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Question Number 220562 by SdC355 last updated on 15/May/25
∫_0 ^( ∞) ((ln(z^2 +1))/(z^2 +1)) dz=I I(t)=∫_0 ^( ∞) ((ln(z^2 +1))/(z^2 +1))e^(−zt) dz I′(t)=−∫_0 ^( ∞) ((z∙ln(z^2 +1))/(z^2 +1))e^(−zt) dz I′(t)=−∫_0 ^( ∞) ((z^2 ln(z^2 +1)+ln(z^2 +1)−ln(z^2 +1))/(z(z^2 +1)))e^(−zt) dz I′(t)=−∫_0 ^( ∞) ((ln(z^2 +1))/z)e^(−zt ) dz+∫_0 ^( ∞) ((ln(z^2 +1))/(z(z^2 +1)))e^(−zt) dz I′′(t)=∫_0 ^∞ ln(z^2 +1)e^(−zt) dz−∫_0 ^( ∞) ((ln(z^2 +1))/(z^2 +1))e^(−zt) dz I^((2)) (t)+I(t)=∫_0 ^( ∞) ln(z^2 +1)e^(−zt) dz........(A) ∫_0 ^( ∞) ln(z^2 +1)e^(−zt) dz cos^2 (α)+sin^2 (α)=1 → 1+tan^2 (α)=sec^2 (α) z=tan(u) → u=tan^(−1) (z) (dz/du)=sec^2 (u) → dz=sec^2 (u)du ∫_0 ^( ∞) ln(z^2 +1)e^(−st) dz=∫_0 ^( (π/2)) sec^2 (u)∙ln(sec^2 (u))e^(−s∙tan(u)) du i can′t Calculate anymore.... can′t solve ODE....(A) that Equation(A) Seems to Weird cus Feynman trick ...... How can i solve ∫_0 ^( ∞) ((ln(z^2 +1))/(z^2 +1)) dz or....Should i do Complex integral.. ∮_( C) f(z) dz.....??
$$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{z}={I} \\ $$$${I}\left({t}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}{e}^{−{zt}} \:\mathrm{d}{z} \\ $$$${I}'\left({t}\right)=−\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{{z}\centerdot\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}{e}^{−{zt}} \:\mathrm{d}{z} \\ $$$${I}'\left({t}\right)=−\int_{\mathrm{0}} ^{\:\infty} \:\:\:\frac{{z}^{\mathrm{2}} \mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{e}^{−{zt}} \:\mathrm{d}{z} \\ $$$${I}'\left({t}\right)=−\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}}{e}^{−{zt}\:} \mathrm{d}{z}+\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{e}^{−{zt}} \:\mathrm{d}{z} \\ $$$${I}''\left({t}\right)=\int_{\mathrm{0}} ^{\infty} \:\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right){e}^{−{zt}} \mathrm{d}{z}−\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}{e}^{−{zt}} \:\mathrm{d}{z} \\ $$$${I}^{\left(\mathrm{2}\right)} \left({t}\right)+{I}\left({t}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right){e}^{−{zt}} \mathrm{d}{z}……..\left(\mathrm{A}\right) \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right){e}^{−{zt}} \mathrm{d}{z} \\ $$$$\mathrm{cos}^{\mathrm{2}} \left(\alpha\right)+\mathrm{sin}^{\mathrm{2}} \left(\alpha\right)=\mathrm{1}\:\rightarrow\:\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\alpha\right)=\mathrm{sec}^{\mathrm{2}} \left(\alpha\right) \\ $$$${z}=\mathrm{tan}\left({u}\right)\:\rightarrow\:{u}=\mathrm{tan}^{−\mathrm{1}} \left({z}\right) \\ $$$$\frac{\mathrm{d}{z}}{\mathrm{d}{u}}=\mathrm{sec}^{\mathrm{2}} \left({u}\right)\:\rightarrow\:\mathrm{d}{z}=\mathrm{sec}^{\mathrm{2}} \left({u}\right)\mathrm{d}{u} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right){e}^{−{st}} \mathrm{d}{z}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\mathrm{sec}^{\mathrm{2}} \left({u}\right)\centerdot\mathrm{ln}\left(\mathrm{sec}^{\mathrm{2}} \left({u}\right)\right){e}^{−{s}\centerdot\mathrm{tan}\left({u}\right)} \mathrm{d}{u} \\ $$$$\mathrm{i}\:\mathrm{can}'\mathrm{t}\:\mathrm{Calculate}\:\mathrm{anymore}…. \\ $$$$\mathrm{can}'\mathrm{t}\:\mathrm{solve}\:\mathrm{ODE}….\left(\mathrm{A}\right) \\ $$$$\mathrm{that}\:\mathrm{Equation}\left(\mathrm{A}\right)\:\mathrm{Seems}\:\mathrm{to}\:\mathrm{Weird}\:\mathrm{cus}\:\mathrm{Feynman}\:\mathrm{trick}\: \\ $$$$…… \\ $$$$\mathrm{How}\:\mathrm{can}\:\mathrm{i}\:\mathrm{solve}\:\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{z} \\ $$$$\mathrm{or}….\mathrm{Should}\:\mathrm{i}\:\mathrm{do}\:\mathrm{Complex}\:\mathrm{integral}.. \\ $$$$\oint_{\:{C}} \:{f}\left({z}\right)\:\mathrm{d}{z}…..?? \\ $$
Answered by breniam last updated on 15/May/25
∫_0 ^∞ ln(z^2 +1)e^(−zt) dz=−(1/t)∫_0 ^∞ (e^(−zt) )′ln(z^2 +1)dz=(2/t)∫_0 ^∞ ((ze^(−zt) )/(z^2 +1))dz J(t)=∫_0 ^∞ ((ze^(−tz) )/(z^2 +1))dz lim_(t→∞) J(t)=0 J′(z)=−∫_0 ^∞ ((z^2 e^(−tz) )/(z^2 +1))dz=−∫_0 ^∞ e^(−tz) dz+∫_0 ^∞ (e^(−zt) /(z^2 +1))dz= −(1/t)+∫_0 ^∞ (e^(−zt) /(z^2 +1))dz J′′(t)=(1/t^2 )−J(t) J(t)+J′′(t)=(1/t^2 ) J(t)=J_1 (t)sin(t) J_1 ′′(t)sin(t)+2J_1 ′(t)cos(t)=(1/t^2 ) J_2 (t)=J_1 ′(t) J_2 ′(t)sin(t)+2J_2 (t)cos(t)=(1/t^2 ) J_2 (t)=((J_3 (t))/(sin^2 (t))) ((J_3 ′(t))/(sin(t)))=(1/t^2 ) J_3 ′(t)=((sin(t))/t^2 ) J_3 (t)=∫((sin(t))/t^2 )dt=−∫((1/t))′sin(t)dt=−((sin(t))/t)+Ci(t)+A J_2 (t)=J_1 ′(t)=−(1/(tsin(t)))+((Ci(t))/(sin^2 (t)))+(A/(sin^2 (t))) J_1 (t)=∫[−(1/(tsin(t)))+((Ci(t))/(sin^2 (t)))]dt+Acot(t) =−∫(dt/(tsin(t)))−∫cot′(t)Ci(t)dt+Acot(t)= =−∫(dt/(tsin(t)))−cot(t)Ci(t)+∫((cos^2 (t))/(tsin(t)))dt+Acot(t)= −Si(t)−cot(t)Ci(t)+Acot(t)+B J(t)=−sin(t)Si(t)−cos(t)Ci(t)+Asin(t)+Bcos(t) lim_(t→∞) J(t)=0⇒B=0∧A=(π/2) J(t)=sin(t)((π/2)−Si(t)) I^((2)) (t)+I(t)=(2/t)sin(t)((π/2)−Si(t)) Did this help?
$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right){e}^{−{zt}} \mathrm{d}{z}=−\frac{\mathrm{1}}{{t}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\left({e}^{−{zt}} \right)'\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{d}{z}=\frac{\mathrm{2}}{{t}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{ze}^{−{zt}} }{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z} \\ $$$${J}\left({t}\right)=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{ze}^{−{tz}} }{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z} \\ $$$$\underset{{t}\rightarrow\infty} {\mathrm{lim}}{J}\left({t}\right)=\mathrm{0} \\ $$$${J}'\left({z}\right)=−\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{z}^{\mathrm{2}} {e}^{−{tz}} }{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z}=−\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{tz}} \mathrm{d}{z}+\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{e}^{−{zt}} }{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z}= \\ $$$$−\frac{\mathrm{1}}{{t}}+\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{e}^{−{zt}} }{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z} \\ $$$${J}''\left({t}\right)=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }−{J}\left({t}\right) \\ $$$${J}\left({t}\right)+{J}''\left({t}\right)=\frac{\mathrm{1}}{{t}^{\mathrm{2}} } \\ $$$${J}\left({t}\right)={J}_{\mathrm{1}} \left({t}\right)\mathrm{sin}\left({t}\right) \\ $$$${J}_{\mathrm{1}} ''\left({t}\right)\mathrm{sin}\left({t}\right)+\mathrm{2}{J}_{\mathrm{1}} '\left({t}\right)\mathrm{cos}\left({t}\right)=\frac{\mathrm{1}}{{t}^{\mathrm{2}} } \\ $$$${J}_{\mathrm{2}} \left({t}\right)={J}_{\mathrm{1}} '\left({t}\right) \\ $$$${J}_{\mathrm{2}} '\left({t}\right)\mathrm{sin}\left({t}\right)+\mathrm{2}{J}_{\mathrm{2}} \left({t}\right)\mathrm{cos}\left({t}\right)=\frac{\mathrm{1}}{{t}^{\mathrm{2}} } \\ $$$${J}_{\mathrm{2}} \left({t}\right)=\frac{{J}_{\mathrm{3}} \left({t}\right)}{\mathrm{sin}^{\mathrm{2}} \left({t}\right)} \\ $$$$\frac{{J}_{\mathrm{3}} '\left({t}\right)}{\mathrm{sin}\left({t}\right)}=\frac{\mathrm{1}}{{t}^{\mathrm{2}} } \\ $$$${J}_{\mathrm{3}} '\left({t}\right)=\frac{\mathrm{sin}\left({t}\right)}{{t}^{\mathrm{2}} } \\ $$$${J}_{\mathrm{3}} \left({t}\right)=\int\frac{\mathrm{sin}\left({t}\right)}{{t}^{\mathrm{2}} }\mathrm{d}{t}=−\int\left(\frac{\mathrm{1}}{{t}}\right)'\mathrm{sin}\left({t}\right)\mathrm{d}{t}=−\frac{\mathrm{sin}\left({t}\right)}{{t}}+\mathrm{Ci}\left(\mathrm{t}\right)+{A} \\ $$$${J}_{\mathrm{2}} \left({t}\right)={J}_{\mathrm{1}} '\left({t}\right)=−\frac{\mathrm{1}}{{t}\mathrm{sin}\left({t}\right)}+\frac{\mathrm{Ci}\left({t}\right)}{\mathrm{sin}^{\mathrm{2}} \left({t}\right)}+\frac{{A}}{\mathrm{sin}^{\mathrm{2}} \left({t}\right)} \\ $$$${J}_{\mathrm{1}} \left({t}\right)=\int\left[−\frac{\mathrm{1}}{{t}\mathrm{sin}\left({t}\right)}+\frac{\mathrm{Ci}\left({t}\right)}{\mathrm{sin}^{\mathrm{2}} \left({t}\right)}\right]\mathrm{d}{t}+{A}\mathrm{cot}\left({t}\right) \\ $$$$=−\int\frac{\mathrm{d}{t}}{{t}\mathrm{sin}\left({t}\right)}−\int\mathrm{cot}'\left({t}\right)\mathrm{Ci}\left({t}\right)\mathrm{d}{t}+{A}\mathrm{cot}\left({t}\right)= \\ $$$$=−\int\frac{\mathrm{d}{t}}{{t}\mathrm{sin}\left({t}\right)}−\mathrm{cot}\left({t}\right)\mathrm{Ci}\left({t}\right)+\int\frac{\mathrm{cos}^{\mathrm{2}} \left({t}\right)}{{t}\mathrm{sin}\left({t}\right)}\mathrm{d}{t}+{A}\mathrm{cot}\left({t}\right)= \\ $$$$−\mathrm{Si}\left({t}\right)−\mathrm{cot}\left({t}\right)\mathrm{Ci}\left({t}\right)+{A}\mathrm{cot}\left({t}\right)+{B} \\ $$$${J}\left({t}\right)=−\mathrm{sin}\left({t}\right)\mathrm{Si}\left({t}\right)−\mathrm{cos}\left({t}\right)\mathrm{Ci}\left({t}\right)+{A}\mathrm{sin}\left({t}\right)+{B}\mathrm{cos}\left({t}\right) \\ $$$$\underset{{t}\rightarrow\infty} {\mathrm{lim}}{J}\left({t}\right)=\mathrm{0}\Rightarrow{B}=\mathrm{0}\wedge{A}=\frac{\pi}{\mathrm{2}} \\ $$$${J}\left({t}\right)=\mathrm{sin}\left({t}\right)\left(\frac{\pi}{\mathrm{2}}−\mathrm{Si}\left(\mathrm{t}\right)\right) \\ $$$${I}^{\left(\mathrm{2}\right)} \left({t}\right)+{I}\left({t}\right)=\frac{\mathrm{2}}{{t}}\mathrm{sin}\left({t}\right)\left(\frac{\pi}{\mathrm{2}}−\mathrm{Si}\left({t}\right)\right) \\ $$$$\mathrm{Did}\:\mathrm{this}\:\mathrm{help}? \\ $$
Commented by SdC355 last updated on 16/May/25
mmm...... thxs
$$\mathrm{mmm}……\:\mathrm{thxs} \\ $$

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