Evaluate integral: ∫0∞ln⁡x1+x2dx

Question

Evaluate integral:  ∫0∞ln⁡x1+x2dx

Pademagk71 · Accepted Answer

In general I(α)=∫0∞ln⁡xx2+α2dx can be evaluated by using substitution u=α2x⇒x=α2u⇒dx=−α2u2du, then I(α)=∫0∞ln⁡(α2u)(α2u)2+α2⋅α2u2du =∫0∞2ln⁡α−ln⁡uα2+u2du =2ln⁡α∫0∞1α2+u2du−∫0∞ln⁡uu2+α2du =2ln⁡α∫0∞1α2+u2du−I(α) I(α)=ln⁡α∫0∞1α2+u2du The last integral can easily be evaluated since it is a common integral. Using substitution u=tan⁡θ, the integral turns out to be I(α)=ln⁡αα∫0π2dθ =πln⁡α2α Thus I(1)=∫0∞ln⁡xx2+1dx=0

Ethan Sanders · Accepted Answer

Here&#039;s another general result, consider for |a|≤1: I(a,b)=∫0∞xab2+x2dx =ba−12∫0∞ta−121+tdt (1) =ba−12B(1+a2,1−a2) (2) =ba−12πcos⁡(π2a) (3) (1) is by subst. x=bt (2) is by the definition of Beta function: B(a,b)=∫0∞xa−1(1+x)a+bdx (3) is by using B(a,b)=Γ(a)Γ(b)Γ(a+b) and Euler&#039;s reflection formula Γ(a)Γ(1−a)=πsin⁡(πa) Hence, ∫0∞xa(log⁡x)nb2+x2dx=dndan(ba−12πcos⁡(π2a)) and when b=1,n=1,a=0, the result is 0

karton · Accepted Answer

Another general approach : Consider ∫0∞xb−1a2+x2dx (1) Rewrite (1) as ∫0∞xb−1a2+x2dx=1a2∫0∞xb−11+(xa)2dx (2) Putting x=ay⇒dx=ady yields ∫0∞xb−1a2+x2dx=ab−2∫0∞yb−11+y2dy where ∫0∞yb−11+y2dy=π2sin⁡(bπ2) Hence ∫0∞xb−1a2+x2dx=π2⋅ab−2sin⁡(bπ2) (3) Differentiating (3) with respect to b and setting b=1 yields ∫0∞ddb[xb−1a2+x2]b=1dx=π2ddb[ab−2sin⁡(bπ2)]b=1 ∫0∞ln⁡xx2+a2dx=πln⁡a2α Thus ∫0∞ln⁡xx2+1dx=0

Evaluate integral: \int_0^\infty \frac{\ln x}{1+x^2}dx

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