find-the-value-of-0-t-1-t-2-dt-

Question Number 36188 by prof Abdo imad last updated on 30/May/18

f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{\sqrt{t}}{1 + t^{2}} d t

Commented by maxmathsup by imad last updated on 15/Aug/18

l e t I = \int_{0}^{\infty} \frac{\sqrt{t}}{1 + t^{2}} d t c h a n g e m e n t \sqrt{t} = x g i v e

I = \int_{0}^{\infty} \frac{x}{1 + x^{4}} (2 x) d x = 2 \int_{0}^{\infty} \frac{x^{2}}{1 + x^{4}} d x = \int_{- \infty}^{+ \infty} \frac{x^{2}}{x^{4} + 1} d x l e t c o n s i d e r

φ (z) = \frac{z^{2}}{z^{4} + 1} w e h a v e φ (z) = \frac{z^{2}}{(z^{2} - i) (z^{2} + i)}

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

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

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

R e s (φ, z_{i}) = \frac{z_{i}^{2}}{4 z_{i}^{3}} = \frac{z_{i}^{3}}{- 4} = - \frac{1}{4} z_{i}^{3} \Rightarrow R e s (φ, e^{\frac{i π}{4}}) = - \frac{1}{4} e^{i \frac{3 π}{4}}

R e s (φ, - e^{- \frac{i π}{4}}) = \frac{{(- e^{- \frac{i π}{4}})}^{3}}{- 4} = \frac{1}{4} e^{- \frac{i 3 π}{4}}

\int_{- \infty}^{+ \infty} φ (z) d z = \frac{2 i π}{4} {- e^{\frac{i 3 π}{4}} + e^{- \frac{i 3 π}{4}}} = - \frac{i π}{2} (e^{\frac{i 3 π}{4}} - e^{- \frac{i 3 π}{4}})

= - \frac{i π}{2} (2 i s i n (\frac{3 π}{4})) = \frac{π \sqrt{2}}{2} \Rightarrow I = \frac{π}{\sqrt{2}} .