calculate-0-3x-2-2-x-2-1-x-2-2i-2-dx-

Question Number 62805 by mathmax by abdo last updated on 25/Jun/19

c a l c u l a t e \int_{0}^{+ \infty} \frac{3 x^{2} - 2}{(x^{2} + 1) {(x^{2} - 2 i)}^{2}} d x

Commented by mathmax by abdo last updated on 26/Jun/19

l e t I = \int_{0}^{\infty} \frac{3 x^{2} - 2}{(x^{2} + 1) {(x^{2} - 2 i)}^{2}} d x \Rightarrow 2 I = \int_{- \infty}^{+ \infty} \frac{3 x^{2} - 2}{(x^{2} + 1) {(x^{2} - 2 i)}^{2}} d x l e t

w (z) = \frac{3 z^{2} - 2}{(z^{2} + 1) {(z^{2} - 2 i)}^{2}} \Rightarrow w (z) = \frac{3 z^{2} - 2}{(z - i) (z + i) {(z - \sqrt{2 i})}^{2} {(z + \sqrt{2 i})}^{2}}

= \frac{3 z^{2} - 2}{(z - i) (z + i) {(z - \sqrt{2} e^{\frac{i π}{4}})}^{2} {(z + \sqrt{2} e^{\frac{i π}{4}})}^{2}} s o t h e p o l e s o f w a r e \bar{+} i a n d \bar{+} \sqrt{2} e^{\frac{i π}{4}}

r e s i d u s t h e o r e m g i v e \int_{- \infty}^{+ \infty} w (z) d z = 2 i π {R e s (w, i) + R e s (w, \sqrt{2} e^{\frac{i π}{4}})}

R e s (w, i) = {l i m}_{z \to i} (z - i) w (z) = \frac{- 5}{2 i {(- 1 - 2 i)}^{2}} = \frac{- 5}{2 i {(2 i + 1)}^{2}} = \frac{- 5}{2 i (- 4 + 4 i + 1)}

= \frac{- 5}{2 i (- 3 + 4 i)}

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

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

= {l i m}_{z \to \sqrt{2} e^{\frac{i π}{4}}} {\frac{6 z (z^{2} + 1) {(z + \sqrt{2} e^{\frac{i π}{4}})}^{2} - (3 z^{2} - 2) {2 z {(z + \sqrt{2} e^{\frac{i π}{4}}}}^{2} + 2 (z^{2} + 1) (z + \sqrt{2} e^{\frac{i π}{4}})}{{(z^{2} + 1)}^{2} {(z + \sqrt{2} e^{\frac{i π}{4}})}^{4}}}

= {l i m}_{z \to \sqrt{2} e^{\frac{i π}{4}}} {\frac{6 z (z^{2} + 1) (z + \sqrt{2} e^{\frac{i π}{4}}) - (3 z^{2} - 2) {2 z (z + \sqrt{2} e^{\frac{i π}{4}}) + 2 (z^{2} + 1)}}{{(z^{2} + 1)}^{2} {(z + \sqrt{2} e^{\frac{i π}{4}})}^{3}}}

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