calculate-ln-1-x-2-1-x-2-dx-

Question Number 59277 by Mr X pcx last updated on 07/May/19

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

Commented by maxmathsup by imad last updated on 18/May/19

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

= - 4 \int_{0}^{\frac{π}{2}} l n (c o s θ) d θ l e t f i n d \int_{0}^{\frac{π}{2}} l n (c o s θ) d θ = I l e t u s e h i s f r i e n d J = \int_{0}^{\frac{π}{2}} l n (s i n θ) d θ

w e h a v e J =_{θ = \frac{π}{2} - t} - \int_{0}^{\frac{π}{2}} l n (c o s t) (- d t) = I a l s o

I + J = 2 I = \int_{0}^{\frac{π}{2}} l n (c o s θ . s i n θ) d θ = \int_{0}^{\frac{π}{2}} l n (\frac{s i n (2 θ)}{2}) d θ = \int_{0}^{\frac{π}{2}} l n (s i n (2 θ)) d θ

- \frac{π}{2} l n (2)

\int_{0}^{\frac{π}{2}} l n (s i n (2 θ)) d θ =_{2 θ = t} \frac{1}{2} \int_{0}^{π} l n (s i n t) d t = \frac{1}{2} {\int_{0}^{\frac{π}{2}} l n (s i n t) d t + \int_{\frac{π}{2}}^{π} l n (s i n t) d t}

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

= \frac{1}{2} I + \frac{1}{2} I = I \Rightarrow 2 I = - \frac{π}{2} l n (2) + I \Rightarrow I = - \frac{π}{2} l n (2) \Rightarrow A = - 4 (- \frac{π}{2}) l n (2)

\Rightarrow A = 2 π l n (2) .