pi-pi-sin-1-1-x-2-dx-

Question Number 60881 by aliesam last updated on 26/May/19

\underset{- π}{\int^{π}} s i n (\frac{1}{1 - x^{2}}) d x

Commented by MJS last updated on 26/May/19

I {don}^{'} t think we can solve this, not even

approximate . {it}^{'} s undefined at x = \pm 1 and

{it}^{'} s oszillating very fast around these values

of x

\underset{- π}{\int^{π}} \sin \frac{1}{1 - x^{2}} d x = 2 \underset{0}{\int^{π}} \sin \frac{1}{1 - x^{2}} d x

Commented by aliesam last updated on 26/May/19

y e s {t h a t}^{'} s r i g h t a n d i p o s t e d i t b e c a u s e i t i s i m p r o p e r i n t e g r a l s

Commented by maxmathsup by imad last updated on 27/May/19

l e t I = \int_{- π}^{π} s i n (\frac{1}{1 - x^{2}}) d x \Rightarrow 2 I = \int_{0}^{π} s i n (\frac{1}{1 - x^{2}}) d x

= \int_{0}^{1} s i n (\frac{1}{1 - x^{2}}) d x + \int_{1}^{π} s i n (\frac{1}{1 - x^{2}}) d x = H + K

H =_{x = s i n θ} \int_{0}^{\frac{π}{2}} s i n (\frac{1}{{c o s}^{2} θ}) c o s θ d θ w e h a v e x - \frac{x^{3}}{6} ⩽ s i n x ⩽ x \Rightarrow

\frac{1}{{c o s}^{2} θ} - \frac{1}{6 {c o s}^{6} θ} ⩽ s i n (\frac{1}{{c o s}^{2} θ}) ⩽ \frac{1}{{c o s}^{2} θ} \Rightarrow \int_{0}^{\frac{π}{2}} c o s θ s i n (\frac{1}{{c o s}^{2} θ}) d θ

⩾ \int_{0}^{\frac{π}{2}} \frac{d θ}{c o s θ} - \frac{1}{6} \int_{0}^{\frac{π}{2}} \frac{d θ}{{c o s}^{5} θ} l e t t a k e \int_{0}^{\frac{π}{2}} \frac{d θ}{c o s θ}

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

= \int_{0}^{1} (\frac{1}{1 + u} + \frac{1}{1 - u}) d u = {[l n ∣ \frac{1 + u}{1 - u} ∣]}_{0}^{1} = \infty s o t h i s i n t e g r a l d i v e r g e \dots d o n t

w a s t e t i m e t o f i n d i t \dots!