Question Number 112642 by mathdave last updated on 09/Sep/20
$${question}\:{proposed}\:{by}\:{A}\mathrm{8};\mathrm{15}: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{ln}\frac{\mathrm{1}}{{x}}\right)}{\mathrm{1}+{x}}{dx} \\ $$
Answered by mathdave last updated on 09/Sep/20
![my solution goes first we need to convert from malsten′s integral to verdi′s integral let I=∫_0 ^1 ((ln(ln(1/x)))/(1+x))dx let t=ln((1/x)) x=e^(−t) and dx=−e^(−t) I=∫_∞ ^0 ((lnt)/(1+e^(−t) ))×−e^(−t) dt=∫_0 ^∞ ((lnt)/(1+e^(−t) ))e^(−t) dt I=∫_0 ^∞ ((lnt)/(1+e^t ))dt I=(∂/∂a)∣_(a=0) ∫_0 ^∞ (t^a /(1+e^t ))dt but note ∫_0 ^∞ (t^s /(1+e^t ))dt=η(s+1)Γ(s+1) I=(∂/∂a)∣_(a=0) [η(a+1)Γ(a+1)] I=[η′(a+1)Γ(a+1)+η(a+1)Γ^′ (a+1)]_(a=0) but Γ(a+1)=Γ(a+1)ψ(a+1) I=[η^′ (a+1)Γ(a+1)+η(a+1)Γ(a+1)ψ(a+1)]_(a=0) I=[η^′ (1)Γ(1)+η(1)Γ(1)ψ(1)] but η^′ (1)=−γln2−(1/2)ln^2 (2),Γ(1)=1,η(1)=ln2,ψ(1)=γ I=−γln2−(1/2)ln^2 (2)+γln2=−(1/2)ln^2 (2) ∵∫_0 ^1 ((ln(ln(1/x)))/(1+x))dx=−(1/2)ln^2 (2) by mathdave(09/08/2020)](https://www.tinkutara.com/question/Q112646.png)
$${my}\:{solution}\:{goes}\: \\ $$$${first}\:{we}\:{need}\:{to}\:{convert}\:{from} \\ $$$${malsten}'{s}\:{integral}\:{to}\:{verdi}'{s}\:{integral} \\ $$$${let}\:{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{ln}\frac{\mathrm{1}}{{x}}\right)}{\mathrm{1}+{x}}{dx}\:\:\:\:{let}\:{t}=\mathrm{ln}\left(\frac{\mathrm{1}}{{x}}\right) \\ $$$${x}={e}^{−{t}} \:\:{and}\:{dx}=−{e}^{−{t}} \\ $$$${I}=\int_{\infty} ^{\mathrm{0}} \frac{\mathrm{ln}{t}}{\mathrm{1}+{e}^{−{t}} }×−{e}^{−{t}} {dt}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}{t}}{\mathrm{1}+{e}^{−{t}} }{e}^{−{t}} {dt} \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}{t}}{\mathrm{1}+{e}^{{t}} }{dt} \\ $$$${I}=\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{0}} \int_{\mathrm{0}} ^{\infty} \frac{{t}^{{a}} }{\mathrm{1}+{e}^{{t}} }{dt} \\ $$$${but}\:{note}\:\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{s}} }{\mathrm{1}+{e}^{{t}} }{dt}=\eta\left({s}+\mathrm{1}\right)\Gamma\left({s}+\mathrm{1}\right) \\ $$$${I}=\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{0}} \left[\eta\left({a}+\mathrm{1}\right)\Gamma\left({a}+\mathrm{1}\right)\right] \\ $$$${I}=\left[\eta'\left({a}+\mathrm{1}\right)\Gamma\left({a}+\mathrm{1}\right)+\eta\left({a}+\mathrm{1}\right)\Gamma^{'} \left({a}+\mathrm{1}\right)\right]_{{a}=\mathrm{0}} \\ $$$${but}\:\Gamma\left({a}+\mathrm{1}\right)=\Gamma\left({a}+\mathrm{1}\right)\psi\left({a}+\mathrm{1}\right) \\ $$$${I}=\left[\eta^{'} \left({a}+\mathrm{1}\right)\Gamma\left({a}+\mathrm{1}\right)+\eta\left({a}+\mathrm{1}\right)\Gamma\left({a}+\mathrm{1}\right)\psi\left({a}+\mathrm{1}\right)\right]_{{a}=\mathrm{0}} \\ $$$${I}=\left[\eta^{'} \left(\mathrm{1}\right)\Gamma\left(\mathrm{1}\right)+\eta\left(\mathrm{1}\right)\Gamma\left(\mathrm{1}\right)\psi\left(\mathrm{1}\right)\right] \\ $$$${but} \\ $$$$\eta^{'} \left(\mathrm{1}\right)=−\gamma\mathrm{ln2}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right),\Gamma\left(\mathrm{1}\right)=\mathrm{1},\eta\left(\mathrm{1}\right)=\mathrm{ln2},\psi\left(\mathrm{1}\right)=\gamma \\ $$$${I}=−\gamma\mathrm{ln2}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+\gamma\mathrm{ln2}=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$$$\because\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{ln}\frac{\mathrm{1}}{{x}}\right)}{\mathrm{1}+{x}}{dx}=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$$${by}\:{mathdave}\left(\mathrm{09}/\mathrm{08}/\mathrm{2020}\right) \\ $$
Commented by mnjuly1970 last updated on 09/Sep/20
$${very}\:{nice}\:\:{mr}\:\:{bathdave}… \\ $$
Commented by Tawa11 last updated on 06/Sep/21
$$\mathrm{great}\:\mathrm{sir} \\ $$