calculate-0-lnx-x-4-1-dx-

Question Number 116097 by mathmax by abdo last updated on 30/Sep/20

calculate \int_{0}^{\infty} \frac{lnx}{x^{4} + 1} dx

Answered by mnjuly1970 last updated on 01/Oct/20

s o l u t i o n : Ω = \int_{0}^{\infty} \frac{l n (x)}{1 + x^{4}} = ? ? ?

x^{4} = t \Rightarrow x = t^{\frac{1}{4}} \Rightarrow d x = \frac{1}{4} t^{\frac{- 3}{4}}

Ω = \frac{1}{16} \int_{0}^{\infty} \frac{l n (t) t^{- \frac{3}{4}}}{1 + t} d t

f (λ) = \int_{0}^{\infty} \frac{t^{- λ}}{1 + t} d t \Rightarrow Ω = \frac{- 1}{16} f^{'} (\frac{3}{4})

b u t :: f (λ) = Γ (λ) Γ (1 - λ); 0 < λ < 1

f (λ) = \frac{π}{s i n (λ π)} \Rightarrow f^{'} (λ) = {[- \frac{π^{2} c o s (π λ)}{{s i n}^{2} (π λ)}]}_{λ = \frac{3}{4}}

f^{'} (\frac{3}{4}) = π^{2} * \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}} = π^{2} \sqrt{2}

Ω := - \frac{1}{16} f^{'} (\frac{3}{4}) = - \frac{π^{2} \sqrt{2}}{16} ✓

\dots m . n . j u l y . 1970 \dots

Commented by 1549442205PVT last updated on 01/Oct/20

third line \overset{?}{\to} fourth line (lnt missed ?)

Commented by mnjuly1970 last updated on 01/Oct/20

m e t h o d : {f e y n m a n}^{'} s t r i c k

p l e a s e l o o k a t a g a i n .

Commented by 1549442205PVT last updated on 01/Oct/20

Thank Sir, i understanded . {(a^{x})}^{'} = a^{x} lna

and put t = \frac{1}{x} - 1 \Rightarrow f (λ) = \int_{0}^{1} x^{- λ} {(1 - x)}^{1 - λ} = β (λ, 1 - λ) = Γ (λ) Γ (1 - λ) = \frac{π}{\sin (λ π)}

Great Sir thanh you very much

Commented by mnjuly1970 last updated on 01/Oct/20

y o u a r e w e l c o m e .

Commented by Tawa11 last updated on 06/Sep/21

grest sir

Answered by mathdave last updated on 01/Oct/20

s o l u t i o n

l e t I = \int_{0}^{1} \frac{\ln x}{x^{4} + 1} d x + \int_{1}^{\infty} \frac{\ln x}{1 + x^{4}} d x = A + B

p u t t i n g x = \frac{1}{x} i n t o B

I = \int_{0}^{1} \frac{\ln x}{1 + x^{4}} d x + \int_{1}^{0} \frac{\ln (\frac{1}{x})}{1 + \frac{1}{x^{4}}} \times - \frac{1}{x^{2}} d x = \int_{0}^{1} \frac{\ln x}{1 + x^{4}} d x - \int_{0}^{1} \frac{x^{2} \ln x}{1 + x^{4}} d x

I = {(-)}^{n} \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} x^{4 n} \ln x d x - {(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} x^{2} . x^{4 n} \ln x d x

I = {(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} \frac{\partial}{\partial a} ∣_{a = 1} \int_{0}^{1} x^{4 n} . x^{a - 1} d x - {(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} \frac{\partial}{\partial a} ∣_{a = 1} \int_{0}^{1} x^{4 n + 2} . x^{a - 1} d x

I = \underset{n = 0}{\sum^{\infty}} \frac{\partial}{\partial a} ∣_{a = 1} \frac{{(- 1)}^{n}}{(4 n + a)} - \underset{n = 0}{\sum^{\infty}} \frac{\partial}{\partial a} ∣_{a = 1} \frac{{(- 1)}^{n}}{(4 n + 2 + a)}

I = - \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(4 n + 1)}^{2}} + \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(4 n + 3)}^{2}} = - \underset{n = - \infty}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(4 n + 1)}^{2}}

n o t e \underset{n = - \infty}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(a x + 1)}^{n}} = - \frac{π}{a^{n}} \lim_{z \to - \frac{1}{a}} \frac{1}{(n - 1)!} ∙ \frac{d^{n - 1}}{{d z}^{n - 1}} (cosec (π z))

I = - \underset{n = - \infty}{\sum^{\infty}} \frac{1}{{(4 n + 1)}^{2}} = - (- \frac{π}{4^{2}}) \lim_{z \to - \frac{1}{4}} \frac{1}{(2 - 1)!} ∙ \frac{d}{d z} (c o s e c (π z))

I = \frac{π}{16} \lim_{z \to - \frac{1}{4}} \frac{1}{1!} (- π cosec (π z) \cot (π z))

I = - \frac{π^{2}}{16} cosec (- \frac{π}{4}) \cot (- \frac{π}{4}) = - \frac{π^{2}}{16} (- \sqrt{2}) (- 1) = - \frac{π^{2}}{16} \sqrt{2}

∵ \int_{0}^{\infty} \frac{\ln x}{1 + x^{4}} d x = - \frac{π^{2}}{8 \sqrt{2}}

b y m a t h d a v e (01 / 10 / 2020)

Commented by Tawa11 last updated on 06/Sep/21

great sir

Answered by Bird last updated on 01/Oct/20

l e t f (a) = \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} d t w i t h 0 < a < 1

w e h a v e f (a) = \frac{π}{s i n (π a)}

a n d f (a) = \int_{0}^{\infty} \frac{e^{(a - 1) l n t}}{1 + t} d t \Rightarrow

f^{^{'}} (a) = \int_{0}^{\infty} \frac{l n t t^{a - 1}}{1 + t} d t = \int_{0}^{\infty} \frac{t^{a - 1} l n t}{1 + t} d t

\int_{0}^{\infty} \frac{l n x}{x^{4} + 1} d x =_{x = t^{\frac{1}{4}}} \frac{1}{4} \int_{0}^{\infty} \frac{l n t}{1 + t} . \frac{1}{4} t^{\frac{1}{4} - 1} d t

= \frac{1}{16} \int_{0}^{\infty} \frac{t^{\frac{1}{4} - 1} l n t}{1 + t} d t

= \frac{1}{16} f^{^{'}} (\frac{1}{4}) b u t f^{^{'}} (a) = - \frac{π^{2} c o s (π a)}{{s i n}^{2} (π a)}

\Rightarrow f^{^{'}} (\frac{1}{4}) = - π^{2} \times \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}} = \frac{- 2 π^{2}}{\sqrt{2}}

= - \sqrt{2} π^{2} \Rightarrow

\int_{0}^{\infty} \frac{l n x}{1 + x^{4}} d x = - \frac{π^{2} \sqrt{2}}{16}

Commented by mnjuly1970 last updated on 01/Oct/20

e x c e l l e n t s i r . .

Commented by Bird last updated on 01/Oct/20

y o u a r e w e l c o m e

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