let-g-x-arctan-x-1-t-2-1-t-2-dt-with-x-gt-0-find-a-simple-form-of-g-x-

Question Number 39023 by maxmathsup by imad last updated on 01/Jul/18

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

f i n d a s i m p l e f o r m o f g (x) .

Commented by math khazana by abdo last updated on 08/Jul/18

w e h a v e g^{^{'}} (x) = \int_{- \infty}^{+ \infty} \frac{1}{1 + x^{2} {(1 + t^{2})}^{2}} d t

= \int_{- \infty}^{+ \infty} \frac{d t}{(x (1 + t^{2}) - i) (x (1 + t^{2}) + i)}

= \int_{- \infty}^{+ \infty} \frac{d t}{({x t}^{2} + x - i) ({x t}^{2} + x + i)}

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

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

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

w e h a v e ∣ 1 - \frac{i}{x} ∣ = \sqrt{1 + \frac{1}{x^{2}}} = \frac{\sqrt{x^{2} + 1}}{x}

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

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

s i n θ = - \frac{1}{\sqrt{x^{2} + 1}} \Rightarrow t a n θ = - \frac{1}{x} \Rightarrow

θ = - a r c t a n (x) \Rightarrow 1 - \frac{i}{x} = r (x) e^{- i a r c t a n (\frac{1}{x})}

\sqrt{1 - \frac{i}{x}} = \sqrt{r (x)} e^{- \frac{i}{2} a r c t a n (\frac{1}{x})} a n d

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

i \sqrt{1 - \frac{i}{x}} = \sqrt{r (x)} e^{i \frac{π}{2} - \frac{i}{2} a r c t a n (\frac{1}{x})}

= \sqrt{r (x)} = \sqrt{r (x)} e^{i (\frac{π}{2} - \frac{1}{2} (\frac{π}{2} - a r c t a n x))}

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

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

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

= \sqrt{r (x)} e^{i (\frac{3 π}{4} - \frac{1}{2} a c t a n (x))} \dots . b e c o n t i n u e d \dots