find the value of integrals I =∫_0 ^∞ (dx/((x^2 +1)^3 )) , J =∫


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 58770 by Mr X pcx last updated on 29/Apr/19

$f i n d t h e v a l u e o f i n t e g r a l s$ $I = \int_{0}^{\infty} \frac{d x}{{(x^{2} + 1)}^{3}}, J = \int_{0}^{\infty} \frac{d x}{{(x^{2} + 1)}^{5}}$
Commented by maxmathsup by imad last updated on 15/May/19
$method of residus we have 2I =∫_(−∞) ^(+∞) (dx/((x^2 +1)^3 )) let f(z) =(1/((z^2 +1)^3 )) =(1/((z−i)^3 (z+i)^3 )) so the poles of f are +^− i (triples) residus theorem give ∫_(−∞) ^(+∞) f(z)dz =2iπ Res(f,i) Res(f,i) =lim_(z→i) (1/((3−1)!)){ (z−i)^3 f(z)}^((2)) =lim_(z→i) (1/2){ (1/((z+i)^3 ))}^((2)) =lim_(z→i) (1/2){((−3(z+i)^2 )/((z+i)^6 ))}^((1)) =lim_(z→i) ((−3)/2){ (1/((z+i)^4 ))}^((1)) =lim_(z→i) ((−3)/2){((−4(z+i)^3 )/((z+i)^8 ))} =lim_(z→i) (6/((z+i)^5 )) =(6/((2i)^5 )) =(6/(2^5 i)) ⇒∫_(−∞) ^(+∞) f(z)dz =2iπ(6/(2^5 i)) =((6π)/2^4 ) =((6π)/(16)) =((3π)/8) ⇒ I =((3π)/(16)) .$
$m e t h o d o f r e s i d u s w e h a v e 2 I = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + 1)}^{3}} l e t f (z) = \frac{1}{{(z^{2} + 1)}^{3}}$ $= \frac{1}{{(z - i)}^{3} {(z + i)}^{3}} s o t h e p o l e s o f f a r e \bar{+} i (t r i p l e s) r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} f (z) d z = 2 i π R e s (f, i)$ $R e s (f, i) = {l i m}_{z \to i} \frac{1}{(3 - 1)!} {{(z - i)}^{3} f (z)}^{(2)}$ $= {l i m}_{z \to i} \frac{1}{2} {\frac{1}{{(z + i)}^{3}}}^{(2)} = {l i m}_{z \to i} \frac{1}{2} {\frac{- 3 {(z + i)}^{2}}{{(z + i)}^{6}}}^{(1)}$ $= {l i m}_{z \to i} \frac{- 3}{2} {\frac{1}{{(z + i)}^{4}}}^{(1)} = {l i m}_{z \to i} \frac{- 3}{2} {\frac{- 4 {(z + i)}^{3}}{{(z + i)}^{8}}}$ $= {l i m}_{z \to i} \frac{6}{{(z + i)}^{5}} = \frac{6}{{(2 i)}^{5}} = \frac{6}{2^{5} i} \Rightarrow \int_{- \infty}^{+ \infty} f (z) d z = 2 i π \frac{6}{2^{5} i} = \frac{6 π}{2^{4}} = \frac{6 π}{16} = \frac{3 π}{8} \Rightarrow$ $I = \frac{3 π}{16} .$
	Answered by tanmay last updated on 30/Apr/19

	$I = \int_{0}^{\infty} \frac{d x}{{(x^{2} + 1)}^{3}}$ $x = t a n a d x = {s e c}^{2} a d a$ $\int_{0}^{\frac{π}{2}} \frac{{s e c}^{2} a d a}{{s e c}^{6} a}$ $\int_{0}^{\frac{π}{2}} {c o s}^{4} a d a$ $\int_{0}^{\frac{π}{2}} {(s i n a)}^{2 \times \frac{1}{2} - 1} \times {(c o s a)}^{2 \times \frac{5}{2} - 1} d a$ $= \frac{⌈ (\frac{1}{2}) ⌈ (\frac{5}{2})}{2 ⌈ (\frac{1}{2} + \frac{5}{2})}$ $n o w ⌈ (\frac{1}{2}) = \sqrt{π}$ $⌈ (\frac{5}{2}) = ⌈ (\frac{3}{2} + 1)$ $= \frac{3}{2} ⌈ (\frac{3}{2})$ $= \frac{3}{2} \times ⌈ (\frac{1}{2} + 1)$ $= \frac{3}{2} \times \frac{1}{2} \times \sqrt{π}$ $⌈ (\frac{1}{2} + \frac{5}{2})$ $= ⌈ (2 + 1)$ $= 2 ⌈ (1 + 1)$ $= 2 \times 1$ $= 2$ $s o a n s w e r i s = \frac{\sqrt{π} \times \frac{3}{4} \times \sqrt{π}}{2 \times 2} = \frac{3}{16} π$ $J = \int_{0}^{\infty} \frac{d x}{{(x^{2} + 1)}^{5}}$ $= \int_{0}^{\frac{π}{2}} \frac{{s e c}^{2} a d a}{{s e c}^{10} a}$ $= \int_{0}^{\frac{π}{2}} {c o s}^{8} a d a$ $= \int_{0}^{\frac{π}{2}} {(s i n a)}^{2 \times \frac{1}{2} - 1} {(c o s a)}^{2 \times \frac{9}{2} - 1} d a$ $= \frac{⌈ (\frac{1}{2}) ⌈ (\frac{9}{2})}{2 ⌈ (\frac{1 + 9}{2})}$ $= \frac{\sqrt{π} \times \frac{7}{2} \times \frac{5}{2} \times \frac{3}{2} \times \frac{1}{2} \times \sqrt{π}}{2 \times 4 \times 3 \times 2} [⌈ 5) = ⌈ (4 + 1) = 4!$ $= \frac{π \times \frac{7 \times 5 \times 3}{2 \times 2 \times 2 \times 2}}{2 \times 4 \times 3 \times 2}$ $= π \times \frac{7 \times 5 \times 3}{16 \times 4 \times 3 \times 2 \times 2}$ $= \frac{35 π}{128} \times \frac{1}{2} = \frac{35 π}{256}$ $p l s c h e c k$ $\int_{0}^{\frac{π}{2}} {s i n}^{2 p - 1} a \times {c o s}_{}^{2 q - 1} a d a = \frac{⌈ (p) ⌈ q)}{2 ⌈ (p + q)}$ $i f o r g o t t o p u t 2 o f d e n o m i n a t o r . . . l a t e r c o r r e c t e d$
	Commented by Mr X pcx last updated on 30/Apr/19

	$t h a n k s s i r$
	Commented by tanmay last updated on 30/Apr/19

	$m o s t w e l c o m e s i r . . .$
	Answered by Smail last updated on 30/Apr/19

	$A_{n} = \int_{0}^{\infty} \frac{d x}{{(x^{2} + 1)}^{n}}$ $l e t x = t a n t \Rightarrow d x = (1 + {t a n}^{2} t) d t$ $A_{n} = \int_{0}^{π / 2} \frac{1 + {t a n}^{2} t}{{(1 + {t a n}^{2} t)}^{n}} d t = \int_{0}^{π / 2} \frac{d t}{{(1 + {t a n}^{2} t)}^{n - 1}}$ $= \int_{0}^{π / 2} {c o s}^{2 (n - 1)} t d t = \int_{0}^{π / 2} (1 - {s i n}^{2} t) {c o s}^{2 n - 4} t d t$ $= \int_{0}^{π / 2} {c o s}^{2 (n - 2)} t d t - \int_{0}^{π / 2} {s i n t s i n t c o s}^{2 n - 4} t d t$ $= A_{n - 1} - \int_{0}^{π / 2} s i n t \times \frac{d (- {c o s}^{2 n - 3} t)}{(2 n - 3)}$ $= A_{n - 1} + {[\frac{1}{2 n - 3} s i n t \times {c o s}^{2 n - 3} t]}_{0}^{π / 2} - \int_{0}^{π / 2} c o s t \times \frac{{c o s}^{2 n - 3} t}{2 n - 3} d t$ $= A_{n - 1} - \frac{A_{n}}{2 n - 3}$ $A_{n} = \frac{2 n - 3}{2 (n - 1)} A_{n - 1} = \frac{2 n - 3}{2 (n - 1)} (\frac{2 n - 5}{2 (n - 2)} A_{n - 2})$ $. . .$ $A_{n} = \frac{(2 (n - 1) - 1) (2 (n - 2) - 1) . . . (2 (n - (n - 1)) - 1)}{2^{n - 1} (n - 1) (n - 2) . . . (n - (n - 1))} A_{n - (n - 1)}$ $= \frac{(2 n - 3) (2 n - 4) (2 n - 5) (2 n - 6) . . . 1}{2^{n - 1} (n - 1)! \times (2 n - 4) (2 n - 6) . . . 2} A_{1}$ $= \frac{(2 n - 3)!}{2^{n - 1} (n - 1)! \times 2^{n - 2} (n - 2)!} \int_{0}^{π / 2} d t$ $= \frac{(2 n - 3)!}{2^{2 n - 3} (n - 1) {((n - 2)!)}^{2}} \times \frac{π}{2}$ $A_{n} = \frac{π (n - 1) (2 n - 3)!}{2^{2 n - 2} {((n - 1)!)}^{2}}$ $F o r n = 3$ $A_{3} = I = \int_{0}^{\infty} \frac{d x}{{(x^{2} + 1)}^{3}} = \frac{π (3 - 1) (6 - 3)!}{2^{6 - 2} {((3 - 1)!)}^{2}} = \frac{3 π}{16}$ $F o r n = 5$ $A_{5} = J = \int_{0}^{\infty} \frac{d x}{{(x^{2} + 1)}^{5}} = \frac{π (5 - 1) (10 - 3)!}{2^{10 - 2} {((5 - 1)!)}^{2}}$ $= \frac{7! π}{2^{12} \times 3^{2}} = \frac{7 \times 5 π}{2^{8}} = \frac{35 π}{256}$
	Commented by Mr X pcx last updated on 30/Apr/19

	$t h a n k s s i r$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum