let-pi-pi-1-1-x-2-arctan-x-dx-with-gt-0-1-find-a-simple-form-of-2-calculate-

Question Number 36429 by prof Abdo imad last updated on 02/Jun/18

l e t φ (λ) = \int_{\frac{λ}{π}}^{\frac{π}{λ}} (1 + \frac{1}{x^{2}}) a r c t a n (x) d x w i t h λ > 0

1) f i n d a s i m p l e f o r m o f φ (λ)

2) c a l c u l a t e φ^{^{'}} (λ) .

Commented by abdo mathsup 649 cc last updated on 03/Jun/18

l e t i n t e g r a t e b y p a r t s u^{^{'}} = 1 + \frac{1}{x^{2}} a n d

v^{^{'}} = a r c t a n (x)

φ (λ) = {[(1 - \frac{1}{x}) a r c t a n x]}_{\frac{λ}{π}}^{\frac{π}{λ}}

- \int_{\frac{λ}{π}}^{\frac{π}{λ}} (1 - \frac{1}{x}) \frac{d x}{1 + x^{2}}

= (1 - \frac{λ}{π}) a r c t a n (\frac{π}{λ}) - (1 - \frac{π}{λ}) a r c t a n (\frac{λ}{π})

- \int_{\frac{λ}{π}}^{\frac{π}{λ}} \frac{(x - 1) d x}{x (1 + x^{2})} b u t

\int_{\frac{λ}{π}}^{\frac{π}{λ}} \frac{x - 1}{x (1 + x^{2})} d x = \int_{\frac{λ}{π}}^{\frac{π}{λ}} \frac{d x}{1 + x^{2}} - \int_{\frac{λ}{π}}^{\frac{π}{λ}} \frac{d x}{x (1 + x^{2})}

= a r c t a n (\frac{π}{λ}) - a r c t a n (\frac{λ}{π}) - \int_{\frac{λ}{π}}^{\frac{π}{λ}} \frac{d x}{x (1 + x^{2})} b u t

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

a = 1 a n d {l i m}_{x \to + \infty} x F (x) = 0 = a + b \Rightarrow b = - 1

c = 0 \Rightarrow \int_{\frac{λ}{π}}^{\frac{π}{λ}} \frac{d x}{x (1 + x^{2})} = \int_{\frac{λ}{π}}^{\frac{π}{λ}} {\frac{1}{x} - \frac{x}{1 + x^{2}}} d x

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

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

φ (λ) = (1 - \frac{λ}{π}) a r c t a n (\frac{π}{λ}) - (1 - \frac{π}{λ}) a r c t a n (\frac{λ}{π})

- l n ∣ \frac{π}{λ \sqrt{1 + {(\frac{π}{λ})}^{2}}} ∣ + l n ∣ \frac{λ}{π \sqrt{1 + {(\frac{λ}{π})}^{2}}} ∣ .

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jun/18

\int_{\frac{λ}{Π}}^{\frac{Π}{λ}} (\frac{1 + x^{2}}{x^{2}}) {t a n}^{- 1} x d x

I = \int (1 + \frac{1}{x^{2}}) {t a n}^{- 1} x d x

= {t a n}^{- 1} x (x + \frac{- 1}{x}) - \int \frac{1}{1 + x^{2}} \times (\frac{x^{2} - 1}{x}) d x

= (\frac{x^{2} - 1}{x}) {t a n}^{- 1} x - \int \frac{x^{2} - 1}{x (x^{2} + 1)}

= (\frac{x^{2} - 1}{x}) {t a n}^{- 1} x - \int \frac{x^{2} + 1 - 2}{x (x^{2} + 1)} d x

= (\frac{x^{2} - 1}{x}) {t a n}^{- 1} x - \int \frac{d x}{x} + 2 \int \frac{x d x}{x^{2} (x^{2} + 1)}

= (\frac{x^{2} - 1}{x}) {t a n}^{- 1} x - l n x + \int \frac{d t}{t (t + 1)}

= (\frac{x^{2} - 1}{x}) {t a n}^{- 1} x - l n x + \int \frac{t + 1 - t}{(t + 1) t} d t

= (\frac{x^{2} - 1}{x}) {t a n}^{- 1} x - l n x + l n (\frac{t}{t + 1})

= (\frac{x^{2} - 1}{x}) {t a n}^{- 1} x - l n x + l n (\frac{x^{2}}{x^{2} + 1})

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