Question Number 153430 by mnjuly1970 last updated on 07/Sep/21
Commented by mnjuly1970 last updated on 07/Sep/21
$${thanks}\:{alot}.... \\ $$
Commented by mindispower last updated on 08/Sep/21
$${pleasur} \\ $$
Answered by mindispower last updated on 07/Sep/21
![J=−(1/2)∫_0 ^∞ ((e^(−x) sin(x)ln(x))/x)dx f(a)=∫_0 ^∞ e^(−x(1+i)) x^a dx ∫_D e^(−z) z^a dz,D_R =[0,R]∪[Re^(iθ) ,θ∈[0,(π/4)]]∪[x(1+i),x∈[0,(R/( (√2)))]] ∫_D_R e^(−z) z^a =0=∫_0 ^R e^(−z) z^a dz+∫_(CR) f(z)dz+(1+i)^(a+1) ∫_(R/( (√2))) ^0 e^(−x(1+i)) x^a =0 lim_(R→∞) ∫_C_R f(z)dz=0⇒ ⇒(1∫_0 ^∞ e^(−x(1+i)) x^a dx=(1/((1+i)^(a+1) ))Γ(1+a)) f(a)= (2^((−a−1)/2) )e^(−((iπ(a+1))/4)) Γ(1+a)) Imf(a)=−(1/2^((1+a)/2) )sin(((a+1)/4)π)Γ(1+a) J=−(1/2).((d imf(a))/da)∣_(a=−1) =(1/2)(−(1/2)(ln(2)2^(−((1+a)/2)) sin(((a+1)/4)π)Γ(1+a)+(π/4)cos(((a+1)/4)π)Γ(1+a)2^(−((1+a)/2)) +Ψ(1+a)Γ(1+a)sin(((a+1)/4)π).2^(−((1+a)/2)π) ) using Ψ(1+a)=Ψ(2+a)−(1/(a+1))) cos(((a+1)/4)π)=1+o(a+1),a→−1 sin(((a+1)/4)π)=((a+1)/4)π+o(a+1) we get J=−(1/2)(−(1/2)ln(2).(π/4)lim_(a→−1) (1+a)Γ(1+a)+(π/4)Γ(1+a) (Ψ(2+a)−(1/(a+1)))Γ(1+a).(π/4)(1+a))) =((ln(2)π)/(16))+lim_(a→−1) (π/4)(Γ(a+1)+Ψ(2+a)Γ(2+a)−((Γ(2+a))/(1+a))) =(π/(16))ln(2)(π/4)−(π/8)lim_(a→−1) ((((1+a)Γ(1+a)−Γ(2+a))/(1+a))+Ψ(1)) Γ(2+a)=Γ(1+a).(1+a) we get (π/8)(((ln(2))/2)+γ)=(π/8)(ln((√2))+γ)](Q153497.png)
$${J}=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}} {sin}\left({x}\right){ln}\left({x}\right)}{{x}}{dx} \\ $$$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{x}\left(\mathrm{1}+{i}\right)} {x}^{{a}} {dx} \\ $$$$\int_{{D}} {e}^{−{z}} {z}^{{a}} {dz},{D}_{{R}} =\left[\mathrm{0},{R}\right]\cup\left[{Re}^{{i}\theta} ,\theta\in\left[\mathrm{0},\frac{\pi}{\mathrm{4}}\right]\right]\cup\left[{x}\left(\mathrm{1}+{i}\right),{x}\in\left[\mathrm{0},\frac{{R}}{\:\sqrt{\mathrm{2}}}\right]\right] \\ $$$$\int_{{D}_{{R}} } {e}^{−{z}} {z}^{{a}} =\mathrm{0}=\int_{\mathrm{0}} ^{{R}} {e}^{−{z}} {z}^{{a}} {dz}+\int_{{CR}} {f}\left({z}\right){dz}+\left(\mathrm{1}+{i}\right)^{{a}+\mathrm{1}} \int_{\frac{{R}}{\:\sqrt{\mathrm{2}}}} ^{\mathrm{0}} {e}^{−{x}\left(\mathrm{1}+{i}\right)} {x}^{{a}} =\mathrm{0} \\ $$$$\underset{{R}\rightarrow\infty} {\mathrm{lim}}\:\int_{{C}_{{R}} } {f}\left({z}\right){dz}=\mathrm{0}\Rightarrow \\ $$$$\Rightarrow\left(\mathrm{1}\int_{\mathrm{0}} ^{\infty} {e}^{−{x}\left(\mathrm{1}+{i}\right)} {x}^{{a}} {dx}=\frac{\mathrm{1}}{\left(\mathrm{1}+{i}\right)^{{a}+\mathrm{1}} }\Gamma\left(\mathrm{1}+{a}\right)\right) \\ $$$$\left.{f}\left({a}\right)=\:\left(\mathrm{2}^{\frac{−{a}−\mathrm{1}}{\mathrm{2}}} \right){e}^{−\frac{{i}\pi\left({a}+\mathrm{1}\right)}{\mathrm{4}}} \Gamma\left(\mathrm{1}+{a}\right)\right) \\ $$$${Imf}\left({a}\right)=−\frac{\mathrm{1}}{\mathrm{2}^{\frac{\mathrm{1}+{a}}{\mathrm{2}}} }{sin}\left(\frac{{a}+\mathrm{1}}{\mathrm{4}}\pi\right)\Gamma\left(\mathrm{1}+{a}\right) \\ $$$${J}=−\frac{\mathrm{1}}{\mathrm{2}}.\frac{{d}\:{imf}\left({a}\right)}{{da}}\mid_{{a}=−\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\mathrm{2}\right)\mathrm{2}^{−\frac{\mathrm{1}+{a}}{\mathrm{2}}} {sin}\left(\frac{{a}+\mathrm{1}}{\mathrm{4}}\pi\right)\Gamma\left(\mathrm{1}+{a}\right)+\frac{\pi}{\mathrm{4}}{cos}\left(\frac{{a}+\mathrm{1}}{\mathrm{4}}\pi\right)\Gamma\left(\mathrm{1}+{a}\right)\mathrm{2}^{−\frac{\mathrm{1}+{a}}{\mathrm{2}}} \right.\right. \\ $$$$\left.+\Psi\left(\mathrm{1}+{a}\right)\Gamma\left(\mathrm{1}+{a}\right){sin}\left(\frac{{a}+\mathrm{1}}{\mathrm{4}}\pi\right).\mathrm{2}^{−\frac{\mathrm{1}+{a}}{\mathrm{2}}\pi} \right) \\ $$$$\left.{using}\:\Psi\left(\mathrm{1}+{a}\right)=\Psi\left(\mathrm{2}+{a}\right)−\frac{\mathrm{1}}{{a}+\mathrm{1}}\right) \\ $$$${cos}\left(\frac{{a}+\mathrm{1}}{\mathrm{4}}\pi\right)=\mathrm{1}+{o}\left({a}+\mathrm{1}\right),{a}\rightarrow−\mathrm{1} \\ $$$${sin}\left(\frac{{a}+\mathrm{1}}{\mathrm{4}}\pi\right)=\frac{{a}+\mathrm{1}}{\mathrm{4}}\pi+{o}\left({a}+\mathrm{1}\right) \\ $$$${we}\:{get} \\ $$$${J}=−\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right).\frac{\pi}{\mathrm{4}}\underset{{a}\rightarrow−\mathrm{1}} {\mathrm{lim}}\left(\mathrm{1}+{a}\right)\Gamma\left(\mathrm{1}+{a}\right)+\frac{\pi}{\mathrm{4}}\Gamma\left(\mathrm{1}+{a}\right)\right. \\ $$$$\left.\left(\left.\Psi\left(\mathrm{2}+{a}\right)−\frac{\mathrm{1}}{{a}+\mathrm{1}}\right)\Gamma\left(\mathrm{1}+{a}\right).\frac{\pi}{\mathrm{4}}\left(\mathrm{1}+{a}\right)\right)\right) \\ $$$$=\frac{{ln}\left(\mathrm{2}\right)\pi}{\mathrm{16}}+\underset{{a}\rightarrow−\mathrm{1}} {\mathrm{lim}}\frac{\pi}{\mathrm{4}}\left(\Gamma\left({a}+\mathrm{1}\right)+\Psi\left(\mathrm{2}+{a}\right)\Gamma\left(\mathrm{2}+{a}\right)−\frac{\Gamma\left(\mathrm{2}+{a}\right)}{\mathrm{1}+{a}}\right) \\ $$$$ \\ $$$$=\frac{\pi}{\mathrm{16}}{ln}\left(\mathrm{2}\right)\frac{\pi}{\mathrm{4}}−\frac{\pi}{\mathrm{8}}\underset{{a}\rightarrow−\mathrm{1}} {\mathrm{lim}}\left(\frac{\left(\mathrm{1}+{a}\right)\Gamma\left(\mathrm{1}+{a}\right)−\Gamma\left(\mathrm{2}+{a}\right)}{\mathrm{1}+{a}}+\Psi\left(\mathrm{1}\right)\right) \\ $$$$ \\ $$$$ \\ $$$$\Gamma\left(\mathrm{2}+{a}\right)=\Gamma\left(\mathrm{1}+{a}\right).\left(\mathrm{1}+{a}\right) \\ $$$${we}\:{get}\:\frac{\pi}{\mathrm{8}}\left(\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+\gamma\right)=\frac{\pi}{\mathrm{8}}\left({ln}\left(\sqrt{\mathrm{2}}\right)+\gamma\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by puissant last updated on 07/Sep/21
$${Sir}\:{mindispower}\:{you}\:{are}\:{a}\:{boss}.. \\ $$
Commented by mnjuly1970 last updated on 08/Sep/21
$${thank}\:{you}\:{so}\:{much}\:{mr}\:{power} \\ $$
Commented by mindispower last updated on 08/Sep/21
$${withe}\:{pleasur}\:{have}\:{a}\:{nice}\:{day} \\ $$