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Question Number 60727 by Mr X pcx last updated on 25/May/19
calculate ∫_1 ^(+∞) ((ln(lnx))/(1+x^2 ))dx
$${calculate}\:\int_{\mathrm{1}} ^{+\infty} \:\frac{{ln}\left({lnx}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$
Commented by maxmathsup by imad last updated on 26/May/19
let A =∫_1 ^(+∞) ((ln(lnx))/(1+x^2 )) dx changement ln(x)=t give A =∫_0 ^(+∞) ((ln(t))/(1+e^(2t) )) e^t dt =∫_0 ^∞ ((e^(−t) ln(t))/(1+e^(−2t) )) dt =∫_0 ^∞ e^(−t) ln(t)(Σ_(n=0) ^∞ (−1)^n e^(−2nt) )dt =Σ_(n=0) ^∞ (−1)^n ∫_0 ^∞ e^(−(2n+1)t) ln(t)dt =Σ_(n=0) ^∞ (−1)^n A_n A_n =∫_0 ^∞ e^(−(2n+1)t) ln(t)dt =_((2n+1)t =u) ∫_0 ^∞ e^(−u) ln((u/(2n+1)))(du/(2n+1)) =(1/(2n+1)) ∫_0 ^∞ e^(−u) {ln(u)−ln(2n+1))du =(1/(2n+1)){ ∫_0 ^∞ e^(−u) ln(u)du −ln(2n+1)∫_0 ^∞ e^(−u) du} ∫_0 ^∞ e^(−u) ln(u)du =−γ (result proved) ∫_0 ^∞ e^(−u) du =[−e^(−u) ]_0 ^∞ =1 ⇒A_n =((−γ)/(2n+1)) −((ln(2n+1))/(2n+1)) ⇒ A =Σ_(n=0) ^∞ (−1)^n {−(γ/(2n+1)) −((ln(2n+1))/(2n+1))} =−γ Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) −Σ_(n=0) ^∞ (((−1)^n )/(2n+1))ln(2n+1) =−((γπ)/4) −Σ_(n=0) ^∞ (−1)^n ((ln(2n+1))/(2n+1)) with γ constant of Euler ...be continued...
$${let}\:{A}\:=\int_{\mathrm{1}} ^{+\infty} \:\frac{{ln}\left({lnx}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:\:{changement}\:{ln}\left({x}\right)={t}\:{give} \\ $$$${A}\:=\int_{\mathrm{0}} ^{+\infty} \:\frac{{ln}\left({t}\right)}{\mathrm{1}+{e}^{\mathrm{2}{t}} }\:{e}^{{t}} \:{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−{t}} {ln}\left({t}\right)}{\mathrm{1}+{e}^{−\mathrm{2}{t}} }\:{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−{t}} {ln}\left({t}\right)\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{e}^{−\mathrm{2}{nt}} \right){dt} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right){t}} \:{ln}\left({t}\right){dt}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:{A}_{{n}} \\ $$$${A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right){t}} {ln}\left({t}\right){dt}\:=_{\left(\mathrm{2}{n}+\mathrm{1}\right){t}\:={u}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−{u}} {ln}\left(\frac{{u}}{\mathrm{2}{n}+\mathrm{1}}\right)\frac{{du}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{u}} \left\{{ln}\left({u}\right)−{ln}\left(\mathrm{2}{n}+\mathrm{1}\right)\right){du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\left\{\:\:\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{u}} {ln}\left({u}\right){du}\:−{ln}\left(\mathrm{2}{n}+\mathrm{1}\right)\int_{\mathrm{0}} ^{\infty} \:{e}^{−{u}} \:{du}\right\} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−{u}} {ln}\left({u}\right){du}\:=−\gamma\:\:\left({result}\:{proved}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−{u}} \:{du}\:=\left[−{e}^{−{u}} \right]_{\mathrm{0}} ^{\infty} \:=\mathrm{1}\:\Rightarrow{A}_{{n}} =\frac{−\gamma}{\mathrm{2}{n}+\mathrm{1}}\:−\frac{{ln}\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{2}{n}+\mathrm{1}}\:\Rightarrow \\ $$$${A}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \left\{−\frac{\gamma}{\mathrm{2}{n}+\mathrm{1}}\:−\frac{{ln}\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{2}{n}+\mathrm{1}}\right\} \\ $$$$=−\gamma\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}{ln}\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$$=−\frac{\gamma\pi}{\mathrm{4}}\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:\frac{{ln}\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{2}{n}+\mathrm{1}}\:\:\:\:{with}\:\gamma\:{constant}\:{of}\:{Euler}\:…{be}\:{continued}… \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 26/May/19
let try another way by use of chang. x =(1/t) ⇒ I = ∫_1 ^(+∞) ((ln(lnx))/(1+x^2 )) dx =−∫_0 ^1 ((ln(ln((1/t))))/(1+(1/t^2 ))) (−(dt/t^2 )) = ∫_0 ^1 ((ln(ln((1/t))))/(1+t^2 )) dt =∫_0 ^1 ((ln(ln((1/t))))/((1+it)(1−it)))dt = (1/2)∫_0 ^1 ((1/(1+it)) +(1/(1−it)))ln(ln((1/t)))dt =(1/2) ∫_0 ^1 ((ln{ln((1/t))})/(1+it))dt +(1/2) ∫_0 ^1 ((ln{ln((1/t))})/(1−it)) dt let W =∫_0 ^1 ((ln{ln((1/t))})/(1+it)) dt chang . ln((1/t)) =u give (1/t) =e^u ⇒t =e^(−u) ⇒ W =−∫_0 ^∞ ((lnu)/(1+ie^(−u) )) (−e^(−u) )du = ∫_0 ^∞ ((e^(−u) ln(u))/(1+ie^(−u) )) du =∫_0 ^∞ e^(−u) ln(u)(Σ_(n=0) ^∞ (−ie^(−u) )^n )du =Σ_(n=0) ^∞ (−i)^n ∫_0 ^∞ e^(−(n+1)u) ln(u)du ∫_0 ^∞ e^(−(n+1)u) ln(u)du =_((n+1)u=t) ∫_0 ^∞ e^(−t) ln((t/(n+1)))(dt/(n+1)) =(1/(n+1)) ∫_0 ^∞ { e^(−t) ln(t)−e^(−t) ln(n+1)}dt =(1/(n+1)) ∫_0 ^∞ e^(−t) ln(t) −((ln(n+1))/(n+1)) ∫_0 ^∞ e^(−t) dt =((−γ)/(n+1)) −((ln(n+1))/(n+1)) ⇒ W =Σ_(n=0) ^∞ ((−γ(−i)^n )/(n+1)) −Σ_(n=0) ^∞ (((−i)^n ln(n+1))/(n+1)) H =∫_0 ^1 ((ln{ln((1/t))})/(1−it ))dt =_(ln((1/t))=u) −∫_0 ^∞ ((ln(u))/(1−i e^(−u) ))(−e^(−u) )du =∫_0 ^∞ ((e^(−u) ln(u))/(1−ie^(−u) )) du =∫_0 ^∞ e^(−u) lnu {Σ_(n=0) ^∞ i^n e^(−nu) }du =Σ_(n=0) ^∞ i^n ∫_0 ^∞ e^(−(n+1)u) ln(u)du =....be continued....
$${let}\:{try}\:{another}\:{way}\:\:\:{by}\:{use}\:{of}\:{chang}.\:{x}\:=\frac{\mathrm{1}}{{t}}\:\Rightarrow \\ $$$${I}\:=\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{ln}\left({lnx}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({ln}\left(\frac{\mathrm{1}}{{t}}\right)\right)}{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}\:\left(−\frac{{dt}}{{t}^{\mathrm{2}} }\right) \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{ln}\left({ln}\left(\frac{\mathrm{1}}{{t}}\right)\right)}{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left({ln}\left(\frac{\mathrm{1}}{{t}}\right)\right)}{\left(\mathrm{1}+{it}\right)\left(\mathrm{1}−{it}\right)}{dt} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\left(\frac{\mathrm{1}}{\mathrm{1}+{it}}\:+\frac{\mathrm{1}}{\mathrm{1}−{it}}\right){ln}\left({ln}\left(\frac{\mathrm{1}}{{t}}\right)\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left\{{ln}\left(\frac{\mathrm{1}}{{t}}\right)\right\}}{\mathrm{1}+{it}}{dt}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left\{{ln}\left(\frac{\mathrm{1}}{{t}}\right)\right\}}{\mathrm{1}−{it}}\:{dt}\:\:{let}\:{W}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left\{{ln}\left(\frac{\mathrm{1}}{{t}}\right)\right\}}{\mathrm{1}+{it}}\:{dt} \\ $$$${chang}\:.\:{ln}\left(\frac{\mathrm{1}}{{t}}\right)\:={u}\:{give}\:\frac{\mathrm{1}}{{t}}\:={e}^{{u}} \:\Rightarrow{t}\:={e}^{−{u}} \:\Rightarrow \\ $$$${W}\:=−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{lnu}}{\mathrm{1}+{ie}^{−{u}} }\:\left(−{e}^{−{u}} \right){du}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{u}} {ln}\left({u}\right)}{\mathrm{1}+{ie}^{−{u}} }\:{du} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{u}} {ln}\left({u}\right)\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−{ie}^{−{u}} \right)^{{n}} \right){du} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−{i}\right)^{{n}} \:\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−\left({n}+\mathrm{1}\right){u}} {ln}\left({u}\right){du} \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−\left({n}+\mathrm{1}\right){u}} {ln}\left({u}\right){du}\:=_{\left({n}+\mathrm{1}\right){u}={t}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}} {ln}\left(\frac{{t}}{{n}+\mathrm{1}}\right)\frac{{dt}}{{n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\infty} \:\:\left\{\:{e}^{−{t}} {ln}\left({t}\right)−{e}^{−{t}} {ln}\left({n}+\mathrm{1}\right)\right\}{dt} \\ $$$$=\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−{t}} {ln}\left({t}\right)\:−\frac{{ln}\left({n}+\mathrm{1}\right)}{{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−{t}} \:{dt} \\ $$$$=\frac{−\gamma}{{n}+\mathrm{1}}\:−\frac{{ln}\left({n}+\mathrm{1}\right)}{{n}+\mathrm{1}}\:\Rightarrow\:{W}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{−\gamma\left(−{i}\right)^{{n}} }{{n}+\mathrm{1}}\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−{i}\right)^{{n}} {ln}\left({n}+\mathrm{1}\right)}{{n}+\mathrm{1}} \\ $$$${H}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left\{{ln}\left(\frac{\mathrm{1}}{{t}}\right)\right\}}{\mathrm{1}−{it}\:}{dt}\:=_{{ln}\left(\frac{\mathrm{1}}{{t}}\right)={u}} \:\:\:\:−\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left({u}\right)}{\mathrm{1}−{i}\:{e}^{−{u}} }\left(−{e}^{−{u}} \right){du} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−{u}} {ln}\left({u}\right)}{\mathrm{1}−{ie}^{−{u}} }\:{du}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:{e}^{−{u}} {lnu}\:\left\{\sum_{{n}=\mathrm{0}} ^{\infty} \:{i}^{{n}} \:{e}^{−{nu}} \right\}{du} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:{i}^{{n}} \:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left({n}+\mathrm{1}\right){u}} \:{ln}\left({u}\right){du}\:=….{be}\:{continued}…. \\ $$

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