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Question Number 68243 by mathmax by abdo last updated on 07/Sep/19

l e t f (x) = a r c t a n (a x + 1) w i t h a r e a l

1) c a l c u l a t e f^{(n)} (x) a n d f^{(n)} (0)

2) d e v e l o p p f a t i n t e g r s e r i e

3) c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{f (x)}{x^{2} + 4} d x

Commented by mathmax by abdo last updated on 08/Sep/19

1) f (x) = a r c t a n (a x + 1) \Rightarrow f^{^{'}} (x) = \frac{a}{1 + {(a x + 1)}^{2}}

= \frac{a}{1 + a^{2} x^{2} + 2 a x + 1} = \frac{a}{a^{2} x^{2} + 2 a x + 2} \Rightarrow

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

a^{2} x^{2} + 2 a x + 2 = 0 \to Δ^{^{'}} = a^{2} - 2 a^{2} = - a^{2} = {(i a)}^{2} \Rightarrow

x_{1} = \frac{- a + i a}{a^{2}} = \frac{- 1 + i}{a} (w e s u p p o s e a \neq 0)

x_{2} = \frac{- 1 - i}{a} \Rightarrow \frac{1}{a^{2} x^{2} + 2 a x + 2} = \frac{1}{a^{2} (x - \frac{- 1 + i}{a}) (x + \frac{1 + i}{a})}

= \frac{1}{a^{2} (x + \frac{\sqrt{2}}{a} e^{\frac{i π}{4}}) (x + \frac{\sqrt{2}}{a} e^{- \frac{i π}{4}})} = \frac{1}{a^{2} \frac{\sqrt{2}}{2} (2 i \frac{\sqrt{2}}{2})} {\frac{1}{x + \frac{\sqrt{2}}{a} e^{- \frac{i π}{4}}} - \frac{1}{x + \frac{\sqrt{2}}{a} e^{+ \frac{i π}{4}}}}

= \frac{1}{{i a}^{2}} {\frac{1}{x + \frac{\sqrt{2}}{a} e^{- \frac{i π}{4}}} - \frac{1}{x + \frac{\sqrt{2}}{a} e^{\frac{i π}{4}}}} \Rightarrow

f^{(n)} (x) = \frac{1}{i a} {{(\frac{1}{x + \frac{\sqrt{2}}{a} e^{- \frac{i π}{4}}})}^{(n - 1)} - {(\frac{1}{x + \frac{\sqrt{2}}{a} e^{\frac{i π}{4}}})}^{) n - 1)}}

= \frac{1}{i a} {\frac{{(- 1)}^{n - 1}) (n - 1)!}{{(x + \frac{\sqrt{2}}{a} e^{- \frac{i π}{4}})}^{n}} - \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x + \frac{\sqrt{2}}{a} e^{\frac{i π}{4}})}^{n}}} \Rightarrow

f^{(n)} (x) = \frac{{(- 1)}^{n - 1} (n - 1)!}{i a} {\frac{{(x + \frac{\sqrt{2}}{a} e^{\frac{i π}{4}})}^{n} - {(x + \frac{\sqrt{2}}{a} e^{- \frac{i π}{4}})}^{n}}{{(x^{2} + \frac{2 x}{a} + \frac{2}{a^{2}})}^{n}}}

x = 0 \Rightarrow f^{(n)} (0) = \frac{{(- 1)}^{n - 1} (n - 1)!}{i a} {\frac{2 i I m {(\frac{\sqrt{2}}{a} e^{\frac{i π}{4}})}^{n}}{{(\frac{2}{a^{2}})}^{n}}}

w e h a v e {(\frac{\sqrt{2}}{a})}^{n} e^{\frac{i n π}{4}} = \frac{2^{\frac{n}{2}}}{a^{n}} (c o s (\frac{n π}{4}) + i s i n (\frac{n π}{4})) \Rightarrow

f^{(n)} (0) = \frac{{(- 1)}^{n - 1} (n - 1)!}{a} \times 2 \times \frac{2^{\frac{n}{2}}}{a^{n}} s i n (\frac{n π}{4})

= \frac{{(- 1)}^{n - 1} (n - 1)!}{a^{n + 1}} \times 2^{\frac{n}{2} + 1} s i n (\frac{n π}{4})

Commented by mathmax by abdo last updated on 08/Sep/19

2) f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} = f (0) + \sum_{n = 1}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n}

= \frac{π}{4} + \sum_{n = 1}^{\infty} {\frac{{(- 1)}^{n - 1}}{{n a}^{n + 1}} \times 2^{\frac{n}{2} + 1} s i n (\frac{n π}{4})} x^{n}

Commented by mathmax by abdo last updated on 08/Sep/19

3) l e t I = \int_{- \infty}^{+ \infty} \frac{f (x)}{x^{2} + 4} d x \Rightarrow I = \int_{- \infty}^{+ \infty} \frac{a r c t a n (a x + 1)}{x^{2} + 4} d x = φ (a)

w e h a v e φ^{^{'}} (a) = \int_{- \infty}^{+ \infty} \frac{a}{(x^{2} + 4) (1 + {(a x + 1)}^{2})} d x

= a \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} + 4) (a^{2} x^{2} + 2 a x + 2)} l e t W (z) = \frac{1}{(z^{2} + 4) (a^{2} z^{2} + 2 a z + 2)}

p o l e s o f W ? a^{2} x^{2} + 2 a x + 2 = 0 \to Δ^{^{'}} = - a^{2} = {(i a)}^{2} \Rightarrow z_{1} = \frac{- a + i a}{a^{2}}

= \frac{- 1 + i}{a} = - \frac{1 - i}{a} = - \frac{\sqrt{2}}{a} e^{- \frac{i π}{4}} {a n d z}_{2} = \frac{- a - i a}{a^{2}} = \frac{- 1 - i}{a}

= - \frac{\sqrt{2}}{a} e^{\frac{i π}{4}} (w e s u p p o s e a > 0) \Rightarrow

W (z) = \frac{1}{(z - 2 i) (z + 2 i) a^{2} (z + \frac{\sqrt{2}}{a} e^{- \frac{i π}{4}}) (z + \frac{\sqrt{2}}{a} e^{\frac{i π}{4}})} = \frac{1}{a^{2} (z^{2} + 4) (z^{2} + \frac{2 z}{a} + \frac{2}{a^{2}})}

\int_{- \infty}^{+ \infty} W (z) d z = 2 i π {R e s (W, 2 i) + R e s (W, - \frac{\sqrt{2}}{a} e^{- \frac{i π}{4}})}

R e s (W, 2 i) = {l i m}_{z \to 2 i} (z - 2 i) W (z) = \frac{1}{(4 i) a^{2} (- 4 + \frac{4 i}{a} + \frac{2}{a^{2}})}

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