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Question Number 57900 by maxmathsup by imad last updated on 13/Apr/19

l e t f (x) = \int_{0}^{\infty} \frac{c o s (π x t)}{{(t^{2} + 3 x^{2})}^{2}} d t w i t h x > 0

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

2) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{c o s (π t)}{{(t^{2} + 3)}^{2}} d t

3) l e t U_{n} = f (n) f i n d n a t u r e o f Σ U_{n}

Commented by maxmathsup by imad last updated on 17/Apr/19

1) w e h a v e 2 f (x) = \int_{- \infty}^{+ \infty} \frac{c o s (π x t)}{{(t^{2} + 3 x^{2})}^{2}} d t = R e (\int_{- \infty}^{+ \infty} \frac{e^{i π x t}}{{(t^{2} + 3 x^{2})}^{2}} d t) l e t

φ (z) = \frac{e^{i π x z}}{{(z^{2} + 3 x^{2})}^{2}} \Rightarrow φ (z) = \frac{e^{i π x z}}{{(z - i x \sqrt{3})}^{2} {(z + i x \sqrt{3})}^{2}} s o t h e p o l e s o f φ

a r e \bar{+} i x \sqrt{3} (d o u b l e s) r e s i d u s t h e o r e m g i v e

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i x \sqrt{3})

R e s (φ, i x \sqrt{3}) = {l i m}_{z \to i x \sqrt{3}} \frac{1}{(2 - 1)!} {{(z - i x \sqrt{3})}^{2} φ (z)}^{(1)}

= {l i m}_{z \to i x \sqrt{x}} {\frac{e^{i π x z}}{{(z + i x \sqrt{3})}^{2}}}^{(1)}

= {l i m}_{z \to i x \sqrt{3}} \frac{i π {x e}^{i π x z} {(z + i x \sqrt{3})}^{2} - 2 (z + i x \sqrt{3}) e^{i π x z}}{{(z + i x \sqrt{3})}^{4}}

= {l i m}_{z \to i x \sqrt{3}} \frac{{i π x (z + i x \sqrt{3}) - 2} e^{i π x z}}{{(z + i x \sqrt{3})}^{3}}

\frac{{i π x (2 i x \sqrt{3}) - 2) e^{i π x (i x \sqrt{3})}}{{(2 i x \sqrt{3})}^{3}} = \frac{{- 2 π \sqrt{3} x^{2} - 2) e^{- π \sqrt{3} x^{2}}}{- 8 i (3 \sqrt{3})} = \frac{(π \sqrt{3} x^{2} + 1) e^{- π \sqrt{3} x^{2}}}{12 i \sqrt{3}} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{(π \sqrt{3} x^{2} + 1) e^{- π \sqrt{3} x^{2}}}{12 i \sqrt{3}} = \frac{π}{6 \sqrt{3}} (π \sqrt{3} x^{2} + 1) e^{- π \sqrt{3} x^{2}} \Rightarrow

f (x) = \frac{π}{12 \sqrt{3}} (π \sqrt{3} x^{2} + 1) e^{- π \sqrt{3} x^{2}} .

2) \int_{0}^{\infty} \frac{c o s (π t)}{{(t^{2} + 3)}^{2}} d t = f (1) = \frac{π}{12 \sqrt{3}} (π \sqrt{3} + 1) e^{- π \sqrt{3}}