study-the-convergence-of-0-1-1-x-x-dx-

Question Number 34221 by abdo imad last updated on 03/May/18

s t u d y t h e c o n v e r g e n c e o f \int_{0}^{1} \frac{\sqrt{1 - x}}{x} d x .

Commented by math khazana by abdo last updated on 04/May/18

l e t p u t x = {s i n}^{2} θ \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{c o s θ}{{s i n}^{2} θ} . 2 s i n θ c o s θ d θ

= \int_{0}^{\frac{π}{2}} \frac{2 {c o s}^{2} θ}{s i n θ} d θ = 2 \int_{0}^{\frac{π}{2}} \frac{1 - {s i n}^{2} θ}{s i n θ} d θ

= 2 \int_{0}^{\frac{π}{2}} \frac{d θ}{s i n θ} - 2 \int_{0}^{\frac{π}{2}} s i n θ d θ b u t w e h a v e

\int_{0}^{\frac{π}{2}} s i n θ d θ = {[- c o s θ]}_{0}^{\frac{π}{2}} = 1 a n d c h . t a n (\frac{θ}{2}) = x g i v e

\int_{ξ}^{\frac{π}{2}} \frac{d θ}{s i n θ} = \int_{ξ}^{\frac{π}{4}} \frac{1}{\frac{2 x}{1 + x^{2}}} \frac{2 d x}{1 + x^{2}} = \int_{ξ}^{\frac{π}{4}} \frac{d x}{x}

= l n (\frac{π}{4}) - l n ξ s o {l i m}_{ξ \to 0^{+}} \int_{ξ}^{\frac{π}{4}} \frac{d θ}{s i n θ} = + \infty

\Rightarrow I i s d i v e r g e n t .

Answered by Joel578 last updated on 03/May/18

I = \lim_{t \to 0^{+}} (\int_{t}^{1} \frac{\sqrt{1 - x}}{x} d x)

\int_{t}^{1} \frac{\sqrt{1 - x}}{x} d x (u = \sqrt{1 - x} \to d u = - \frac{1}{2 u} d x)

= - \int_{\sqrt{1 - t}}^{0} \frac{2 u^{2}}{1 - u^{2}} d u = 2 \int_{\sqrt{1 - t}}^{0} \frac{u^{2}}{u^{2} - 1} d u

= 2 \int_{\sqrt{1 - t}}^{0} (1 + \frac{1}{u^{2} - 1}) d u

= 2 \int_{\sqrt{1 - t}}^{0} (1 + \frac{1}{2 (u - 1)} - \frac{1}{2 (u + 1)}) d u

= 2 {[u + \frac{1}{2} \ln (\frac{u - 1}{u + 1})]}_{\sqrt{1 - t}}^{0}

= 2 [0 + \frac{1}{2} \ln (- 1) - \sqrt{1 - t} - \frac{1}{2} \ln (\frac{\sqrt{1 - t} - 1}{\sqrt{1 - t} + 1})]

= i π - 2 \sqrt{1 - t} - \ln (\frac{\sqrt{1 - t} - 1}{\sqrt{1 - t} + 1})

I = \lim_{t \to 0^{+}} [i π - 2 \sqrt{1 - t} - \ln (\frac{\sqrt{1 - t} - 1}{\sqrt{1 - t} + 1})]

= i π - 2 - \ln 0

= \infty

The integral doesnt converge

Commented by math khazana by abdo last updated on 04/May/18

s i r j o e l f r o m w h e r e c o m e t h e c o m p l e x i π t h e

i n t e g r a l i s r e a l \dots .