Integration Questions

Question Number 189489 by mnjuly1970 last updated on 17/Mar/23

$$\\$$ $$\:\:\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\infty} {e}^{\:−{x}} {cos}\left({x}\right){ln}\left({x}\right){dx}=? \\$$ $$\:\:\:\:\:−−− \\$$ $$\:\:\:\:\:\:{f}\:\left({a}\:\right)=\:\int_{\mathrm{0}} ^{\:\infty} {e}^{\:−{x}} {cos}\left({x}\right){x}^{\:{a}} \:{dx} \\$$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\$$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{Re}\:\int_{\mathrm{0}} ^{\:\infty} {e}^{\:−{x}} .{e}^{\:−{ix}} .{x}^{\:{a}} {dx} \\$$ $$\:\:\:\:\:\:\:\:\:\:\:=\:{Re}\:\int_{\mathrm{0}} ^{\:\infty} \:{e}^{\:−{x}\:\left(\mathrm{1}+{i}\right)} .{x}^{\:{a}} \:{dx} \\$$ $$\:\:\:\:\:\:\:\:\:\:=\:{Re}\left(\mathscr{L}\:\:\left\{\:{x}^{\:{a}} \:\right\}\mid_{\:{s}=\:{i}+\mathrm{1}} \right) \\$$ $$\:\:\:\:\:\:\:\:\:=\:{Re}\left(\:\frac{\Gamma\:\left(\mathrm{1}+{a}\right)}{{s}^{\:{a}+\mathrm{1}} }\:\mid_{\:\mathrm{1}+{i}} =\:\frac{\Gamma\:\left(\mathrm{1}+{a}\right)}{\left(\mathrm{1}+{i}\right)^{\:{a}+\mathrm{1}} }\:\right) \\$$ $$\:\:\:\:\:\:\:\:{Re}\:\left(\Gamma\left(\mathrm{1}+{a}\right).\mathrm{2}^{\:\frac{\mathrm{1}+{a}}{\mathrm{2}}} .\:{e}^{\:−\frac{{i}\pi}{\mathrm{4}}\:\left(\mathrm{1}+{a}\right)} \right) \\$$ $$\:\:\:\:\Omega=\:{f}\:'\left(\mathrm{0}\right)=....... \\$$ $$\:\: \\$$

Commented byGbenga last updated on 19/Mar/23

$$\boldsymbol{\mathrm{here}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{my}}\:\boldsymbol{\mathrm{solution}}....... \\$$ $$\boldsymbol{\Omega}=\int_{\mathrm{0}} ^{\infty} \boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{x}}} \boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{dx}} \\$$ $$\boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{ix}}} =\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)+\boldsymbol{\mathrm{isin}}\left(\boldsymbol{\mathrm{x}}\right) \\$$ $$\boldsymbol{\mathrm{R}{e}}\left(\boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{ix}}} \right)=\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right) \\$$ $$\boldsymbol{\Omega}=\boldsymbol{\mathrm{R}{e}}\int_{\mathrm{0}} ^{\infty} \boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{x}}} \boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{ix}}} \boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{dx}} \\$$ $$\boldsymbol{\Omega}=\boldsymbol{\mathrm{Re}}\int_{\mathrm{0}} ^{\infty} \boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{x}}\left(\mathrm{1}+\boldsymbol{\mathrm{i}}\right)} \boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{dx}} \\$$ $$\boldsymbol{\Omega}=\boldsymbol{\mathrm{R}{e}}\left[\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\int_{\mathrm{0}} ^{\infty} \boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{x}}\left(\mathrm{1}+\boldsymbol{\mathrm{i}}\right)} \boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{a}}} \boldsymbol{\mathrm{dx}}\right]_{\boldsymbol{\mathrm{a}}=\mathrm{0}} \\$$ $$\boldsymbol{\mathrm{let}}\:\boldsymbol{\mathrm{x}}\left(\mathrm{1}+\boldsymbol{\mathrm{i}}\right)=\boldsymbol{\mathrm{u}}\:\boldsymbol{\mathrm{x}}=\frac{\boldsymbol{\mathrm{u}}}{\mathrm{1}+\boldsymbol{\mathrm{i}}\:}\:\boldsymbol{\mathrm{dx}}=\frac{\boldsymbol{\mathrm{du}}}{\mathrm{1}+\boldsymbol{\mathrm{i}}} \\$$ $$\boldsymbol{\Omega}=\boldsymbol{\mathrm{Re}}\left[\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\left[\frac{\mathrm{1}}{\left(\mathrm{1}+\boldsymbol{\mathrm{i}}\right)^{\boldsymbol{\mathrm{a}}+\mathrm{1}} }\left[\int_{\mathrm{0}} ^{\infty} \boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{u}}} \boldsymbol{\mathrm{u}}^{\boldsymbol{\mathrm{a}}} \boldsymbol{\mathrm{du}}\right]_{\boldsymbol{\mathrm{a}}=\mathrm{0}} \right.\right. \\$$ $$\boldsymbol{\Omega}=\boldsymbol{\mathrm{R}{e}}\left[\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\left[\frac{\boldsymbol{\Gamma}\left(\boldsymbol{\mathrm{a}}+\mathrm{1}\right)}{\left(\mathrm{1}+\boldsymbol{\mathrm{i}}\right)^{\boldsymbol{\mathrm{a}}+\mathrm{1}} }\right]_{\boldsymbol{\mathrm{a}}=\mathrm{0}} \right. \\$$ $$\boldsymbol{\Omega}=\boldsymbol{\mathrm{Re}}\left(−\frac{\boldsymbol{\Gamma}\left(\boldsymbol{\mathrm{a}}+\mathrm{1}\right)}{\left(\mathrm{1}+\boldsymbol{\mathrm{i}}\right)^{\boldsymbol{\mathrm{a}}+\mathrm{1}} }\left[−\boldsymbol{\psi}^{\mathrm{0}} \left(\boldsymbol{\mathrm{a}}+\mathrm{1}\right)+\boldsymbol{\mathrm{log}}\left(\mathrm{1}+\boldsymbol{\mathrm{i}}\right)\right]\right)_{\boldsymbol{\mathrm{a}}=\mathrm{0}} \\$$ $$\boldsymbol{\mathrm{Re}}\left(−\frac{\mathrm{1}}{\left(\mathrm{1}+\boldsymbol{\mathrm{i}}\right)}\left[−\boldsymbol{\psi}^{\mathrm{0}} \left(\mathrm{1}\right)+\boldsymbol{\mathrm{log}}\left(\mathrm{1}+\boldsymbol{\mathrm{i}}\right)\right]\right)=−\frac{\boldsymbol{\gamma}}{\mathrm{2}}+\boldsymbol{\mathrm{Re}}\left(\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\boldsymbol{\mathrm{i}}}{\mathrm{2}}\right)\boldsymbol{\mathrm{log}}\left(\mathrm{1}+\boldsymbol{\mathrm{i}}\right)\right)=−\frac{\boldsymbol{\gamma}}{\mathrm{2}}−\frac{\boldsymbol{\pi}}{\mathrm{8}}−\frac{\boldsymbol{\mathrm{log}}\left(\mathrm{2}\right)}{\mathrm{4}} \\$$ $$\therefore\:\boldsymbol{\mathrm{our}}\:\boldsymbol{\mathrm{answer}}\:\boldsymbol{\mathrm{is}}\:−\frac{\boldsymbol{\gamma}}{\mathrm{2}}−\frac{\boldsymbol{\pi}}{\mathrm{8}}−\frac{\boldsymbol{\mathrm{log}}\left(\mathrm{2}\right)}{\mathrm{4}} \\$$ $$\bigstar\boldsymbol{{Small}}\:\boldsymbol{{Laplace}}\bigstar \\$$

Answered by leodera last updated on 17/May/23

$${my}\:{solution} \\$$ $${Re}\int_{\mathrm{0}} ^{\infty} {e}^{−\left(\mathrm{1}−{i}\right){x}} {In}\left({x}\right){dx}\:=\:{Re}\left(\mathscr{L}\left\{\mathrm{ln}\:\left({x}\right)\right\}_{{s}=\mathrm{1}−{i}} \right) \\$$ $$=\:{Re}\left(−\frac{\mathrm{1}}{\mathrm{1}−{i}}\left(\gamma\:+\:\mathrm{ln}\:\left(\mathrm{1}−{i}\right)\right)\right) \\$$ $$\:=\:{Re}\left(−\frac{\mathrm{1}+{i}}{\mathrm{2}}\left(\gamma\:\:+\:\mathrm{ln}\:\left(\sqrt{\mathrm{2}}{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\right)\right) \\$$ $$=\:{Re}\left(−\frac{\mathrm{1}+{i}}{\mathrm{2}}\left(\gamma\:\:+\:\mathrm{ln}\left(\sqrt{\mathrm{2}}\right)\:+\:{i}\frac{\pi}{\mathrm{4}}\right)\right) \\$$ $$=\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\gamma+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{4}}\right) \\$$ $$=\:−\frac{\gamma}{\mathrm{2}}\:−\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\left(\mathrm{2}\right)\:−\:\frac{\pi}{\mathrm{8}}\: \\$$ $$\\$$ $$\\$$