calculate-dx-x-2-1-x-2-2-x-2-3-

Question Number 64150 by mathmax by abdo last updated on 14/Jul/19

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

Commented by mathmax by abdo last updated on 14/Jul/19

r e s i d u s m e t h o d l e t f (z) = \frac{1}{(z^{2} + 1) (z^{2} + 2) (z^{2} + 3)}

w e h a v e f (z) = \frac{1}{(z - i) (z + i) (z - i \sqrt{2}) (z + i \sqrt{2}) (z - i \sqrt{3}) (z + i \sqrt{3})}

s o t h e p o l e s o f f a r e \bar{+} i, \bar{+} i \sqrt{2} a n d \bar{+} i \sqrt{3} r e s i d u s t h e o r e m g i v e

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

R e s (f, i) = {l i m}_{z \to i} (z - i) f (z) = \frac{1}{2 i (- 1 + 2) (- 1 + 3)} = \frac{1}{4 i}

R e s (f, i \sqrt{2}) = {l i m}_{z \to i \sqrt{2}} (z - i \sqrt{2}) f (z) = \frac{1}{2 i \sqrt{2} (- 2 + 1) (- 2 + 3)}

= \frac{1}{- 2 i \sqrt{2}}

R e s (f, i \sqrt{3}) = {l i m}_{z \to i \sqrt{3}} (z - i \sqrt{3}) f (z) = \frac{1}{2 i \sqrt{3} (- 3 + 1) (- 3 + 2)}

= \frac{1}{4 i \sqrt{3}} \Rightarrow \int_{- \infty}^{+ \infty} f (z) d z = 2 i π {\frac{1}{4 i} - \frac{1}{2 i \sqrt{2}} + \frac{1}{4 i \sqrt{3}}}

= \frac{π}{2} - \frac{π}{\sqrt{2}} + \frac{π}{2 \sqrt{3}} \Rightarrow I = \frac{π}{2} - \frac{π}{\sqrt{2}} + \frac{π}{\sqrt{3}} .

Commented by mathmax by abdo last updated on 14/Jul/19

e r r o r o f t y p o I = \frac{π}{2} - \frac{π}{\sqrt{2}} + \frac{π}{2 \sqrt{3}} .

Answered by ajfour last updated on 14/Jul/19

l e t x^{2} + 2 = t

\frac{1}{(t - 1) t (t + 1)} = \frac{1 / 2}{t - 1} - \frac{1}{t} + \frac{1 / 2}{t + 1}

I = \frac{1}{2} \int \frac{d x}{x^{2} + 1} - \int \frac{d x}{x^{2} + 2} + \frac{1}{2} \int \frac{d x}{x^{2} + 3}

= \frac{1}{2} \tan^{- 1} x - \frac{1}{\sqrt{2}} \tan^{- 1} \frac{x}{\sqrt{2}} + \frac{1}{2 \sqrt{3}} \tan^{- 1} \frac{x}{\sqrt{3}} + c

w i t h l i m i t s

I = \frac{π}{2} - \frac{π}{\sqrt{2}} + \frac{π}{2 \sqrt{3}}

= (\frac{3 - 3 \sqrt{2} + \sqrt{3}}{6}) π .

Commented by mathmax by abdo last updated on 14/Jul/19

t h a n k y o u s i r a j f o u r y o u r a n s w e r i s c o r r e c t .