calculate-arctan-2x-1-x-2-3-2-dx-

Question Number 77995 by mathmax by abdo last updated on 12/Jan/20

c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{a r c t a n (2 x + 1)}{{(x^{2} + 3)}^{2}} d x

Commented by msup trace by abdo last updated on 13/Jan/20

I = \int_{- \infty}^{+ \infty} \frac{a r c t a n (2 x + 1)}{{(x^{2} + 3)}^{2}} d x \Rightarrow

I =_{x = \sqrt{3} t} \int_{- \infty}^{+ \infty} \frac{a r c t a n (2 \sqrt{3} t + 1)}{9 {(t^{2} + 1)}^{2}} \sqrt{3} d t

= \frac{\sqrt{3}}{9} \int_{- \infty}^{+ \infty} \frac{a r c t a n (2 \sqrt{3} t + 1)}{{(t^{2} + 1)}^{2}} d t

l e t φ (z) = \frac{a r c t a n (2 \sqrt{3} z + 1)}{{(z^{2} + 1)}^{2}} \Rightarrow

φ (z) = \frac{a r c t a n (2 \sqrt{3} z + 1)}{{(z - i)}^{2} {(z + i)}^{2}}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i)

R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} φ (z)}^{(1)}

= {l i m}_{z \to i} {\frac{a r c t a n (2 \sqrt{3} z + 1)}{{(z + i)}^{2}}}^{(1)}

= {l i m}_{z \to i} \frac{\frac{2 \sqrt{3}}{1 + {(2 \sqrt{3} z + 1)}^{2}} {(z + i)}^{2} - 2 (z + i) a r c t a n (2 \sqrt{3} z + 1)}{{(z + i)}^{4}}

= {l i m}_{z \to i} \frac{\frac{2 \sqrt{3} (z + i)}{1 + {(2 \sqrt{3} z + 1)}^{2}} - 2 a r c t a n (2 \sqrt{3} z + 1)}{{(z + i)}^{3}}

= \frac{\frac{4 i \sqrt{3}}{1 + {(2 \sqrt{3} i + 1)}^{2}} - 2 a r c t a n (2 \sqrt{3} i + 1)}{- 8 i}

= - \frac{\sqrt{3}}{2} \times \frac{1}{1 - 12 + 4 \sqrt{3} i + 1} + \frac{1}{4 i} a r c t a n (2 \sqrt{3} i + 1)

= \frac{- \sqrt{3}}{2 (- 10 + 4 \sqrt{3} i)} + \frac{1}{4 i} a r c t a n (2 \sqrt{3} i + 1)

= \frac{\sqrt{3}}{4 (5 - 2 \sqrt{3} i)} - \frac{i}{4} a r c t a n (2 \sqrt{3} i + 1)

Commented by msup trace by abdo last updated on 13/Jan/20

w e h a v e a r c t a n (z) = \frac{1}{2 i} l n (\frac{1 + i z}{1 - i z}) \Rightarrow

a r c t a n (2 \sqrt{3} i + 1) = \frac{1}{2 i} l n (\frac{1 + i (2 \sqrt{3} i + 1)}{1 - i (2 \sqrt{3} i + 1)})

= \frac{1}{2 i} l n (\frac{1 - 2 \sqrt{3} + i}{1 + 2 \sqrt{3} - i}) a l s o

1 - 2 \sqrt{3} + i = \sqrt{{(1 - 2 \sqrt{3})}^{2} + 1} e^{i a r c t a n (\frac{1}{1 - 2 \sqrt{3}})}

1 + 2 \sqrt{3} - i = \sqrt{{(1 + 2 \sqrt{3})}^{2} + 1} e^{- i a r c t a n (\frac{1}{1 + 2 \sqrt{3}})}

\Rightarrow l n (\frac{\dots}{\dots}) = \frac{1}{2} l n ({(1 - 2 \sqrt{3})}^{2} + 1) + i a r c t a n (\frac{1}{1 - 2 \sqrt{3}})

- \frac{1}{2} l n ({(1 + 2 \sqrt{3})}^{2} + 1) + i a r c t a n (\frac{1}{1 + 2 \sqrt{3}})

Commented by mathmax by abdo last updated on 13/Jan/20

\Rightarrow a r c t a n (2 \sqrt{3} i + 1) = \frac{1}{4 i} l n (\frac{1 - 4 \sqrt{3} + 12 + 1}{1 + 4 \sqrt{3} + 12 + 1})

+ \frac{1}{2} {a r c t a n (\frac{1}{1 - 2 \sqrt{3}}) + a r c t a n (\frac{1}{1 + 2 \sqrt{3}})}

= \frac{1}{4 i} l n (\frac{7 - 2 \sqrt{3}}{7 + 2 \sqrt{3}}) + \frac{1}{2} {\frac{π}{2} - a r c t a n (2 \sqrt{3} + 1) - (\frac{π}{2} - a r c t a n (2 \sqrt{3} - 1)}

= \frac{1}{4 i} l n (\frac{7 - 2 \sqrt{3}}{7 + 2 \sqrt{3}}) + \frac{1}{2} (a r c t a n (2 \sqrt{3} - 1) - a r c t a n (2 \sqrt{3} + 1))