Question Number 148498 by mathmax by abdo last updated on 28/Jul/21
Answered by mathmax by abdo last updated on 29/Jul/21
![Ψ=∫_0 ^∞ ((arctan(2x))/(1+x^2 ))dx let f(a)=∫_0 ^∞ ((arctan(ax))/(1+x^2 ))dx ⇒ f^′ (a)=∫_0 ^∞ (x/((1+a^2 x^2 )(x^2 +1)))dx =_(ax=y) ∫_0 ^∞ (y/(a(1+y^2 )((y^2 /a^2 ) +1)))(dy/a) =∫_0 ^∞ ((ydy)/((y^2 +1)(y^2 +a^2 )))=(1/(a^2 −1))∫_0 ^∞ y((1/(y^2 +1))−(1/(y^2 +a^2 )))dy =(1/(2(a^2 −1))){∫_0 ^∞ ((2ydy)/(y^2 +1))−∫_0 ^∞ ((2ydy)/(y^2 +a^2 ))} =(1/(2(a^2 −1))){log∣((y^2 +1)/(y^2 +a^2 ))∣]_0 ^(∞ ) =(1/(2(a^2 −1)))(−log((1/a^2 ))) =((2aloga)/(2(a^2 −1)))=((aloga)/(a^2 −1)) ⇒∫_0 ^2 f^′ (a)da =f(2)=∫_0 ^2 ((aloga)/(a^2 −1))da =(1/2)∫_0 ^2 ((1/(a−1))+(1/(a+1)))loga da =(1/2)∫_0 ^2 ((loga)/(a−1)) +(1/2)∫_0 ^2 ((loga)/(a+1))da we have ∫_0 ^2 ((loga)/(a−1))da =∫_0 ^1 ((loga)/(a−1))da +∫_1 ^2 ((loga)/(a−1))da(→a−1=t) =−∫_0 ^1 ((loga)/(1−a))da +∫_0 ^1 ((log(t+1))/t)dt but ∫_0 ^1 ((loga)/(1−a))da =∫_0 ^1 logaΣ_(n=0) ^∞ a^(n ) da =Σ_(n=0) ^∞ ∫_0 ^1 a^n logada and A_n =∫_0 ^1 a^n loga da =[(a^(n+1) /(n+1))loga]_0 ^1 −∫_0 ^1 (a^n /(n+1))da =−(1/((n+1)^2 )) ⇒ ∫_0 ^1 ((loga)/(1−a))da=−Σ_(n=0) ^∞ (1/((n+1)^2 ))=−(π^2 /6) log^′ (t+1)=(1/(1+t))=Σ_(n=0) ^∞ (−1)^n t^n ⇒log(1+t)=Σ_(n=0) ^∞ (((−1)^n )/(n+1))t^(n+1) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n)t^n ⇒((log(t+1))/t)=Σ_(n=1) ^∞ (((−1)^(n−1) )/n)t^(n−1) ⇒ ∫_0 ^1 ((log(1+t))/t)dt =Σ_(n=1) ^∞ (((−1)^(n−1) )/n)∫_0 ^1 t^(n−1) dt =Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ) =(1−2^(1−2) )ξ(2)=(1/2).(π^2 /6)=(π^2 /(12)) ∫_0 ^2 ((loga)/(a+1))da =∫_0 ^1 ((loga)/(a+1))da +∫_1 ^2 ((loga)/(a+1))da =∫_0 ^1 logaΣ_(n=0) ^∞ (−a)^n +∫_1 ^2 ((loga)/(a+1))da....be clntinued...](Q148542.png)
Commented bypuissant last updated on 29/Jul/21
Question Number 148498 by mathmax by abdo last updated on 28/Jul/21 | ||
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Answered by mathmax by abdo last updated on 29/Jul/21 | ||
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Commented bypuissant last updated on 29/Jul/21 | ||
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