calculste-0-ln-x-x-2-x-1-dx-

Question Number 108750 by mathmax by abdo last updated on 19/Aug/20

calculste \int_{0}^{\infty} \frac{\ln (x)}{x^{2} - x + 1} dx

Answered by mnjuly1970 last updated on 19/Aug/20

sol \dots . : put : x = \frac{1}{t} \Rightarrow Ω = \int_{0}^{\infty} \frac{l n (x)}{x^{2} - x + 1} d x = - Ω

2 Ω = 0 \Rightarrow Ω = \int_{0}^{\infty} \frac{l n (x)}{1 - x + x^{2}} d x = 0

\dots . . M . N \dots . .

Answered by mathmax by abdo last updated on 19/Aug/20

if q is a fraction with no real poles we have

\int_{0}^{\infty} q (x) lnx dx = - \frac{1}{2} Re (\sum_{i} Re (q (z) \ln^{2} z, a_{i}) let use this

we have q (z) \ln^{2} z = \frac{\ln^{2} z}{z^{2} - z + 1} = w (z) poles ?

z^{2} - z + 1 = 0 \to Δ = - 3 \Rightarrow z_{1} = \frac{1 + i \sqrt{3}}{2} = e^{\frac{i 2 π}{3}} and z_{2} = \frac{1 - i \sqrt{3}}{2} = e^{- \frac{i 2 π}{3}}

w (z) = \frac{\ln^{2} z}{(z - z_{1}) (z - z_{2})} \Rightarrow Res (w, z_{1}) = \frac{\ln^{2} (z_{1})}{z_{1} - z_{2}} = \frac{\ln^{2} (e^{\frac{i 2 π}{3}})}{i \sqrt{3}}

= \frac{1}{i \sqrt{3}} {(\frac{i 2 π}{3})}^{2} = \frac{- 4 π^{2}}{3 i \sqrt{3}} also Res (w, z_{2}) = \frac{\ln^{2} (z_{2})}{z_{2} - z_{1}} = \frac{4 π^{2}}{3 i \sqrt{3}} \Rightarrow

Σ Re (q (z) \ln^{2} (z), a_{i}) = 0 \Rightarrow \int_{0}^{\infty} \frac{\ln (x)}{x^{2} - x + 1} dx = 0