}} \boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{from}}\:\boldsymbol{\mathrm{euler}}\:\boldsymbol{\mathrm{formula}} \\ $$$$\boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{ix}}} =\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)+\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right) \\ $$$$\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)=\boldsymbol{\mathcal{I}{m}}\left(\boldsymbol{{e}}^{−\boldsymbol{\mathrm{ix}}} \right) \\ $$$$\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\mid_{\boldsymbol{\mathrm{a}}=\mathrm{1}} \boldsymbol{\mathcal{I}{m}}\int_{\mathrm{0}} ^{\infty} \boldsymbol{{e}}^{−\boldsymbol{\mathrm{ix}}} \boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{a}}−\mathrm{1}} \boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{let}}\:\boldsymbol{\mathrm{ix}}=\boldsymbol{\mathrm{u}}\:\boldsymbol{\mathrm{x}}=\frac{\boldsymbol{\mathrm{u}}}{\boldsymbol{\mathrm{i}}}\:\boldsymbol{\mathrm{dx}}=\frac{\boldsymbol{\mathrm{du}}}{\boldsymbol{\mathrm{i}}} \\ $$$$\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\mid_{\boldsymbol{\mathrm{a}}=\mathrm{1}\:} \boldsymbol{\mathcal{I}{m}}\int_{\mathrm{0}} ^{\infty} \boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{u}}} \left(\frac{\boldsymbol{\mathrm{u}}}{\boldsymbol{\mathrm{i}}}\right)^{\boldsymbol{\mathrm{a}}−\mathrm{1}} \frac{\boldsymbol{\mathrm{du}}}{\boldsymbol{\mathrm{i}}} \\ $$$$\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\mid_{\boldsymbol{{a}}=\mathrm{1}} \:\boldsymbol{\mathcal{I}{m}}\left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{i}}^{{a}} }\right)\int_{\mathrm{0}} ^{\infty} \boldsymbol{{e}}^{−\boldsymbol{{u}}} \boldsymbol{\mathrm{u}}^{\boldsymbol{\mathrm{a}}−\mathrm{1}} \boldsymbol{\mathrm{du}} \\ $$$$\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\mid_{\boldsymbol{{a}}=\mathrm{1}} \:\boldsymbol{\mathcal{I}{m}}\left(\frac{\mathrm{1}}{\boldsymbol{{i}}^{{a}} }\right)\int_{\mathrm{0}} ^{\infty} \boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{u}}} \boldsymbol{\mathrm{u}}^{\left(\boldsymbol{\mathrm{a}}−\mathrm{1}+\mathrm{1}\right)−\mathrm{1}} \boldsymbol{\mathrm{du}} \\ $$$$\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\mid_{\boldsymbol{{a}}=\mathrm{1}} \:\boldsymbol{\mathcal{I}{m}}\left(\frac{\boldsymbol{\Gamma}\left({a}\right)}{{i}^{{a}} }\right)=−\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\mid_{\boldsymbol{{a}}=\mathrm{1}} \left(−\boldsymbol{\Gamma}\left(\boldsymbol{{a}}\right)\boldsymbol{{sin}}\left(\frac{\boldsymbol{\pi{a}}}{\mathrm{2}}\right)\right)=−\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\Gamma}\left(\boldsymbol{{a}}\right)\left(\boldsymbol{\pi{cos}}\left(\frac{\boldsymbol{\pi{a}}}{\mathrm{2}}\right)+\mathrm{2}\boldsymbol{\mathrm{sin}}\left(\frac{\boldsymbol{\pi{a}}}{\mathrm{2}}\right)\boldsymbol{\psi}\left(\boldsymbol{{a}}\right)\right)\mid_{\boldsymbol{{a}}=\mathrm{1}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\Gamma}\left(\mathrm{1}\right)\left(\boldsymbol{\pi{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}\right)+\mathrm{2}\boldsymbol{\pi{sin}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}\right)\boldsymbol{\psi}\left(\mathrm{1}\right)\right)=−\frac{\boldsymbol{\gamma\pi}}{\mathrm{2}} \\ $$

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