prove-that-0-ln-1-x-x-4-17x-2-16-dx-pi-60-ln-2-

Question Number 162535 by mnjuly1970 last updated on 30/Dec/21

prove that

Ω = \int_{0}^{\infty} \frac{\ln (\frac{1}{x})}{x^{4} + 17 x^{2} + 16} d x \overset{?}{=} \frac{π}{60} \ln (2)

Commented by Ar Brandon last updated on 30/Dec/21

Synthax error, Sir . Your d x is missing .

Haha

Commented by amin96 last updated on 30/Dec/21

very big mistake

Commented by mnjuly1970 last updated on 30/Dec/21

y e s y o u a r e r i g h t s i r \dots

Answered by Ar Brandon last updated on 24/Mar/22

Ω = \int_{0}^{\infty} \frac{\ln (\frac{1}{x})}{x^{4} + 17 x^{2} + 16} d x = - \int_{0}^{\infty} \frac{\ln x}{x^{4} + 17 x^{2} + 16} d x

φ (z) = \frac{{(\ln z)}^{2}}{z^{4} + 17 z^{2} + 16}, poles : (z^{2} + 16) (z^{2} + 1) = 0 \Rightarrow z_{1} = 4 e^{\frac{π}{2} i}, z_{2} = 4 e^{- \frac{π}{2} i}, z_{3} = e^{\frac{π}{2} i}, z_{4} = e^{- \frac{π}{2} i}

Ω = - \frac{1}{2} R e (R e s (φ, z_{1}) + R e s (φ, z_{2}) + R e s (φ, z_{3}) + R e s (φ, z_{4}))

R e s (φ, z_{1}) = \lim_{z \to z_{1}} (\frac{{(\ln z)}^{2}}{(z - z_{2}) (z - z_{3}) (z - z_{4})}) = \frac{{(\ln 4 + \frac{π}{2} i)}^{2}}{(8 i) (3 i) (5 i)} = \frac{i}{120} (\ln^{2} 4 - \frac{π^{2}}{4} + i π \ln 4)

R e s (φ, z_{2}) = \lim_{z \to z_{2}} (\frac{{(\ln z)}^{2}}{(z - z_{1}) (z - z_{3}) (z - z_{4})}) = \frac{{(\ln 4 - \frac{π}{2} i)}^{2}}{(- 8 i) (- 5 i) (- 3 i)} = - \frac{i}{120} (\ln^{2} 4 + \frac{π^{2}}{4} - i π \ln 4)

R e s (φ, z_{3}) = \lim_{z \to z_{3}} (\frac{{(\ln z)}^{2}}{(z - z_{1}) (z - z_{2}) (z - z_{4})}) = \frac{{(\frac{π}{2} i)}^{2}}{(- 3 i) (5 i) (2 i)} = \frac{i}{30} (\frac{π^{2}}{4}) = i \frac{π^{2}}{120}

R e s (φ, z_{4}) = \lim_{z \to z_{4}} (\frac{{(\ln z)}^{2}}{(z - z_{1}) (z - z_{2}) (z - z_{3})}) = \frac{{(- \frac{π}{2} i)}^{2}}{(- 5 i) (3 i) (- 2 i)} = - \frac{i}{30} (\frac{π^{2}}{4}) = - i \frac{π^{2}}{120}

Ω = - \frac{1}{2} (- \frac{π \ln 4}{120} - \frac{π \ln 4}{120}) = \frac{π \ln 4}{120} = \frac{π}{60} \ln (2)

Commented by mnjuly1970 last updated on 30/Dec/21

v e r y n i c e s o l u t i o n . . r e s i d u e s t h e o r y . .

t h a n k s a l o t s i r b r a n d o n . .

Answered by mindispower last updated on 30/Dec/21

= - \int_{0}^{\infty} - \frac{l n (x)}{(x^{2} + 16) (x^{2} + 1)} d x = - \frac{1}{15} \int_{0}^{\infty} \frac{l n (x)}{x^{2} + 1} + \frac{l n (x)}{x^{2} + 16} d x

= \frac{1}{15} (- \int_{0}^{\infty} \frac{l n (x)}{x^{2} + 1} d x + \int \frac{l n (4 y)}{16 (y^{2} + 1)} 4 y

= - \frac{3}{60} \int_{0}^{\infty} \frac{l n (x)}{1 + x^{2}} + \frac{l n (4)}{60} \int_{0}^{\infty} \frac{d x}{1 + x^{2}}

\int_{0}^{\infty} \frac{l n (x)}{1 + x^{2}} d x, x \to \frac{1}{x} \Rightarrow= - \int_{0}^{\infty} \frac{l n (x)}{1 + x^{2}} d x

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

Ω = \frac{\ln (4)}{60} \int_{0}^{\infty} \frac{d x}{1 + x^{2}} = \frac{l n (4)}{60} \lim_{x \to \infty} {[\tan^{- 1} (z)]}_{0}^{x}

= \frac{l n (4)}{60} . \frac{π}{2} = \frac{π l n (2)}{60}