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Question Number 139524 by mnjuly1970 last updated on 28/Apr/21
......advanced calculus..... prove that: i::𝛗=∫_0 ^( ∞) ((1/2)e^(βˆ’2x) βˆ’(1/(1+e^x )))(1/x) dx=log((1/( (βˆšΟ€))) ) ii::∫_0 ^( ∞) ((1/2)βˆ’(1/(1+e^(βˆ’x) )))(e^(βˆ’2x) /x)dx=log(((βˆšΟ€)/2))
$$\:\:\:\:\:\:\:\:\:\:\:\:……{advanced}\:\:{calculus}….. \\ $$$$\:\:\:\:\:\:\:\:\:{prove}\:\:{that}: \\ $$$$\:{i}::\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \left(\frac{\mathrm{1}}{\mathrm{2}}{e}^{βˆ’\mathrm{2}{x}} βˆ’\frac{\mathrm{1}}{\mathrm{1}+{e}^{{x}} }\right)\frac{\mathrm{1}}{{x}}\:{dx}={log}\left(\frac{\mathrm{1}}{\:\sqrt{\pi}}\:\right) \\ $$$$\:{ii}::\int_{\mathrm{0}} ^{\:\infty} \left(\frac{\mathrm{1}}{\mathrm{2}}βˆ’\frac{\mathrm{1}}{\mathrm{1}+{e}^{βˆ’{x}} }\right)\frac{{e}^{βˆ’\mathrm{2}{x}} }{{x}}{dx}={log}\left(\frac{\sqrt{\pi}}{\mathrm{2}}\right) \\ $$
Answered by mindispower last updated on 30/Apr/21
i..ii are similar in way this can solve by let βˆ…=∫_0 ^∞ ((e^(βˆ’2x) /2)βˆ’(e^(βˆ’x) /(1+e^(βˆ’x) ))).(dx/x) e^(βˆ’x) =tβ‡’Ο†=∫_0 ^1 ((1/(1+t))βˆ’(t/2))(dt/(ln(t))) let f(z)=∫_0 ^1 ((1/(1+t))βˆ’(t/2))(t^z /(ln(t)))dt fβ€²(z)=∫_0 ^1 ((t^z /(1+t))βˆ’(t^(1+z) /2))dt=βˆ’(1/(2(2+z)))+∫_0 ^1 ((t^z βˆ’t^(1+z) )/(1βˆ’t^2 ))dt t^2 β†’t in 2nd ⇔fβ€²(z)=βˆ’(1/(2(2+z)))+(1/2)(βˆ’Ξ³+∫_0 ^1 ((1βˆ’t^(z/2) )/(1βˆ’t)).dt) βˆ’(1/2)(βˆ’Ξ³+∫_0 ^1 ((1βˆ’t^((zβˆ’1)/2) )/(1βˆ’t))dt) Ξ¨(x+1)=βˆ’Ξ³+∫_0 ^1 ((1βˆ’t^x )/(1βˆ’t))dtβ‡’ fβ€²(z)=βˆ’(1/(2(2+z)))+(1/2)(Ξ¨((z/2)+1)βˆ’Ξ¨(((z+1)/2))) f(z)=βˆ’(1/2)ln(2+z)+log(((Ξ“((z/2)+1))/(Ξ“(((z+1)/2)))))+c Ξ“(z+a)∼(√(2Ο€)).z^(z+aβˆ’(1/2)) e^(βˆ’z) ((Ξ“((z/2)+1))/(Ξ“(((z+1)/2))))∼((((z/2))^((z/2)+(1/2)) )/((z)^(z/2) )) f(z)∼ln((1/( (√(2+z)))).(√(z/2)))+cβ‡’lim_(zβ†’βˆž) f(z)=ln((1/( (√2))))+c back to f(z)=∫_0 ^1 ((1/(1+t))βˆ’(t/2))(t^z /(ln(t)))dtβ‡’lim_(zβ†’βˆž) f(z)=0 β‡’c=ln((√2))⇔f(z)=∫_0 ^1 ((1/(1+t))βˆ’(t/2)).(t^z /(ln(t)))=ln((1/( (√(2+z))))) +log(((Ξ“((z/2)+1))/(Ξ“(((z+1)/2)))))+ln((√2)) f(0)=Ο†=ln((1/( (√2))))+ln((√2))+log(((Ξ“(1))/(Ξ“((1/2)))))=log((1/( (βˆšΟ€)))) ⇔φ=∫_0 ^∞ ((e^(βˆ’2x) /2)βˆ’(e^(βˆ’x) /(1+e^(βˆ’x) ))).(dx/x)=ln((1/( (βˆšΟ€))))
$${i}..{ii}\:{are}\:{similar}\:{in}\:{way}\:{this}\:{can}\:{solve}\:{by} \\ $$$${let}\:\emptyset=\int_{\mathrm{0}} ^{\infty} \left(\frac{{e}^{βˆ’\mathrm{2}{x}} }{\mathrm{2}}βˆ’\frac{{e}^{βˆ’{x}} }{\mathrm{1}+{e}^{βˆ’{x}} }\right).\frac{{dx}}{{x}} \\ $$$${e}^{βˆ’{x}} ={t}\Rightarrow\phi=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+{t}}βˆ’\frac{{t}}{\mathrm{2}}\right)\frac{{dt}}{{ln}\left({t}\right)} \\ $$$${let}\:{f}\left({z}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+{t}}βˆ’\frac{{t}}{\mathrm{2}}\right)\frac{{t}^{{z}} }{{ln}\left({t}\right)}{dt} \\ $$$${f}'\left({z}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{{t}^{{z}} }{\mathrm{1}+{t}}βˆ’\frac{{t}^{\mathrm{1}+{z}} }{\mathrm{2}}\right){dt}=βˆ’\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}+{z}\right)}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{z}} βˆ’{t}^{\mathrm{1}+{z}} }{\mathrm{1}βˆ’{t}^{\mathrm{2}} }{dt} \\ $$$${t}^{\mathrm{2}} \rightarrow{t}\:{in}\:\mathrm{2}{nd}\:\Leftrightarrow{f}'\left({z}\right)=βˆ’\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}+{z}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\left(βˆ’\gamma+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}βˆ’{t}^{\frac{{z}}{\mathrm{2}}} }{\mathrm{1}βˆ’{t}}.{dt}\right) \\ $$$$βˆ’\frac{\mathrm{1}}{\mathrm{2}}\left(βˆ’\gamma+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}βˆ’{t}^{\frac{{z}βˆ’\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}βˆ’{t}}{dt}\right) \\ $$$$\Psi\left({x}+\mathrm{1}\right)=βˆ’\gamma+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}βˆ’{t}^{{x}} }{\mathrm{1}βˆ’{t}}{dt}\Rightarrow \\ $$$${f}'\left({z}\right)=βˆ’\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}+{z}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\left(\Psi\left(\frac{{z}}{\mathrm{2}}+\mathrm{1}\right)βˆ’\Psi\left(\frac{{z}+\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$${f}\left({z}\right)=βˆ’\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}+{z}\right)+{log}\left(\frac{\Gamma\left(\frac{{z}}{\mathrm{2}}+\mathrm{1}\right)}{\Gamma\left(\frac{{z}+\mathrm{1}}{\mathrm{2}}\right)}\right)+{c} \\ $$$$\Gamma\left({z}+{a}\right)\sim\sqrt{\mathrm{2}\pi}.{z}^{{z}+{a}βˆ’\frac{\mathrm{1}}{\mathrm{2}}} {e}^{βˆ’{z}} \\ $$$$\frac{\Gamma\left(\frac{{z}}{\mathrm{2}}+\mathrm{1}\right)}{\Gamma\left(\frac{{z}+\mathrm{1}}{\mathrm{2}}\right)}\sim\frac{\left(\frac{{z}}{\mathrm{2}}\right)^{\frac{{z}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}} }{\left({z}\right)^{\frac{{z}}{\mathrm{2}}} } \\ $$$${f}\left({z}\right)\sim{ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}+{z}}}.\sqrt{\frac{{z}}{\mathrm{2}}}\right)+{c}\Rightarrow\underset{{z}\rightarrow\infty} {\mathrm{lim}}{f}\left({z}\right)={ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)+{c} \\ $$$${back}\:{to}\:{f}\left({z}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+{t}}βˆ’\frac{{t}}{\mathrm{2}}\right)\frac{{t}^{{z}} }{{ln}\left({t}\right)}{dt}\Rightarrow\underset{{z}\rightarrow\infty} {\mathrm{lim}}{f}\left({z}\right)=\mathrm{0} \\ $$$$\Rightarrow{c}={ln}\left(\sqrt{\mathrm{2}}\right)\Leftrightarrow{f}\left({z}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+{t}}βˆ’\frac{{t}}{\mathrm{2}}\right).\frac{{t}^{{z}} }{{ln}\left({t}\right)}={ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}+{z}}}\right) \\ $$$$+{log}\left(\frac{\Gamma\left(\frac{{z}}{\mathrm{2}}+\mathrm{1}\right)}{\Gamma\left(\frac{{z}+\mathrm{1}}{\mathrm{2}}\right)}\right)+{ln}\left(\sqrt{\mathrm{2}}\right) \\ $$$${f}\left(\mathrm{0}\right)=\phi={ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)+{ln}\left(\sqrt{\mathrm{2}}\right)+{log}\left(\frac{\Gamma\left(\mathrm{1}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}\right)={log}\left(\frac{\mathrm{1}}{\:\sqrt{\pi}}\right) \\ $$$$\Leftrightarrow\phi=\int_{\mathrm{0}} ^{\infty} \left(\frac{{e}^{βˆ’\mathrm{2}{x}} }{\mathrm{2}}βˆ’\frac{{e}^{βˆ’{x}} }{\mathrm{1}+{e}^{βˆ’{x}} }\right).\frac{{dx}}{{x}}={ln}\left(\frac{\mathrm{1}}{\:\sqrt{\pi}}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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