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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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