Question Number 220562 by SdC355 last updated on 15/May/25 | ||
![]() | ||
$$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{z}={I} \\ $$$${I}\left({t}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}{e}^{−{zt}} \:\mathrm{d}{z} \\ $$$${I}'\left({t}\right)=−\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{{z}\centerdot\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}{e}^{−{zt}} \:\mathrm{d}{z} \\ $$$${I}'\left({t}\right)=−\int_{\mathrm{0}} ^{\:\infty} \:\:\:\frac{{z}^{\mathrm{2}} \mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{e}^{−{zt}} \:\mathrm{d}{z} \\ $$$${I}'\left({t}\right)=−\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}}{e}^{−{zt}\:} \mathrm{d}{z}+\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{e}^{−{zt}} \:\mathrm{d}{z} \\ $$$${I}''\left({t}\right)=\int_{\mathrm{0}} ^{\infty} \:\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right){e}^{−{zt}} \mathrm{d}{z}−\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}{e}^{−{zt}} \:\mathrm{d}{z} \\ $$$${I}^{\left(\mathrm{2}\right)} \left({t}\right)+{I}\left({t}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right){e}^{−{zt}} \mathrm{d}{z}........\left(\mathrm{A}\right) \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right){e}^{−{zt}} \mathrm{d}{z} \\ $$$$\mathrm{cos}^{\mathrm{2}} \left(\alpha\right)+\mathrm{sin}^{\mathrm{2}} \left(\alpha\right)=\mathrm{1}\:\rightarrow\:\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\alpha\right)=\mathrm{sec}^{\mathrm{2}} \left(\alpha\right) \\ $$$${z}=\mathrm{tan}\left({u}\right)\:\rightarrow\:{u}=\mathrm{tan}^{−\mathrm{1}} \left({z}\right) \\ $$$$\frac{\mathrm{d}{z}}{\mathrm{d}{u}}=\mathrm{sec}^{\mathrm{2}} \left({u}\right)\:\rightarrow\:\mathrm{d}{z}=\mathrm{sec}^{\mathrm{2}} \left({u}\right)\mathrm{d}{u} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right){e}^{−{st}} \mathrm{d}{z}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\mathrm{sec}^{\mathrm{2}} \left({u}\right)\centerdot\mathrm{ln}\left(\mathrm{sec}^{\mathrm{2}} \left({u}\right)\right){e}^{−{s}\centerdot\mathrm{tan}\left({u}\right)} \mathrm{d}{u} \\ $$$$\mathrm{i}\:\mathrm{can}'\mathrm{t}\:\mathrm{Calculate}\:\mathrm{anymore}.... \\ $$$$\mathrm{can}'\mathrm{t}\:\mathrm{solve}\:\mathrm{ODE}....\left(\mathrm{A}\right) \\ $$$$\mathrm{that}\:\mathrm{Equation}\left(\mathrm{A}\right)\:\mathrm{Seems}\:\mathrm{to}\:\mathrm{Weird}\:\mathrm{cus}\:\mathrm{Feynman}\:\mathrm{trick}\: \\ $$$$...... \\ $$$$\mathrm{How}\:\mathrm{can}\:\mathrm{i}\:\mathrm{solve}\:\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{z} \\ $$$$\mathrm{or}....\mathrm{Should}\:\mathrm{i}\:\mathrm{do}\:\mathrm{Complex}\:\mathrm{integral}.. \\ $$$$\oint_{\:{C}} \:{f}\left({z}\right)\:\mathrm{d}{z}.....?? \\ $$ | ||
Answered by breniam last updated on 15/May/25 | ||
![]() | ||
$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right){e}^{−{zt}} \mathrm{d}{z}=−\frac{\mathrm{1}}{{t}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\left({e}^{−{zt}} \right)'\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{d}{z}=\frac{\mathrm{2}}{{t}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{ze}^{−{zt}} }{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z} \\ $$$${J}\left({t}\right)=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{ze}^{−{tz}} }{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z} \\ $$$$\underset{{t}\rightarrow\infty} {\mathrm{lim}}{J}\left({t}\right)=\mathrm{0} \\ $$$${J}'\left({z}\right)=−\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{z}^{\mathrm{2}} {e}^{−{tz}} }{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z}=−\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{tz}} \mathrm{d}{z}+\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{e}^{−{zt}} }{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z}= \\ $$$$−\frac{\mathrm{1}}{{t}}+\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{e}^{−{zt}} }{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z} \\ $$$${J}''\left({t}\right)=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }−{J}\left({t}\right) \\ $$$${J}\left({t}\right)+{J}''\left({t}\right)=\frac{\mathrm{1}}{{t}^{\mathrm{2}} } \\ $$$${J}\left({t}\right)={J}_{\mathrm{1}} \left({t}\right)\mathrm{sin}\left({t}\right) \\ $$$${J}_{\mathrm{1}} ''\left({t}\right)\mathrm{sin}\left({t}\right)+\mathrm{2}{J}_{\mathrm{1}} '\left({t}\right)\mathrm{cos}\left({t}\right)=\frac{\mathrm{1}}{{t}^{\mathrm{2}} } \\ $$$${J}_{\mathrm{2}} \left({t}\right)={J}_{\mathrm{1}} '\left({t}\right) \\ $$$${J}_{\mathrm{2}} '\left({t}\right)\mathrm{sin}\left({t}\right)+\mathrm{2}{J}_{\mathrm{2}} \left({t}\right)\mathrm{cos}\left({t}\right)=\frac{\mathrm{1}}{{t}^{\mathrm{2}} } \\ $$$${J}_{\mathrm{2}} \left({t}\right)=\frac{{J}_{\mathrm{3}} \left({t}\right)}{\mathrm{sin}^{\mathrm{2}} \left({t}\right)} \\ $$$$\frac{{J}_{\mathrm{3}} '\left({t}\right)}{\mathrm{sin}\left({t}\right)}=\frac{\mathrm{1}}{{t}^{\mathrm{2}} } \\ $$$${J}_{\mathrm{3}} '\left({t}\right)=\frac{\mathrm{sin}\left({t}\right)}{{t}^{\mathrm{2}} } \\ $$$${J}_{\mathrm{3}} \left({t}\right)=\int\frac{\mathrm{sin}\left({t}\right)}{{t}^{\mathrm{2}} }\mathrm{d}{t}=−\int\left(\frac{\mathrm{1}}{{t}}\right)'\mathrm{sin}\left({t}\right)\mathrm{d}{t}=−\frac{\mathrm{sin}\left({t}\right)}{{t}}+\mathrm{Ci}\left(\mathrm{t}\right)+{A} \\ $$$${J}_{\mathrm{2}} \left({t}\right)={J}_{\mathrm{1}} '\left({t}\right)=−\frac{\mathrm{1}}{{t}\mathrm{sin}\left({t}\right)}+\frac{\mathrm{Ci}\left({t}\right)}{\mathrm{sin}^{\mathrm{2}} \left({t}\right)}+\frac{{A}}{\mathrm{sin}^{\mathrm{2}} \left({t}\right)} \\ $$$${J}_{\mathrm{1}} \left({t}\right)=\int\left[−\frac{\mathrm{1}}{{t}\mathrm{sin}\left({t}\right)}+\frac{\mathrm{Ci}\left({t}\right)}{\mathrm{sin}^{\mathrm{2}} \left({t}\right)}\right]\mathrm{d}{t}+{A}\mathrm{cot}\left({t}\right) \\ $$$$=−\int\frac{\mathrm{d}{t}}{{t}\mathrm{sin}\left({t}\right)}−\int\mathrm{cot}'\left({t}\right)\mathrm{Ci}\left({t}\right)\mathrm{d}{t}+{A}\mathrm{cot}\left({t}\right)= \\ $$$$=−\int\frac{\mathrm{d}{t}}{{t}\mathrm{sin}\left({t}\right)}−\mathrm{cot}\left({t}\right)\mathrm{Ci}\left({t}\right)+\int\frac{\mathrm{cos}^{\mathrm{2}} \left({t}\right)}{{t}\mathrm{sin}\left({t}\right)}\mathrm{d}{t}+{A}\mathrm{cot}\left({t}\right)= \\ $$$$−\mathrm{Si}\left({t}\right)−\mathrm{cot}\left({t}\right)\mathrm{Ci}\left({t}\right)+{A}\mathrm{cot}\left({t}\right)+{B} \\ $$$${J}\left({t}\right)=−\mathrm{sin}\left({t}\right)\mathrm{Si}\left({t}\right)−\mathrm{cos}\left({t}\right)\mathrm{Ci}\left({t}\right)+{A}\mathrm{sin}\left({t}\right)+{B}\mathrm{cos}\left({t}\right) \\ $$$$\underset{{t}\rightarrow\infty} {\mathrm{lim}}{J}\left({t}\right)=\mathrm{0}\Rightarrow{B}=\mathrm{0}\wedge{A}=\frac{\pi}{\mathrm{2}} \\ $$$${J}\left({t}\right)=\mathrm{sin}\left({t}\right)\left(\frac{\pi}{\mathrm{2}}−\mathrm{Si}\left(\mathrm{t}\right)\right) \\ $$$${I}^{\left(\mathrm{2}\right)} \left({t}\right)+{I}\left({t}\right)=\frac{\mathrm{2}}{{t}}\mathrm{sin}\left({t}\right)\left(\frac{\pi}{\mathrm{2}}−\mathrm{Si}\left({t}\right)\right) \\ $$$$\mathrm{Did}\:\mathrm{this}\:\mathrm{help}? \\ $$ | ||
Commented by SdC355 last updated on 16/May/25 | ||
![]() | ||
$$\mathrm{mmm}......\:\mathrm{thxs} \\ $$ | ||