1-calculate-0-1-0-pi-2-dxdy-1-xtany-2-2-find-the-value-of-0-pi-2-t-tant-dt-

Question Number 27693 by abdo imad last updated on 12/Jan/18

1) c a l c u l a t e \int \int_{] 0, 1] \times] 0, \frac{π}{2}]} \frac{d x d y}{1 + {(x t a n y)}^{2}}

2) f i n d t h e v a l u e o f \int_{0}^{\frac{π}{2}} \frac{t}{t a n t} d t .

Commented by abdo imad last updated on 13/Jan/18

1) l e t p u t I = \int \int_{] 0, 1] \times] 0, \frac{π}{2}]} \frac{d x d y}{1 + {(x t a n y)}^{2}}

I = \int_{0}^{1} (\int_{0}^{\frac{π}{2}} \frac{d y}{1 + x^{2} {t a n}^{2} y}) d x b u t t h e c h . t a n y = t g i v e

\int_{0}^{\frac{π}{2}} \frac{d y}{1 + x^{2} {t a n}^{2} y} = \int_{0}^{\infty} \frac{1}{(1 + x^{2} t^{2})} \frac{d t}{(1 + t^{2})}

= \frac{1}{2} \int_{R} \frac{d t}{(1 + t^{2}) (1 + x^{2} t^{2})} l e t i n t r o d u c e t h e c o m p l e x f u n c t i o n

f (z) = \frac{1}{(1 + z^{2}) (1 + x^{2} z^{2})}

f (z) = \frac{1}{x^{2} (z - i) (z + i) (z - \frac{i}{x}) (z + \frac{i}{x})} (d o n t f o r g e t t h a t 0 < x ⩽ 1)

t h e p o l e s o f f a r e i a n d - i a n d \frac{i}{x} a n d - \frac{i}{x} s o

\int_{R} f (z) d z = 2 i π (R e (f, i) + R e (f, \frac{i}{x})) b u t

R e s (f, i) = {l i m}_{z - > i} (z - i) f (z) = \frac{1}{x^{2} (2 i) (- 1 + \frac{1}{x^{2}})}

= \frac{1}{2 i (1 - x^{2})}

R e (f, \frac{i}{x}) = {l i m}_{z - > \frac{i}{x}} (z - \frac{i}{x}) f (z) = \frac{1}{x^{2} (- \frac{1}{x^{2}} + 1) (\frac{2 i}{x})}

= \frac{x}{2 i (x^{2} - 1)}

\int_{R} f (z) d z = 2 i π (\frac{1}{2 i (1 - x^{2})} - \frac{x}{2 i (1 - x^{2})})

= π \frac{1 - x}{1 - x^{2}} = \frac{π}{1 + x} a n d \int_{0}^{\frac{π}{2}} \frac{d y}{1 + x^{2} {t a n}^{2} y} = \frac{π}{2 (1 + x)}

\Rightarrow I = \int_{0}^{1} \frac{π}{2 (1 + x)} d x = \frac{π}{2} {[l n / 1 + x /]}_{0}^{1} = \frac{π}{2} l n 2

2) b y f u b i n i t h e r e m w e h a v e a l s o

I = \int_{0}^{\frac{π}{2}} (\int_{0}^{1} \frac{d x}{1 + {(x t a n y)}^{2}}) d y a n d t h e c h . x t a n y = t g i v e

\int_{0}^{1} \frac{d x}{1 + {(x t a n y)}^{2}} = \int_{0}^{t a n y} \frac{1}{1 + t^{2}} \frac{d t}{t a n y}

= \frac{1}{t a n y} a r c t a n (t a n y) = \frac{y}{t a n y} s o

I = \int_{0}^{\frac{π}{2}} \frac{y}{t a n y} d y \Rightarrow \int_{0}^{\frac{π}{2}} \frac{t}{t a n t} d t = \frac{π}{2} l n (2) .

Answered by sma3l2996 last updated on 13/Jan/18

1 : I = \int \int_{] 0, 1] \times] 0, π / 2]} \frac{d x d y}{{(x t a n y)}^{2}} = \underset{b \to 0}{l i m} \int_{b}^{π / 2} (\underset{a \to 0}{l i m} \int_{a}^{1} \frac{d x}{1 + {(x t a n y)}^{2}}) d y

l e t t = x t a n y \Rightarrow d t = t a n y d x

I = \underset{b \to 0}{l i m} \int_{b}^{π / 2} \frac{1}{t a n y} \times \underset{a \to 0}{l i m} (\int_{a}^{t a n y} \frac{d t}{1 + t^{2}}) d y

= \underset{b \to 0}{l i m} \int_{b}^{π / 2} \frac{1}{t a n y} \times \underset{a \to 0}{l i m} {[{t a n}^{- 1} (t)]}_{a}^{t a n y} d y

= \underset{b \to 0}{l i m} \int_{b}^{π / 2} \frac{1}{t a n y} (y - 0) d y = \underset{b \to 0}{l i m} \int_{0}^{π / 2} \frac{y d y}{t a n y}

l e t z = e^{i y}

I = - \int_{Δ} \frac{l n (z) (z^{2} - 1)}{z^{2} + 1} \times \frac{d z}{z}

l e t f (z) = \frac{l n (z) (z^{2} - 1)}{z (z + i) (z - i)}

s o I = - 2 i π R e s (f, i) = - 2 i π (\frac{l n (i) (- 2)}{- 2})

I = π^{2}

Commented by abdo imad last updated on 13/Jan/18

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i n r e s i d u s a p p l i c a t i o n \dots .