nice-calculus-evaluate-0-ln-x-x-2-1-3-dx-

Question Number 127539 by mnjuly1970 last updated on 30/Dec/20

\dots n i c e c a l c u l u s \dots

e v a l u a t e :::

Φ = \int_{0}^{\infty} \frac{l n (x)}{{(x^{2} + 1)}^{3}} d x = ?

Answered by Lordose last updated on 30/Dec/20

u = \ln (x) \Rightarrow du = \frac{dx}{x}

dv = \frac{1}{{(x^{2} + 1)}^{3}} dx \Rightarrow v \overset{O s t r o g r a s k i}{=} \frac{1}{8} (\frac{x ({3 x}^{2} + 5)}{{(x^{2} + 1)}^{2}} + {3 \tan}^{- 1} (x))

Φ = (\frac{1}{8} ∣ \frac{x ({3 x}^{2} + 5) \ln (x)}{{(x^{2} + 1)}^{2}} ∣_{0}^{\infty} - \frac{1}{8} \int_{0}^{\infty} \frac{{3 x}^{2} + 5}{{(x^{2} + 1)}^{2}} dx - \frac{3}{8} \int_{0}^{\infty} \frac{\tan^{- 1} (x)}{x} dx

Φ = - (\frac{1}{8} (2 π) + 0) = - \frac{π}{4}

Answered by mathmax by abdo last updated on 01/Jan/21

Φ = - \frac{1}{2} Re (Σ Res (f, a_{i}) with f (z) = \frac{\ln^{2} z}{{(z^{2} + 1)}^{3}} we have

f (z) = \frac{\ln^{2} z}{{(z - i)}^{3} {(z + i)}^{3}} so the poles of φ are \bar{+} i (triples)

Res (f, i) = \lim_{z \to i} \frac{1}{(3 - 1)!} {{(z - i)}^{3} f (z)}^{(2)}

= \lim_{z \to i} \frac{1}{2} {\frac{\ln^{2} z}{{(z + i)}^{3}}}^{(2)} = \frac{1}{2} \lim_{z \to i} {\frac{\frac{2 lnz}{z} {(z + i)}^{3} - 3 {(z + i)}^{2} \ln^{2} z}{{(z + i)}^{6}}}^{(1)}

= \frac{1}{2} \lim_{z \to i} {\frac{\frac{2 lnz (z + i)}{z} - {3 \ln}^{2} z}{{(z + i)}^{4}}}^{(1)}

= \frac{1}{2} \lim_{z \to i} {\frac{2 (z + i) lnz - {3 zln}^{2} z}{z {(z + i)}^{4}}}^{(1)}

= \frac{1}{2} \lim_{z \to i} \frac{(2 lnz + \frac{2 (z + i)}{z} - {3 \ln}^{2} z - 6 lnz) z {(z + i)}^{4} - ({(z + i)}^{4} + 4 z {(z + i)}^{3}) 2 (z + i) lnz - {2 zln}^{2} z}{z^{2} {(z + i)}^{8}}

\dots . be continued \dots