let-give-g-x-0-arctan-x-1-t-2-1-t-2-dt-find-a-simple-form-of-g-x-without-integral-

Question Number 29077 by abdo imad last updated on 04/Feb/18

l e t g i v e g (x) = \int_{0}^{\infty} \frac{a r c t a n (x (1 + t^{2}))}{1 + t^{2}} d t f i n d a s i m p l e

f o r m o f g^{^{'}} (x) w i t h o u t i n t e g r a l .

Commented by abdo imad last updated on 10/Feb/18

g^{^{'}} (x) = \int_{0}^{\infty} \frac{d t}{1 + x^{2} {(1 + t^{2})}^{2}} = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{d t}{1 + x^{2} {(1 + t^{2})}^{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

φ (z) = \frac{1}{1 + x^{2} {(1 + z^{2})}^{2}} p o l e s o f φ ? w e h a v e

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

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

r o o t s o f z^{2} + 1 + \frac{i}{x} = 0 ? \Leftrightarrow z^{2} = - 1 - \frac{i}{x} = {(i \sqrt{1 + \frac{i}{x}})}^{2}

∣ 1 + \frac{i}{x} ∣= \sqrt{1 + \frac{1}{x^{2}}} = \frac{\sqrt{1 + x^{2}}}{∣ x ∣} a n d 1 + \frac{i}{x}

= \frac{\sqrt{1 + x^{2}}}{∣ x ∣} (\frac{1}{\frac{\sqrt{1 + x^{2}}}{∣ x ∣}} + \frac{1}{x \frac{\sqrt{1 + x^{2}}}{∣ x ∣}} i) = r e^{i θ} \Rightarrow c o s θ = \frac{∣ x ∣}{\sqrt{1 + x^{2}}}

a n d s i n θ = \frac{∣ x ∣}{x \sqrt{1 + x^{2}}} . l e t s u p p o s e x > 0 \Rightarrow c o s θ = \frac{x}{\sqrt{1 + x^{2}}}

a n d s i n θ = \frac{1}{\sqrt{1 + x^{2}}} \Rightarrow t a n θ = \frac{1}{x} \Rightarrow θ = a r c t a n (\frac{1}{x}) a n d

1 + \frac{i}{x} = \frac{\sqrt{1 + x^{2}}}{x} e^{i a r c t a n (\frac{1}{x})} \Rightarrow \sqrt{1 + \frac{i}{x}} = \frac{{(1 + x^{2})}^{\frac{1}{4}}}{\sqrt{x}} e^{\frac{i}{2} a r c t a n (\frac{1}{x})}

\Rightarrow z = + - i \frac{{(1 + x^{2})}^{\frac{1}{4}}}{\sqrt{x}} e^{\frac{i}{2} a r c t a n (\frac{1}{x})} t h e p o l e s a r e

z_{0} = i \frac{{(1 + x^{2})}^{\frac{1}{4}}}{\sqrt{x}} (c o s (\frac{1}{2} a r c t a n (\frac{1}{x}) + i s i n (\frac{1}{2} a r c t a n (\frac{1}{x}))

z_{1} = - i \frac{{(1 + x^{2})}^{\frac{1}{4}}}{\sqrt{x}} (c o s (\frac{1}{2} a r c t a n (\frac{1}{x}) + i s i n (\frac{1}{2} a r c t a n (\frac{1}{x}))

r o o t s o f z^{2} + 1 - \frac{i}{x} = 0 b e c o n t i n u e d \dots . .