find-the-value-of-ln-1-x-2-1-x-2-dx-

Question Number 61386 by maxmathsup by imad last updated on 02/Jun/19

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

Commented by perlman last updated on 02/Jun/19

x = t g (a)

d x = (1 + {t g}^{2} (a)) d a ==> d a = \frac{d x}{1 + x^{2}}

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

I = \int l n (c o s (a)) d a = \int l n (s i n (a)) d a

==> 2 I = \int {l n (c o s (a)) + l n (s i n (a))} d a = \int l n (c o s (a) s i n (a)) d a

= \int (l n (s i n (2 a) - l n (2)) d a = - l n (2) \frac{π}{2} + \int l n (s i n (2 a)) d a

l e t 2 a = x

\int_{0}^{\frac{π}{2}} l n s i n (2 a) d a = \frac{1}{2} \int_{0}^{π} l n s i n x d x = \frac{1}{2} \int_{0}^{\frac{π}{2}} l n s i n x d x + \frac{1}{2} \int_{\frac{π}{2}}^{π} l n s i n x d x

l e t y = \frac{π}{2} - x i n s e c o n d o n e

\int_{0}^{\frac{π}{2}} l n s i n (2 a) d a = \frac{1}{2} \int_{0}^{\frac{π}{2}} l n s i n x d x - \frac{1}{2} \int_{0}^{\frac{π}{2}} l n c o s (a) d a = 0

==> 2 I = \frac{- l n (2) π}{2} ==> I = \frac{- l n (2) π}{4}

Commented by maxmathsup by imad last updated on 02/Jun/19

l e t A = \int_{- \infty}^{+ \infty} \frac{l n (1 + x^{2})}{1 + x^{2}} d x \Rightarrow A = 2 \int_{0}^{\infty} \frac{l n (1 + x^{2})}{1 + x^{2}} d x c h a n g e m e n t x = t a n θ g i v e

A = 2 \int_{0}^{\frac{π}{2}} \frac{l n (1 + {t a n}^{2} θ)}{1 + {t a n}^{2} θ} (1 + {t a n}^{2} θ) d θ = 2 \int_{0}^{\frac{π}{2}} l n (\frac{1}{{c o s}^{2} θ}) d θ

= - 4 \int_{0}^{\frac{π}{2}} l n (c o s θ) d θ w e h a v e \int_{0}^{\frac{π}{2}} l n (c o s θ) d θ = - \frac{π}{2} l n (2) (r e s u l t p r o v e d) \Rightarrow

A = - 4 (\frac{- π}{2}) l n (2) \Rightarrow A = 2 π l n (2) .

Commented by maxmathsup by imad last updated on 02/Jun/19

p a r a m e t r i c w a y l e t f (t) = \int_{- \infty}^{+ \infty} \frac{l n (1 + {t x}^{2})}{1 + x^{2}} d x w i t h t ⩾ 0

f^{^{'}} (t) = \int_{- \infty}^{+ \infty} \frac{x^{2}}{(1 + {t x}^{2}) (x^{2} + 1)} d x l e t φ (z) = \frac{z^{2}}{({t z}^{2} + 1) (z^{2} + 1)} p o l e s o f φ ?

φ (z) = \frac{z^{2}}{(\sqrt{t} z - i) (\sqrt{t} z + i) (z - i) (z + i)} = \frac{z^{2}}{t (z - \frac{i}{\sqrt{t}}) (z + \frac{i}{\sqrt{t}}) (z - i) (z + i)}

t h e p o l e s o f φ a r e \bar{+} i a n d \bar{+} \frac{i}{\sqrt{t}} r e s i d u s t h e o r e m g i v e

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, i) + R e s (φ, \frac{i}{\sqrt{t}})}

R e s (φ, i) = {l i m}_{z \to i} (z - i) φ (z) = \frac{- 1}{(1 - t) 2 i} = \frac{- 1}{2 i (1 - t)}

R e s (φ, \frac{i}{\sqrt{t}}) = {l i m}_{z \to \frac{i}{\sqrt{t}}} (z - \frac{i}{\sqrt{t}}) φ (z) = \frac{- 1}{t^{2} (\frac{2 i}{\sqrt{t}}) (- \frac{1}{t} + 1)} = \frac{- t \sqrt{t}}{2 {i t}^{2} (t - 1)} = - \frac{\sqrt{t}}{2 i t (t - 1)}

\Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {\frac{- 1}{2 i (1 - t)} + \frac{\sqrt{t}}{2 i t (1 - t)}} = π {\frac{\sqrt{t}}{t (1 - t)} - \frac{1}{1 - t}} = f^{^{'}} (t) \Rightarrow

f (t) = π \int \frac{\sqrt{t} d t}{t (1 - t)} - π \int \frac{d t}{1 - t} + c

\int \frac{d t}{1 - t} = - l n ∣ 1 - t ∣ + c_{0}

\int \frac{\sqrt{t} d t}{t (1 - t)} =_{\sqrt{t} = u} \int \frac{u (2 u) d u}{u^{2} (1 - u^{2})} = \int \frac{2 d u}{1 - u^{2}} = \int {\frac{1}{1 - u} + \frac{1}{1 + u}} d u

= l n ∣ \frac{1 + u}{1 - u} ∣ + c_{1} = l n ∣ \frac{1 + \sqrt{t}}{1 - \sqrt{t}} ∣ + c_{1} \Rightarrow

f (t) = π l n ∣ \frac{1 + \sqrt{t}}{1 - \sqrt{t}} ∣ + π l n ∣ 1 - t ∣ + c

f (0) = 0 = c \Rightarrow f (t) = π l n ∣ \frac{1 + \sqrt{t}}{1 - \sqrt{t}} ∣ + π l n ∣ 1 - t ∣ .

\int_{- \infty}^{+ \infty} \frac{l n (1 + x^{2})}{1 + x^{2}} d x = {l i m}_{x \to 1} f (1) l e t f i n d t h i s l i m i t

f (t) = π l n ∣ 1 + \sqrt{t} ∣ - π l n ∣ 1 - \sqrt{t} ∣ + π l n ∣ 1 - t ∣ \Rightarrow

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

{l i m}_{t \to 1} f (t) = {l i m}_{u \to 1} f (u^{2}) = 2 π l n (2) \Rightarrow ★ \int_{- \infty}^{+ \infty} \frac{l n (1 + x^{2})}{1 + x^{2}} d x = 2 π l n (2) ★

a n d g e n e r a l l y \int_{- \infty}^{+ \infty} \frac{l n (1 + {t x}^{2})}{1 + x^{2}} d x = π l n ∣ \frac{1 + \sqrt{t}}{1 - \sqrt{t}} ∣ + π l n ∣ 1 - t ∣ w i t h t ⩾ 0