Question Number 103190 by bemath last updated on 13/Jul/20

∫_(π/4) ^(π/2) ln(ln(tan x)) dx

Answered by abdomathmax last updated on 13/Jul/20

ln(tanx) =t ⇒ I =∫_(π/4) ^(π/2) ln(ln(tanx))dx  tanx =e^(t )  ⇒x =arctan(e^t ) ⇒  I =∫_0 ^(+∞)  ln(t) ×(e^t /(1+e^(2t) )) dt =∫_0 ^∞  ((e^t ln(t))/(1+e^(2t) ))dt  =∫_0 ^∞  ((e^(−t)  ln(t))/(1+e^(−2x) )) dt =∫_0 ^∞  e^(−t) ln(t)(Σ_(n=0) ^(∞ ) (−1)^(n )  e^(−(nt) )dt  =Σ_(n=0) ^∞  (−1)^(n )  ∫_0 ^∞  e^(−(n+1)t) ln(t)dt  A_n =∫_0 ^∞  e^(−(n+1)t)  ln(t)dt =_((n+1)t=u)  ∫_0 ^∞ e^(−u) ln((u/(n+1)))(du/(n+1))  =(1/(n+1)) ∫_0 ^∞  {e^(−u) ln(u)−e^(−u) ln(n+1))du  =−(γ/(n+1))−((ln(n+1))/(n+1)) ∫_0 ^∞ e^(−u ) du  =−(γ/(n+1))−((ln(n+1))/(n+1)) ⇒  I =−γ Σ_(n=0) ^(∞ )  (((−1)^n )/(n+1)) −Σ_(n=0) ^∞  (((−1)^n )/(n+1))ln(n+1)  Σ_(n=0) ^∞ (((−1)^n )/(n+1)) =Σ_(n=1) ^(∞ )  (((−1)^(n−1) )/n) =ln(2)  Σ_(n=0) ^∞  (((−1)^n )/(n+1))ln(n+1) =Σ_(n=1) ^∞  (((−1)^(n−1) )/n)ln(n)  ⇒I =−γln(2) +Σ_(n=1) ^∞  (((−1)^n lnn)/n)  ...be continued...

Commented byabdomathmax last updated on 13/Jul/20

i dont remember the value of Σ_(n=1) ^∞  (((−1)^n ln(n))/n)  take a look at the platform...