calculate ∫_(−∞) ^(+∞) (dx/((x^2 +1)(x^2 +2)(x^2 +3)))


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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
$residus method let f(z)=(1/((z^2 +1)(z^2 +2)(z^2 +3))) we have f(z) =(1/((z−i)(z+i)(z−i(√2))(z+i(√2))(z−i(√3))(z+i(√3)))) so the poles of f are +^− i ,+^− i(√2)and +^− i(√3) residus theorem give ∫_(−∞) ^(+∞) f−z)dz =2iπ {Res(f,i)+Res(f,i(√2)) +Res(f,i(√3))} Res(f,i) =lim_(z→i) (z−i)f(z) =(1/(2i(−1+2)(−1+3))) =(1/(4i)) Res(f,i(√2))=lim_(z→i(√2)) (z−i(√2))f(z) =(1/(2i(√2)(−2+1)(−2+3))) =(1/(−2i(√2))) Res(f,i(√3)) =lim_(z→i(√3)) (z−i(√3))f(z)=(1/(2i(√3)(−3+1)(−3+2))) =(1/(4i(√3))) ⇒∫_(−∞) ^(+∞) f(z)dz =2iπ{(1/(4i)) −(1/(2i(√2))) +(1/(4i(√3)))} =(π/2) −(π/(√2)) +(π/(2(√3))) ⇒ I =(π/2) −(π/(√2)) +(π/(√3)) .$
$r e s i d u s m e t h o d l e t f (z) = \frac{1}{(z^{2} + 1) (z^{2} + 2) (z^{2} + 3)}$ $w e h a v e f (z) = \frac{1}{(z - i) (z + i) (z - i \sqrt{2}) (z + i \sqrt{2}) (z - i \sqrt{3}) (z + i \sqrt{3})}$ $s o t h e p o l e s o f f a r e \bar{+} i, \bar{+} i \sqrt{2} a n d \bar{+} i \sqrt{3} r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} f - z) d z = 2 i π {R e s (f, i) + R e s (f, i \sqrt{2}) + R e s (f, i \sqrt{3})}$ $R e s (f, i) = {l i m}_{z \to i} (z - i) f (z) = \frac{1}{2 i (- 1 + 2) (- 1 + 3)} = \frac{1}{4 i}$ $R e s (f, i \sqrt{2}) = {l i m}_{z \to i \sqrt{2}} (z - i \sqrt{2}) f (z) = \frac{1}{2 i \sqrt{2} (- 2 + 1) (- 2 + 3)}$ $= \frac{1}{- 2 i \sqrt{2}}$ $R e s (f, i \sqrt{3}) = {l i m}_{z \to i \sqrt{3}} (z - i \sqrt{3}) f (z) = \frac{1}{2 i \sqrt{3} (- 3 + 1) (- 3 + 2)}$ $= \frac{1}{4 i \sqrt{3}} \Rightarrow \int_{- \infty}^{+ \infty} f (z) d z = 2 i π {\frac{1}{4 i} - \frac{1}{2 i \sqrt{2}} + \frac{1}{4 i \sqrt{3}}}$ $= \frac{π}{2} - \frac{π}{\sqrt{2}} + \frac{π}{2 \sqrt{3}} \Rightarrow I = \frac{π}{2} - \frac{π}{\sqrt{2}} + \frac{π}{\sqrt{3}} .$
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

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