calculate-0-lnx-x-2-x-1-2-dx-

Question Number 127236 by Bird last updated on 28/Dec/20

c a l c u l a t e \int_{0}^{\infty} \frac{l n x}{{(x^{2} - x + 1)}^{2}} d x

Answered by mindispower last updated on 28/Dec/20

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

= a + b + c

a = 0 : x \to \frac{1}{x} \Rightarrow a = - a

c =

x = {t g}^{\frac{2}{3}} (t) \Rightarrow d x = \frac{2}{3} (1 + {t g}^{2} (t)) {t g}^{- \frac{1}{3}} (t) d t

\Leftrightarrow \int_{0}^{\frac{π}{2}} \frac{2}{3} . \frac{l n (t g (t))}{{(1 + {t g}^{2} (t))}^{2}} . \frac{2}{3} (1 + {t g}^{2} (t)) . {t g}^{- \frac{1}{3}} (t) d t

= \frac{4}{9} \int_{0}^{\frac{π}{2}} \frac{l n (t g (t))}{1 + {t g}^{2} (t)} . {t g}^{- \frac{1}{3}} d t

= \frac{4}{9} \int_{0}^{\frac{π}{2}} \frac{l n (s i n (x)) - l n (c o s (x))}{{c o s}^{- \frac{1}{3}} (t)} {s i n}^{- \frac{1}{3}} (t) . {c o s}^{2} (t) d t

= \frac{4}{9} . \int_{0}^{\frac{π}{2}} l n (s i n (t)) {c o s}^{\frac{7}{3}} (t) {s i n}^{- \frac{1}{3}} (t) - \frac{4}{9} \int_{0}^{\frac{π}{2}} l n (c o s (t)) {c o s}^{\frac{7}{3}} (t) {s i n}^{- \frac{1}{3}} (t) d t

= \frac{1}{9} (A - B)

β (x, y) = 2 \int_{0}^{\frac{π}{2}} c o s {(t)}^{2 x - 1} {s i n}^{2 y - 1} (t) d t

\partial_{x} β (x, y) = 2 . \int_{0}^{\frac{π}{2}} 2 l n (c o s (t)) c o s {(t)}^{2 x - 1} {s i n}^{2 y - 1} (t) d t

A = \frac{1}{9} \partial_{y} β {(x, y)}_{(x, y) = (\frac{5}{3}, \frac{1}{3})}

B = \frac{1}{9} \partial_{x} β {(x, y)}_{(x, y) = (\frac{5}{3}, \frac{1}{3})}

\partial_{x} β (x, y) = β (x, y) [Ψ (x) - Ψ (x + y))

\frac{β (\frac{5}{3}, \frac{1}{3})}{9} (Ψ (\frac{1}{3}) - Ψ (2) - Ψ (\frac{5}{3}) + Ψ (2))

= \frac{Γ (\frac{1}{3}) Γ (\frac{5}{3})}{9 Γ (2)} (Ψ (\frac{1}{3}) - Ψ (\frac{5}{3}))

= \frac{Γ (\frac{1}{3}) . \frac{2}{3} Γ (\frac{2}{3})}{9 Γ (2)} (Ψ (\frac{1}{3}) - \frac{3}{2} - Ψ (\frac{2}{3}))

= \frac{4}{27} . \frac{π}{\sqrt{3}} (- \frac{3}{2} + π c o t (\frac{2 π}{3})) = \frac{4 π}{27 \sqrt{3}} (- \frac{3}{2} - \frac{π}{\sqrt{3}})

= \frac{- 2}{81} π (3 \sqrt{3} + 2 π)

b = \frac{8}{9} \int_{0}^{\frac{π}{2}} l n (t g (t)) \frac{{s i n}^{\frac{2}{3}} (t)}{{c o s}^{\frac{2}{3}} (t)} {c o s}^{2} (t) . {t g}^{- \frac{1}{3}} (t)

= \frac{8}{9} (\int_{0}^{\frac{π}{2}} l n (s i n (t)) {c o s}^{\frac{5}{3}} (t) {s i n}^{\frac{1}{3}} (t) d t - \int_{0}^{\frac{π}{2}} l n (c o s (t)) {c o s}^{\frac{5}{3}} (t) {s i n}^{\frac{1}{3}} (t) d t)

= \frac{2}{9} (Ψ (\frac{2}{3}) - Ψ (2) - Ψ (\frac{4}{3}) - Ψ (2)) β (\frac{2}{3}, \frac{4}{3})

= \frac{2}{9} (Ψ (\frac{2}{3}) - Ψ (\frac{4}{3})) . \frac{Γ (\frac{2}{3}) Γ (\frac{4}{3})}{Γ (2)}

= \frac{2}{9} (Ψ (\frac{2}{3}) - 3 - Ψ (\frac{1}{3})) . \frac{1}{3} \frac{π}{s i n (\frac{π}{3})}

= \frac{2 π}{27} (π c o t (\frac{π}{3}) - 3) . \frac{2}{\sqrt{3}} = \frac{4 π}{27 \sqrt{3}} (\frac{π}{\sqrt{3}} - 3) . = \frac{4 π}{81} (π - 3 \sqrt{3})

\int_{0}^{\infty} \frac{l n (x)}{{(x^{2} - x + 1)}^{2}} = \frac{4 π}{81} (π - 3 \sqrt{3}) - \frac{2 π}{81} (3 \sqrt{3} + 2 π)

= \frac{- 18 π \sqrt{3}}{81} = \frac{- 2 π \sqrt{3}}{9} \approx - 1.2092

Commented by mathmax by abdo last updated on 29/Dec/20

thank you sir for this hardwork

Commented by mindispower last updated on 29/Dec/20

w i t h e p l e a s u r