Question Number 130433 by mnjuly1970 last updated on 25/Jan/21
$$\:\:\:\:\:\:….{nice}\:\:\:{calculus}… \\ $$$$\:\:\:{prove}\:{that}:: \\ $$$$\:\:\:\Psi=\int_{\mathrm{0}} ^{\:\infty} {e}^{−{x}} {ln}\left({x}\right){cos}\left({x}\right){dx}\overset{?} {=}\frac{\mathrm{1}}{\mathrm{8}}\left(−\mathrm{4}\gamma−\pi−\mathrm{2}{ln}\left(\mathrm{2}\right)\right) \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 25/Jan/21
$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} {x}^{{a}} {e}^{−{x}} {cosx}\:{dx}\:=\:\frac{\Gamma\left({a}+\mathrm{1}\right)}{\mathrm{2}^{\frac{{a}+\mathrm{1}}{\mathrm{2}}} }{cos}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right) \\ $$$${I}'\left({a}\right)=\frac{\Gamma'\left({a}+\mathrm{1}\right)}{\mathrm{2}^{\frac{{a}+\mathrm{1}}{\mathrm{2}}} }{cos}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right)−\frac{\pi}{\mathrm{4}}.\frac{\Gamma\left({a}+\mathrm{1}\right)}{\mathrm{2}^{\frac{{a}+\mathrm{1}}{\mathrm{2}}} }{sin}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right)−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\Gamma\left({a}+\mathrm{1}\right)}{\mathrm{2}^{\frac{{a}+\mathrm{1}}{\mathrm{2}}} }{cos}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right){log}\left(\mathrm{2}\right) \\ $$$${I}'\left(\mathrm{0}\right)=\frac{−\gamma}{\mathrm{2}}−\frac{\pi}{\mathrm{8}}−\frac{{log}\left(\mathrm{2}\right)}{\mathrm{4}}=−\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{4}\gamma+\pi+{log}\left(\mathrm{4}\right)\right) \\ $$
Commented by mnjuly1970 last updated on 25/Jan/21
$${very}\:{nice}\:. \\ $$$${Allah}\:{keep}\:{you}… \\ $$
Answered by mnjuly1970 last updated on 26/Jan/21
![Ψ=Re(∫_0 ^( ∞) e^(−x) e^(−ix) ln(x)dx) =Re((d/da)(∫_0 ^( ∞) e^(−x(1+i)) x^a dx=Ω)∣_(a=0) Ω =^(x(1+i)=t) ∫_0 ^( ∞) e^(−t) (t^a /((1+i)^(a+1) ))dt =(1/(2^((a+1)/2) e^((iπ(a+1)/4) )) Γ(a+1) ∴ ((d(Ω))/da)=Γ′(a+1)2^(((−(a+1))/2) ) ∗e^((−iπ(a+1))/4) −(((ln(2))/2)2^((−(a+1))/2) ∗e^((−iπ(a+1))/4) +((iπ)/4)e^((−iπ(a+1))/4) ∗2^((−(a+1))/2) )Γ(a+1) ∴ (((d(Ω))/da))∣_(a=0) =((Γ′(1))/( (√2)))(cos((π/4))−isin((π/4))) −[((ln(2))/( 2(√2)))(cos((π/4))−isin((π/4)))+((iπ)/(4(√2)))(cos((π/4))−isin((π/4)))]Γ(1) ∴ Re((dΩ/da))∣_(a=0) =((−γ)/2)−((ln(2))/4)−(π/8) Ψ= −(1/8)(π+4γ+2ln(2))](https://www.tinkutara.com/question/Q130501.png)
$$\:\:\:\Psi=\mathscr{R}{e}\left(\int_{\mathrm{0}} ^{\:\infty} {e}^{−{x}} {e}^{−{ix}} {ln}\left({x}\right){dx}\right) \\ $$$$\:\:\:\:\:\:=\mathscr{R}{e}\left(\frac{{d}}{{da}}\left(\int_{\mathrm{0}} ^{\:\infty} {e}^{−{x}\left(\mathrm{1}+{i}\right)} {x}^{{a}} {dx}=\Omega\right)\mid_{{a}=\mathrm{0}} \right. \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\Omega\:\overset{{x}\left(\mathrm{1}+{i}\right)={t}} {=}\:\int_{\mathrm{0}} ^{\:\infty} {e}^{−{t}} \frac{{t}^{{a}} }{\left(\mathrm{1}+{i}\right)^{{a}+\mathrm{1}} }{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}^{\frac{{a}+\mathrm{1}}{\mathrm{2}}} {e}^{\frac{{i}\pi\left({a}+\mathrm{1}\right.}{\mathrm{4}}} }\:\Gamma\left({a}+\mathrm{1}\right) \\ $$$$\:\:\:\:\therefore\:\:\frac{{d}\left(\Omega\right)}{{da}}=\Gamma'\left({a}+\mathrm{1}\right)\mathrm{2}^{\frac{−\left({a}+\mathrm{1}\right)}{\mathrm{2}}\:} \ast{e}^{\frac{−{i}\pi\left({a}+\mathrm{1}\right)}{\mathrm{4}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left(\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\mathrm{2}^{\frac{−\left({a}+\mathrm{1}\right)}{\mathrm{2}}} \ast{e}^{\frac{−{i}\pi\left({a}+\mathrm{1}\right)}{\mathrm{4}}} +\frac{{i}\pi}{\mathrm{4}}{e}^{\frac{−{i}\pi\left({a}+\mathrm{1}\right)}{\mathrm{4}}} \ast\mathrm{2}^{\frac{−\left({a}+\mathrm{1}\right)}{\mathrm{2}}} \right)\Gamma\left({a}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\therefore\:\left(\frac{{d}\left(\Omega\right)}{{da}}\right)\mid_{{a}=\mathrm{0}} =\frac{\Gamma'\left(\mathrm{1}\right)}{\:\sqrt{\mathrm{2}}}\left({cos}\left(\frac{\pi}{\mathrm{4}}\right)−{isin}\left(\frac{\pi}{\mathrm{4}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left[\frac{{ln}\left(\mathrm{2}\right)}{\:\mathrm{2}\sqrt{\mathrm{2}}}\left({cos}\left(\frac{\pi}{\mathrm{4}}\right)−{isin}\left(\frac{\pi}{\mathrm{4}}\right)\right)+\frac{{i}\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\left({cos}\left(\frac{\pi}{\mathrm{4}}\right)−{isin}\left(\frac{\pi}{\mathrm{4}}\right)\right)\right]\Gamma\left(\mathrm{1}\right) \\ $$$$\therefore\:\mathscr{R}{e}\left(\frac{{d}\Omega}{{da}}\right)\mid_{{a}=\mathrm{0}} =\frac{−\gamma}{\mathrm{2}}−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{4}}−\frac{\pi}{\mathrm{8}} \\ $$$$\:\:\:\:\:\Psi=\:−\frac{\mathrm{1}}{\mathrm{8}}\left(\pi+\mathrm{4}\gamma+\mathrm{2}{ln}\left(\mathrm{2}\right)\right) \\ $$