Question Number 149986 by mnjuly1970 last updated on 08/Aug/21

Answered by mindispower last updated on 08/Aug/21

ln(1+e^x )=x+ln(e^(−x) +1) I=∫_0 ^∞ (x/(1+e^(2x) ))dx+∫_0 ^∞ ((e^(−2x) ln(1+e^(−x) ))/(1+e^(−2x) ))dx =(1/4)∫_0 ^∞ ((tdt)/(1+e^t ))+∫_0 ^1 ((tln(1+t))/(1+t^2 ))dt =((Γ(2)ζ(2))/4) ∫_0 ^1 ((tln(1+t))/(1+t^2 ))dt done with ∫_0 ^1 ((ln(1+(√x)))/(1+x))dx Quation =posted

Commented bymnjuly1970 last updated on 08/Aug/21

thx alot sir power....mercey ∫_0 ^( ∞) (( x)/(1+e^( x) )) dx =η(2)Γ(2) = (1/2) ζ (2) Γ (2)...