1)find f(α) =∫_0 ^∞ ((cos(αx))/((x^4 +1)^2 ))dx with α real 2) find the value of ∫_0 ^∞ ((cos(2x))/((x^4 +1)^2 ))dx 3) find nature of the serie Σf(n)


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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
$1) f(α) =∫_0 ^∞ ((cos(αx))/((x^4 +1)^2 ))dx ⇒2f(α) =∫_(−∞) ^(+∞) ((cos(αx))/((x^4 +1)^2 ))dx =Re(∫_(−∞) ^(+∞) (e^(iαx) /((x^4 +1)^2 ))dx) let W(z) =(e^(iαz) /((z^4 +1)^2 )) poles of W? W(z) =(e^(iαz) /((z^2 −i)^2 (z^2 +i)^2 )) =(e^(iαz) /((z−e^((iπ)/4) )^2 (z+e^((iπ)/4) )^2 (z−e^(−((iπ)/4)) )^2 (z+e^(−((iπ)/4)) )^2 )) so the poles of W are +^− e^((iπ)/4) and +^− e^(−((iπ)/4)) ∫_(−∞) ^(+∞) W(z)dz =2iπ { Res(W,e^((iπ)/4) ) +Res(W,−e^(−((iπ)/4)) )} Res(W,e^((iπ)/4) ) =lim_(z→e^((iπ)/4) ) (1/((2−1)!)){ (z−e^((iπ)/4) )^2 W(z)}^((1)) =lim_(z→e^((iπ)/4) ) {(e^(iαz) /((z+e^((iπ)/4) )^2 (z^2 +i)^2 ))}^((1)) =lim_(z→e^((iπ)/4) ) ((iα e^(iαz) (z+e^((iπ)/4) )^2 (z^2 +i)^2 −e^(iαz) {(z+e^((iπ)/4) )^2 (z^2 +i)^2 }^((1)) )/((z+e^((iπ)/4) )^4 (z^2 +i)^4 )) but (d/dz){(z+e^((iπ)/4) )^2 (z^2 +i)^2 }=2(z+e^((iπ)/4) )(z^2 +i)^2$
$1) f (α) = \int_{0}^{\infty} \frac{c o s (α x)}{{(x^{4} + 1)}^{2}} d x \Rightarrow 2 f (α) = \int_{- \infty}^{+ \infty} \frac{c o s (α x)}{{(x^{4} + 1)}^{2}} d x$ $= R e (\int_{- \infty}^{+ \infty} \frac{e^{i α x}}{{(x^{4} + 1)}^{2}} d x) l e t W (z) = \frac{e^{i α z}}{{(z^{4} + 1)}^{2}} p o l e s o f W ?$ $W (z) = \frac{e^{i α z}}{{(z^{2} - i)}^{2} {(z^{2} + i)}^{2}} = \frac{e^{i α z}}{{(z - e^{\frac{i π}{4}})}^{2} {(z + e^{\frac{i π}{4}})}^{2} {(z - e^{- \frac{i π}{4}})}^{2} {(z + e^{- \frac{i π}{4}})}^{2}}$ $s o t h e p o l e s o f W a r e \bar{+} e^{\frac{i π}{4}} a n d \bar{+} e^{- \frac{i π}{4}}$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π {R e s (W, e^{\frac{i π}{4}}) + R e s (W, - e^{- \frac{i π}{4}})}$ $R e s (W, e^{\frac{i π}{4}}) = {l i m}_{z \to e^{\frac{i π}{4}}} \frac{1}{(2 - 1)!} {{(z - e^{\frac{i π}{4}})}^{2} W (z)}^{(1)}$ $= {l i m}_{z \to e^{\frac{i π}{4}}} {\frac{e^{i α z}}{{(z + e^{\frac{i π}{4}})}^{2} {(z^{2} + i)}^{2}}}^{(1)}$ $= {l i m}_{z \to e^{\frac{i π}{4}}} \frac{i α e^{i α z} {(z + e^{\frac{i π}{4}})}^{2} {(z^{2} + i)}^{2} - e^{i α z} {{(z + e^{\frac{i π}{4}})}^{2} {(z^{2} + i)}^{2}}^{(1)}}{{(z + e^{\frac{i π}{4}})}^{4} {(z^{2} + i)}^{4}}$ $b u t \frac{d}{d z} {{(z + e^{\frac{i π}{4}})}^{2} {(z^{2} + i)}^{2}} = 2 (z + e^{\frac{i π}{4}}) {(z^{2} + i)}^{2}$
Commented by mathmax by abdo last updated on 29/Sep/19
$(d/dz){(z+e^((iπ)/4) )^2 (z^2 +i)^2 } =2(z+e^((iπ)/4) )(z^2 +i)+4z(z^2 +i)(z+e^((iπ)/4) )^2 Res(W,e^((iπ)/4) )=lim_(z→e^((iπ)/4) ) ((iαe^(iαz) (z+e^((iπ)/4) )^2 (z^2 +i)^2 −e^(iαz) {2(z+e^((iπ)/4) )(z^2 +i)+4z(z^2 +i)(z+e^((iπ)/4) )^2 )/((z+e^((iπ)/4) )^4 (z^2 +i)^4 )) ...be continued...$
$\frac{d}{d z} {{(z + e^{\frac{i π}{4}})}^{2} {(z^{2} + i)}^{2}} = 2 (z + e^{\frac{i π}{4}}) (z^{2} + i) + 4 z (z^{2} + i) {(z + e^{\frac{i π}{4}})}^{2}$ $R e s (W, e^{\frac{i π}{4}}) = {l i m}_{z \to e^{\frac{i π}{4}}} \frac{i α e^{i α z} {(z + e^{\frac{i π}{4}})}^{2} {(z^{2} + i)}^{2} - e^{i α z} {2 (z + e^{\frac{i π}{4}}) (z^{2} + i) + 4 z (z^{2} + i) {(z + e^{\frac{i π}{4}})}^{2}}{{(z + e^{\frac{i π}{4}})}^{4} {(z^{2} + i)}^{4}}$ $. . . b e c o n t i n u e d . . .$
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