Integration Questions

Question Number 130306 by benjo_mathlover last updated on 24/Jan/21

$$\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{x}\:\mathrm{cos}\:\left(\mathrm{x}\right)\:\mathrm{ln}\:\left(\mathrm{x}\right)\mathrm{e}^{−\mathrm{x}} \:\mathrm{dx}\:?\: \\$$

Answered by Dwaipayan Shikari last updated on 24/Jan/21

$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} {x}^{{a}} {cosx}\:{e}^{−{x}} {dx} \\$$$${I}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {x}^{{a}} {e}^{−\left(\mathrm{1}−{i}\right){x}} +{x}^{{a}} {e}^{−\left(\mathrm{1}+{i}\right){a}} {dx} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left({a}+\mathrm{1}\right)\left(\frac{\mathrm{1}}{\left(\mathrm{1}−{i}\right)^{{a}+\mathrm{1}} }+\frac{\mathrm{1}}{\left(\mathrm{1}+{i}\right)^{{a}+\mathrm{1}} }\right)=\frac{\Gamma\left({a}+\mathrm{1}\right)}{\mathrm{2}^{\frac{{a}+\mathrm{1}}{\mathrm{2}}} }{cos}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right) \\$$$${I}'\left({a}\right)=\frac{\Gamma'\left({a}+\mathrm{1}\right)}{\mathrm{2}^{\frac{{a}+\mathrm{1}}{\mathrm{2}}} }{cos}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right)−\frac{\pi}{\mathrm{4}}\:\frac{\Gamma\left({a}+\mathrm{1}\right)}{\mathrm{2}^{\frac{{a}+\mathrm{1}}{\mathrm{2}}} }{sin}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right)−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\Gamma\left({a}+\mathrm{1}\right)}{\mathrm{2}^{\frac{{a}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}} }{log}\left(\mathrm{2}\right){cos}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right) \\$$$${I}'\left(\mathrm{1}\right)=−\frac{\pi}{\mathrm{4}}.\frac{\mathrm{1}}{\mathrm{2}}=−\frac{\pi}{\mathrm{8}} \\$$

Answered by mnjuly1970 last updated on 24/Jan/21

$$\Omega=\mathscr{R}{e}\int_{\mathrm{0}} ^{\:\:\infty} {xe}^{−{ix}} {ln}\left({x}\right){e}^{−{x}} {dx} \\$$$$\:\:\:\:\:=\mathscr{R}{e}\int_{\mathrm{0}} ^{\:\infty} {xe}^{−{x}\left(\mathrm{1}+{i}\right)} {ln}\left({x}\right){dx} \\$$$$\:\:\:\:\:=\mathscr{R}{e}\left(\frac{{d}}{{ds}}\int_{\mathrm{0}} ^{\:\infty} {x}^{{s}} {e}^{−{x}\left(\mathrm{1}+{i}\right)} {dx}\right)\mid_{{s}=\mathrm{1}} \\$$$$\:\:\:=\:\:\mathscr{R}{e}\left(\frac{{d}}{{ds}}\left[\int_{\mathrm{0}} ^{\:\infty} {x}^{{s}} {e}^{−{x}\left(\mathrm{1}+{i}\right)} {dx}\right]\right)\mid_{{s}=\mathrm{1}} \\$$$$\:\:\:\Phi=\int_{\mathrm{0}} ^{\:\infty} {x}^{\mathrm{1}+{s}} {e}^{−{x}\left(\mathrm{1}+{i}\right)} {dx}\overset{{x}\left(\mathrm{1}+{i}\right)={t}} {=}\int_{\mathrm{0}} ^{\:\infty} \frac{{t}^{{s}} }{\left(\mathrm{1}+{i}\right)^{{s}+\mathrm{1}} }{e}^{−{t}} {dt} \\$$$$\:\:\:\:\:\:\:\Phi=\Gamma\left({s}+\mathrm{1}\right)\left(\mathrm{1}+{i}\right)^{−{s}−\mathrm{1}} \\$$$$\:\:\:\:\frac{{d}\Phi}{{ds}}=\Gamma'\left({s}+\mathrm{1}\right)\left(\mathrm{1}+{i}\right)^{−{s}−\mathrm{1}} \mid_{{s}=\mathrm{1}} +\left[{e}^{−\left({s}+\mathrm{1}\right){ln}\left(\mathrm{1}+{i}\right)} \overset{\:\:'} {\right]}\Gamma\left({s}+\mathrm{1}\right)\mid_{{s}=\mathrm{1}} \\$$$$\:\:\:\frac{{d}\Phi}{{ds}}\:\mid_{{s}=\mathrm{1}} =\Gamma'\left(\mathrm{2}\right)\left(\mathrm{1}+{i}\right)^{−\mathrm{2}} +\left(−{ln}\left(\mathrm{1}+{i}\right)\right){e}^{−\mathrm{2}{ln}\left(\mathrm{1}+{i}\right)} \Gamma\left(\mathrm{2}\right) \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\psi\left(\mathrm{2}\right)\left(\mathrm{2}{i}\right)^{−\mathrm{1}} +\left(−{ln}\left(\mathrm{1}+{i}\right)\right){e}^{−{ln}\left(\mathrm{2}{i}\right)} .\mathrm{1} \\$$$$=\left[−\psi\left(\mathrm{2}\right)\ast\frac{{i}}{\mathrm{2}}\:={Imaginary}\right]−{ln}\left(\sqrt{\mathrm{2}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\ast\frac{\mathrm{1}}{\mathrm{2}{i}} \\$$$$\:\:\therefore\:\Omega\:=−\frac{\pi}{\mathrm{8}}\:...\checkmark\:\checkmark \\$$$$\:\:\:\:\:\: \\$$$$\:\:\:\:\:\:\: \\$$$$\:\: \\$$

Commented by benjo_mathlover last updated on 24/Jan/21

$$\mathrm{the}\:\mathrm{ans}\:\mathrm{is}\:−\frac{\pi}{\mathrm{8}} \\$$

Commented by mnjuly1970 last updated on 24/Jan/21

$${thanks}\:{alot}... \\$$$$\: \\$$