1-find-f-0-cos-x-x-4-1-2-dx-with-real-2-find-the-value-of-0-cos-2x-x-4-1-2-dx-3-find-nature-of-the-serie-f-n-

Question Number 69803 by Abdo msup. last updated on 28/Sep/19

1) f i n d f (α) = \int_{0}^{\infty} \frac{c o s (α x)}{{(x^{4} + 1)}^{2}} d x w i t h α r e a l

2) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{c o s (2 x)}{{(x^{4} + 1)}^{2}} d x

3) f i n d n a t u r e o f t h e s e r i e Σ f (n)

Commented by mathmax by abdo last updated on 29/Sep/19

1) f (α) = \int_{0}^{\infty} \frac{c o s (α x)}{{(x^{4} + 1)}^{2}} d x \Rightarrow 2 f (α) = \int_{- \infty}^{+ \infty} \frac{c o s (α x)}{{(x^{4} + 1)}^{2}} d x

= R e (\int_{- \infty}^{+ \infty} \frac{e^{i α x}}{{(x^{4} + 1)}^{2}} d x) l e t W (z) = \frac{e^{i α z}}{{(z^{4} + 1)}^{2}} p o l e s o f W ?

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

s o t h e p o l e s o f W a r e \bar{+} e^{\frac{i π}{4}} a n d \bar{+} e^{- \frac{i π}{4}}

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

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

= {l i m}_{z \to e^{\frac{i π}{4}}} {\frac{e^{i α z}}{{(z + e^{\frac{i π}{4}})}^{2} {(z^{2} + i)}^{2}}}^{(1)}

= {l i m}_{z \to e^{\frac{i π}{4}}} \frac{i α e^{i α z} {(z + e^{\frac{i π}{4}})}^{2} {(z^{2} + i)}^{2} - e^{i α z} {{(z + e^{\frac{i π}{4}})}^{2} {(z^{2} + i)}^{2}}^{(1)}}{{(z + e^{\frac{i π}{4}})}^{4} {(z^{2} + i)}^{4}}

b u t \frac{d}{d z} {{(z + e^{\frac{i π}{4}})}^{2} {(z^{2} + i)}^{2}} = 2 (z + e^{\frac{i π}{4}}) {(z^{2} + i)}^{2}

Commented by mathmax by abdo last updated on 29/Sep/19

\frac{d}{d z} {{(z + e^{\frac{i π}{4}})}^{2} {(z^{2} + i)}^{2}} = 2 (z + e^{\frac{i π}{4}}) (z^{2} + i) + 4 z (z^{2} + i) {(z + e^{\frac{i π}{4}})}^{2}

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

\dots b e c o n t i n u e d \dots