calculus-I-complex-analysis-calculate-0-ln-x-x-2-3x-2-dx-ln-2-2-2-

Question Number 127224 by mnjuly1970 last updated on 28/Dec/20

\dots c a l c u l u s (I) - c o m p l e x a n a l y s i s \dots

c a l c u l a t e ::

Φ = \int_{0}^{\infty} \frac{l n (x)}{x^{2} + 3 x + 2} d x = \frac{{l n}^{2} (2)}{2}

Answered by mindispower last updated on 27/Dec/20

f (z) = \frac{l n (z)}{z^{2} + 2 z + 3}, a = - 1 - i \sqrt{2}

\int_{0}^{\infty} \frac{{l n}^{2} (z)}{z^{2} + 2 z + 3} - \int_{0}^{\infty} \frac{{(l n (z) + 2 i π)}^{2}}{z^{2} + 2 z + 3} d z = 2 i π R e s (f (z), a, \bar{a})

= - 4 i π \int_{0}^{\infty} \frac{l n (z)}{z^{2} + 2 z + 3} + \int_{0}^{\infty} \frac{4 π^{2} d z}{z^{2} + 2 z + 3} = 2 i π R e s (f, \bar{a}, a)

\Rightarrow \int_{0}^{\infty} \frac{l n (z)}{z^{2} + 2 z + 3} d z = - \frac{1}{2} R e (R e s (f, a, \bar{a}))

R e s (f, a, \bar{a}) = \frac{l n (a)}{2 a + 2} + \frac{l n (\bar{a})}{2 \bar{a} + 2}

l n (- 1 - i \sqrt{2}) = l n (\sqrt{3}) + i (a r c t a n (\sqrt{2}) + π)

l n (- 1 + i \sqrt{2}) = l n (\sqrt{3}) + i (π - a r c t a n (\sqrt{2}))

= \frac{(l n (\sqrt{3}) + i {(π + a r c t a n (\sqrt{2}))}^{2}}{- 2 i \sqrt{2}} + \frac{{(l n (\sqrt{3}) + i (π - a r c t a n (\sqrt{2})))}^{2}}{2 i \sqrt{2}}

= 2 \frac{- i l n (\sqrt{3}) a r c t a n (\sqrt{2}) + π a r c t a n (\sqrt{2})}{i \sqrt{2}}

\int_{0}^{\infty} \frac{l n (x)}{x^{2} + 2 x + 3} d x = \frac{l n (\sqrt{3}) a r c t a n (\sqrt{2})}{\sqrt{2}} ≃ 0, 371

Commented by mnjuly1970 last updated on 28/Dec/20

t h a n k y o u s o m u c h . .

Commented by mindispower last updated on 28/Dec/20

a l w a y s p l e a s u r

Answered by Bird last updated on 28/Dec/20

l e t φ (z) = \frac{{l n}^{2} z}{z^{2} + 2 z + 3} w e h a v e

\int_{0}^{\infty} \frac{{l n}^{2} x}{x^{2} + 2 x + 3} d x - \int_{0}^{\infty} \frac{{(l n x + 2 i π)}^{2}}{x^{2} + 2 x + 3} d x

= 2 i π {R e s (φ, z_{1}) + R e s (φ, z_{2})} (1)

z^{2} + 2 z + 3 = 0 \to Δ^{^{'}} = - 2 \Rightarrow

z_{1} = - 1 + i \sqrt{2} a n d z_{2} = - 1 - i \sqrt{2}

z_{1} = i^{2} + i \sqrt{2} = i (\sqrt{2} + i)

= i \sqrt{3} e^{i s r c t a n (\frac{1}{\sqrt{2}})}

= \sqrt{3} e^{\frac{i π}{2}} e^{i (\frac{π}{2} - a r c t a n (\sqrt{2}))}

= \sqrt{3} e^{i (π - a r c t a n (\sqrt{2}))}

z_{2} = i^{2} - i \sqrt{2} = i (- \sqrt{2} + i)

= e^{\frac{i π}{2}} \sqrt{3} e^{i a r c t a n (- \frac{1}{\sqrt{2}})}

= \sqrt{3} e^{\frac{i π}{2}} e^{- i (\frac{π}{2} - a r c t a (\sqrt{2}))}

= \sqrt{3} e^{i a r c t a n (\sqrt{2})} \Rightarrow

R e s (φ, z_{1}) = \frac{{l n}^{2} (z_{1})}{z_{1} - z_{2}}

= \frac{(l n (\sqrt{3}) + i {(π - a r c t a n \sqrt{2})}^{2}}{2 i \sqrt{2}} = \dots

R e s (φ, z_{2}) = \frac{{(l n (\sqrt{3}) + i a r c t a n (\sqrt{2}))}^{2}}{- 2 i \sqrt{2}}

(1) \Rightarrow - \int_{0}^{\infty} \frac{4 i π l n x - 4 π^{2}}{x^{2} + 2 x + 3} d x

= 2 i π Σ R e s (. .) \Rightarrow

\int_{0}^{\infty} \frac{l n x}{x^{2} + 2 x + 3} d x = - \frac{1}{2} R e (Σ R e s)

r e s t t o c a l c u l a t e Σ R e s (f, z_{1})

\dots b e v o n t i n u e d \dots

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