find-the-value-of-the-following-integral-0-ln-2-x-1-x-2-dx-

Question Number 192407 by mnjuly1970 last updated on 17/May/23

f i n d t h e v a l u e o f t h e f o l l o w i n g i n t e g r a l

χ = \int_{0}^{\infty} \frac{\ln^{2} (x)}{1 + x^{2}} d x = ?

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Answered by ARUNG_Brandon_MBU last updated on 17/May/23

χ = \int_{0}^{\infty} \frac{\ln^{2} x}{1 + x^{2}} d x = \frac{1}{8} \int_{0}^{\infty} \frac{t^{- \frac{1}{2}} \ln^{2} t}{1 + t} d t

= \frac{\partial^{2}}{\partial α^{2}} ∣_{α = \frac{1}{2}} \frac{1}{8} \int_{0}^{\infty} \frac{t^{α - 1}}{1 + t} d t = \frac{\partial^{2}}{\partial α^{2}} ∣_{α = \frac{1}{2}} \frac{1}{8} β (α, 1 - α)

= \frac{\partial^{2}}{\partial α^{2}} ∣_{α = \frac{1}{2}} \frac{1}{8} \cdot \frac{π}{\sin (π α)} = \frac{\partial}{\partial α} ∣_{α = \frac{1}{2}} - \frac{1}{8} \cdot \frac{π^{2} \cos (π α)}{\sin^{2} (π α)}

= - \frac{π^{3}}{8} ∣_{α = \frac{1}{2}} (- \frac{1}{\sin (π α)} - \frac{{2 \cos}^{2} (π α)}{\sin^{3} (π α)}) = \frac{π^{3}}{8}

Commented by mnjuly1970 last updated on 17/May/23

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