nice-calculus-calculate-1-ln-x-4-2x-2-2-x-x-2-1-dx-

Question Number 127958 by mnjuly1970 last updated on 03/Jan/21

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

c a l c u l a t e

Ω = \int_{1}^{\infty} \frac{l n (x^{4} - 2 x^{2} + 2)}{x \sqrt{x^{2} - 1}} d x = ?

Answered by Dwaipayan Shikari last updated on 03/Jan/21

\int_{1}^{\infty} \frac{l o g (x^{4} - 2 x^{2} + 2)}{x \sqrt{x^{2} - 1}} d x

= \int_{0}^{\frac{π}{2}} \frac{l o g ({s e c}^{4} θ - 2 {s e c}^{2} θ + 2)}{s e c θ t a n θ} s e c θ t a n θ

= \int_{0}^{\frac{π}{2}} l o g (1 - 2 {c o s}^{2} θ + 2 {c o s}^{4} θ) - 4 \int_{0}^{\frac{π}{2}} l o g (c o s θ) d θ

= \int_{0}^{\frac{π}{2}} l o g (1 - \frac{1}{2} {s i n}^{2} 2 θ) + 2 π l o g (2)

= \int_{0}^{\frac{π}{2}} l o g ({c o s}^{2} 2 θ + \frac{1}{2} {s i n}^{2} 2 θ) + 2 π l o g (2)

= π l o g (\frac{1 + \frac{1}{\sqrt{2}}}{2}) + π l o g (4) = π l o g (2 + \sqrt{2})

L e m m a

\int_{0}^{\frac{π}{2}} l o g (a^{2} {c o s}^{2} θ + b^{2} {s i n}^{2} θ) = π l o g (\frac{a + b}{2})