Question Number 104644 by Rohit@Thakur last updated on 22/Jul/20

∫_0 ^1 ((log(1−x+x^2 −x^3 +x^4 )dx)/x) = −(π^2 /(15))

Answered by mathmax by abdo last updated on 23/Jul/20

I =∫_0 ^1  ((ln(1−x+x^2 −x^3  +x^4 ))/x)dx ⇒I =∫_0 ^1  ((ln(((1−(−x)^5 )/(1−(−x)))))/x)dx  =∫_0 ^1  ((ln(1+x^5 )−ln(1+x))/x)dx =∫_0 ^1  ((ln(1+x^5 ))/x)dx−∫_0 ^1  ((ln(1+x))/x)dx  we have ln^′ (1+u) =(1/(1+u)) =Σ_(n=0) ^∞  (−1)^n  u^n    for ∣u∣<1 ⇒  ln(1+u)=Σ_(n=0) ^∞  (((−1)^n  u^(n+1) )/(n+1)) +c(c=0) =Σ_(n=1) ^∞  (((−1)^(n−1) u^n )/n) ⇒  ((ln(1+x))/x) =Σ_(n=1) ^∞  (((−1)^(n−1) )/n) x^(n−1)  ⇒∫_0 ^(1 )  ((ln(1+x))/x)dx =Σ_(n=1) ^∞  (((−1)^(n−1) )/n^2 )  =−Σ_(n=1) ^∞  (((−1)^n )/n^2 ) =−δ(2) =−(2^(1−2) −1)ξ(2) =−(−(1/2))×(π^2 /6) =(π^2 /(12))  ln(1+x^5 ) =Σ_(n=1) ^∞  (((−1)^(n−1) )/n) x^(5n)  ⇒((ln(1+x^5 ))/x) =Σ_(n=1) ^∞  (((−1)^(n−1) )/n) x^(5n−1)  ⇒  ∫_0 ^1  ((ln(1+x^5 ))/x)dx =Σ_(n=1) ^∞  (((−1)^(n−1) )/(n(5n))) =−(1/5) Σ_(n=1) ^∞  (((−1)^n )/n^2 ) =(1/5)×(π^2 /(12)) =(π^2 /(60))  ⇒ I =(π^2 /(60))−(π^2 /(12)) =((1/(60))−(1/(12)))π^2  =((1−5)/(60))×π^2  =−((4π^2 )/(60)) =−(π^2 /(15))  the result is proved.