let-f-t-0-ln-1-tx-1-x-2-dx-with-t-lt-1-1-determine-a-explicit-form-of-f-t-2-find-the-value-of-0-ln-1-x-1-x-2-dx-

Question Number 63395 by mathmax by abdo last updated on 03/Jul/19

l e t f (t) = \int_{0}^{\infty} \frac{l n (1 + t x)}{1 + x^{2}} d x w i t h ∣ t ∣< 1

1) d e t e r m i n e a e x p l i c i t f o r m o f f (t)

2) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{l n (1 + x)}{1 + x^{2}} d x

Commented by Prithwish sen last updated on 03/Jul/19

f (t) = \int_{0}^{\infty} \frac{\ln (1 + tx)}{1 + x^{2}} dt

f^{'} (t) = \int_{0}^{\infty} \frac{δ}{δ t} {\frac{\ln (1 + tx)}{1 + x^{2}}} dx

= \int_{0}^{\infty} \frac{x}{(1 + x^{2}) (1 + tx)} dx

= \frac{1}{1 + t^{2}} \int_{0}^{B} {\frac{x}{1 + x^{2}} + \frac{t}{1 + x^{2}} - \frac{t}{1 + tx}} dx, B \to \infty

= \frac{1}{1 + t^{2}} {[\ln (\frac{\sqrt{1 + x^{2}}}{1 + tx}) + t (\tan^{- 1} x)]}_{0}^{B} B \to \infty

Now, \lim_{B \to \infty} \ln (\frac{\sqrt{1 + x^{2}}}{1 + tx}) \Rightarrow \lim_{B \to \infty} \ln (\frac{\sqrt{1 + B^{2}}}{1 + Bt})

put B = \frac{1}{n} \Rightarrow \lim_{n \to 0} \ln (\frac{\sqrt{n^{2} + 1}}{n + t}) \Rightarrow - \ln (t)

∴ f (t) = - \int \frac{\ln (t)}{1 + t^{2}} dt + \frac{π}{2} \int \frac{tdt}{1 + t^{2}}

to be continue \dots \dots .

Commented by mathmax by abdo last updated on 04/Jul/19

1) w e h a v e f^{^{'}} (t) = \int_{0}^{\infty} \frac{x}{(x^{2} + 1) (1 + t x)} d x l e t d e c o m p o s e F (x) = \frac{x}{(t x + 1) (x^{2} + 1)}

F (x) = \frac{a}{t x + 1} + \frac{b x + c}{x^{2} + 1}

a = {l i m}_{x \to - \frac{1}{t}} (t x + 1) F (x) = \frac{- 1}{t (\frac{1}{t^{2}} + 1)} = \frac{- t^{2}}{t (1 + t^{2})} = - \frac{t}{1 + t^{2}}

{l i m}_{x \to + \infty} x F (x) = 0 = \frac{a}{t} + b \Rightarrow b = - \frac{a}{t} = \frac{1}{1 + t^{2}} \Rightarrow F (x) = - \frac{t}{(1 + t^{2}) (t x + 1)} + \frac{\frac{x}{1 + t^{2}} + c}{x^{2} + 1}

F (0) = 0 = - \frac{t}{1 + t^{2}} + c \Rightarrow c = \frac{t}{1 + t^{2}} \Rightarrow

F (x) = \frac{- t}{(1 + t^{2}) (t x + 1)} + \frac{1}{1 + t^{2}} \frac{x + t}{x^{2} + 1} \Rightarrow

f^{^{'}} (t) = \frac{- t}{1 + t^{2}} \int_{0}^{\infty} \frac{d x}{t x + 1} + \frac{1}{2 (1 + t^{2})} \int_{0}^{\infty} \frac{2 x}{x^{2} + 1} d x + \frac{t}{1 + t^{2}} \int_{0}^{\infty} \frac{d x}{1 + x^{2}}

= \frac{1}{1 + t^{2}} {{[\frac{1}{2} l n (x^{2} + 1) - l n ∣ t x + 1 ∣]}_{0}^{+ \infty}} + \frac{π t}{2 (1 + t^{2})}

= \frac{1}{1 + t^{2}} {[l n ∣ \frac{\sqrt{x^{2} + 1}}{t x + 1} ∣]}_{0}^{+ \infty} + \frac{π t}{2 (1 + t^{2})} = \frac{1}{1 + t^{2}} l n (\frac{1}{t^{2}}) + \frac{π t}{2 (1 + t^{2})}

= - \frac{l n ∣ t ∣}{1 + t^{2}} + \frac{π t}{2 (1 + t^{2})} \Rightarrow

f (t) = - \int_{0}^{t} \frac{l n x}{1 + x^{2}} d x + \frac{π}{2} \int_{0}^{t} \frac{x}{1 + x^{2}} d x + C

= - \int_{0}^{t} \frac{l n x}{1 + x^{2}} d x + \frac{π}{4} l n (1 + t^{2}) + C

C = f (0) = 0 \Rightarrow f (t) = \frac{π}{4} l n (1 + t^{2}) - \int_{0}^{t} \frac{l n x}{1 + x^{2}} d x (w e s u p p o s e t ⩾ 0)

2) w e h a v e \int_{0}^{\infty} \frac{l n (1 + x)}{1 + x^{2}} d x = f (1) = \frac{π}{4} l n (2) - \int_{0}^{1} \frac{l n (x)}{1 + x^{2}} d x

c h a n g e m e n t x = t a n θ g i v e

\frac{l n (x)}{1 + x^{2}} d x = \int_{0}^{\frac{π}{4}} \frac{l n (t a n θ)}{1 + {t a n}^{2} θ} (1 + {t a n}^{2} θ) d θ = \int_{0}^{\frac{π}{4}} l n (t a n θ) d θ

= {[θ l n (t a n θ)]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} θ \frac{1 + {t a n}^{2} θ}{t a n θ} d θ = - \int_{0}^{\frac{π}{4}} θ \frac{1}{{c o s}^{2} θ \frac{s i n θ}{c o s θ}} d θ

= \int_{0}^{\frac{π}{4}} \frac{θ}{c o s θ s i n θ} d θ = 2 \int_{0}^{\frac{π}{4}} \frac{θ}{s i n (2 θ)} d θ =_{2 θ = u} 2 \int_{0}^{\frac{π}{2}} \frac{u}{2 s i n (u)} \frac{d u}{2}

= \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{u}{s i n u} d u =_{t a n (\frac{u}{2}) = α} \frac{1}{2} \int_{0}^{1} \frac{2 a r c t a n (α)}{\frac{2 α}{1 + α^{2}}} \frac{2 d α}{1 + α^{2}}

= \frac{1}{2} \int_{0}^{1} \frac{a r c t a n (α)}{α} d α \dots b e c o n t i n u e d \dots

Commented by mathmax by abdo last updated on 04/Jul/19

l e t t r y a n o t h e r w a y w e h a v e \int_{0}^{\infty} \frac{l n (1 + x)}{1 + x^{2}} d x = \frac{π}{4} l n (2) - \int_{0}^{1} \frac{l n (x)}{1 + x^{2}} d x

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

= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} x^{2 n} l n (x) d x b y [p a r t s

\int_{0}^{1} x^{2 n} l n (x) d x = {[\frac{1}{2 n + 1} x^{2 n + 1} l n (x)]}_{0}^{1} - \int_{0}^{1} \frac{1}{2 n + 1} x^{2 n + 1} \frac{d x}{x}

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

r e s t t o f i n d t h e v a l u e o f t h i s s e r i e b y f o u r i e r s e r i e s \dots . .