caoculate-0-t-dt-1-t-4-2-

Question Number 40155 by maxmathsup by imad last updated on 16/Jul/18

c a o c u l a t e \int_{0}^{\infty} \frac{t d t}{{(1 + t^{4})}^{2}}

Commented by maxmathsup by imad last updated on 16/Jul/18

R e s i d u s m e t h o d

c h a n g e m e n t t^{2} = x g i v e

I = \frac{1}{2} \int_{0}^{\infty} \frac{2 t}{{(1 + t^{4})}^{2}} d t = \frac{1}{2} \int_{0}^{\infty} \frac{d x}{{(1 + x^{2})}^{2}} = \frac{1}{4} \int_{- \infty}^{+ \infty} \frac{d x}{{(1 + x^{2})}^{2}}

l e t φ (z) = \frac{1}{{(z^{2} + 1)}^{2}} w e h a v e φ (z) = \frac{1}{{(z - i)}^{2} {(z + i)}^{2}}

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

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

= {l i m}_{z \to i} {{(z + i)}^{- 2}}^{(1)} = {l i m}_{z \to i} - 2 {(z + i)}^{- 3} = - 2 {(2 i)}^{- 3} = \frac{- 2}{{(2 i)}^{3}} = \frac{- 2}{- 8 i} = \frac{1}{4 i}

I = \frac{1}{4} (2 i π) \frac{1}{4 i} \Rightarrow I = \frac{π}{8} .

Answered by ajfour last updated on 16/Jul/18

I = \frac{1}{2} \int_{0}^{\infty} \frac{d z}{{(1 + z^{2})}^{2}} w h e r e z = t^{2}

l e t z = \tan θ \Rightarrow d z = \sec^{2} θ d θ

I = \frac{1}{4} \int_{0}^{π / 2} (1 + \cos 2 θ) d θ

= \frac{π}{8} .

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