let-f-n-a-cos-nx-x-2-x-a-2-dx-with-a-1-1-find-a-explicit-form-of-f-n-a-2-study-the-convervenge-of-f-n-a-3-determine-also-g-n-a-cos-nx

Question Number 61884 by maxmathsup by imad last updated on 10/Jun/19

l e t f_{n} (a) = \int_{- \infty}^{+ \infty} \frac{c o s (n x)}{{(x^{2} + x + a)}^{2}} d x w i t h a ⩾ 1

1) f i n d a e x p l i c i t f o r m o f f_{n} (a)

2) s t u d y t h e c o n v e r v e n g e o f Σ f_{n} (a)

3) d e t e r m i n e a l s o g_{n} (a) = \int_{- \infty}^{+ \infty} \frac{c o s (n x)}{{(x^{2} + x + a)}^{3}} d x

s t u d y t h e c o n v e r g e n c e o f Σ g n (a)

Commented by maxmathsup by imad last updated on 11/Jun/19

1) w e h a v e f_{n} (a) = R e (\int_{- \infty}^{+ \infty} \frac{e^{i n x}}{{(x^{2} + x + a)}^{2}}) l e t w (z) = \frac{e^{i n z}}{{(z^{2} + z + a)}^{2}} p o l e s o f w ?

z^{2} + z + a = 0 \to Δ = 1 - 4 a < 0 \Rightarrow Δ = {(i \sqrt{4 a - 1})}^{2} \Rightarrow

z_{1} = \frac{- 1 + i \sqrt{4 a - 1}}{2} a n d z_{2} = \frac{- 1 - i \sqrt{4 a - 1}}{2} t h e p o l e s o f w a r e z_{1} a n d z_{2} (d o u b l e s)

w (z) = \frac{e^{i n z}}{{(z - z_{1})}^{2} {(z - z_{2})}^{2}} r e s i d u s t h e o r e m g i v e \int_{- \infty}^{+ \infty} w (z) d z = 2 i π R e s (w, z_{1})

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

= {l i m}_{z \to z_{1}} {\frac{e^{i n z}}{{(z - z_{2})}^{2}}}^{(1)} = {l i m}_{z \to z_{1}} \frac{i n e^{i n z} {(z - z_{2})}^{2} - 2 (z - z_{2}) e^{i n z}}{{(z - z_{2})}^{4}}

= {l i m}_{z \to z_{1}} \frac{(i n (z - z_{2}) - 2} e^{i n z}}{{(z - z_{2})}^{3}} = \frac{{i n (z_{1} - z_{2}) - 2) e^{{i n z}_{1}}}{{(z_{1} - z_{2})}^{3}}

= \frac{{i n i \sqrt{4 a - 1} - 2} e^{i n (\frac{- 1 + i \sqrt{4 a - 1}}{2})}}{{(i \sqrt{4 a - 1})}^{3}} = - \frac{(n \sqrt{4 a - 1} + 2) e^{- \frac{i n}{2}} e^{- \frac{n}{2} \sqrt{4 a - 1}}}{- i (4 a - 1) \sqrt{4 a - 1}}

= \frac{(n \sqrt{4 a - 1} + 2) e^{- \frac{n}{2} \sqrt{4 a - 1}}}{i (4 a - 1) \sqrt{4 a - 1}} {c o s (\frac{n}{2}) - i s i n (\frac{n}{2})} \Rightarrow

\int_{- \infty}^{+ \infty} w (z) d z = 2 i π \frac{(n \sqrt{4 a - 1} + 2) e^{- \frac{n}{2} \sqrt{4 a - 1}}}{i (4 a - 1) \sqrt{4 a - 1}} {\dots .} \Rightarrow

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

Commented by maxmathsup by imad last updated on 11/Jun/19

2) w e h a v e Σ f_{n} (a) = Σ \frac{2 π \sqrt{4 a - 1}}{(4 a - 1) \sqrt{4 a - 1}} n e^{- \frac{n}{2} \sqrt{4 a - 1}} + Σ \frac{4 π}{(4 a - 1) \sqrt{4 a - 1}} e^{- \frac{n}{2} \sqrt{4 a - 1}}

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

w e h a v e ∣ e^{- \frac{\sqrt{4 a - 1}}{2}} ∣< 1 a n d Σ x^{n}, Σ {n x}^{n} c o n v e r g s \Rightarrow Σ f_{n} c o n v e r g e s

Commented by maxmathsup by imad last updated on 11/Jun/19

3) w e h a v e f_{n}^{^{'}} (a) = \int_{- \infty}^{+ \infty} \frac{\partial}{\partial a} (\frac{c o s (n x)}{{(x^{2} + x + a)}^{2}}) d x

= - \int_{- \infty}^{+ \infty} \frac{2 (x^{2} + x + a)}{{(x^{2} + x + a)}^{4}} c o s (n x) d x = - 2 \int_{- \infty}^{+ \infty} \frac{c o s (n x)}{{(x^{2} + x + a)}^{2}} d x = - 2 g_{n} (a) \Rightarrow

g_{n} (a) = - \frac{1}{2} f_{n}^{^{'}} (a) r e s t t o c a l c u l a t e f^{^{'}} n (a) \dots .