calculate-by-residus-method-the-integral-0-dx-1-x-2-n-with-n-integr-and-n-1-

Question Number 70594 by mathmax by abdo last updated on 06/Oct/19

c a l c u l a t e b y r e s i d u s m e t h o d t h e i n t e g r a l \int_{0}^{\infty} \frac{d x}{{(1 + x^{2})}^{n}}

w i t h n i n t e g r a n d n ⩾ 1

Commented by mathmax by abdo last updated on 23/Oct/19

l e t A_{n} = \int_{0}^{\infty} \frac{d x}{{(x^{2} + 1)}^{n}} \Rightarrow 2 A_{n} = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + 1)}^{n}} l e t W (z) = \frac{1}{{(z^{2} + 1)}^{n}}

\Rightarrow W (z) = \frac{1}{{(z - i)}^{n} {(z + i)}^{n}} s o t h e p o l e s o f W a r e i a n d - i

a t o r d r n 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, i) = {l i m}_{z \to i} \frac{1}{(n - 1)!} {{(z - i)}^{n} W (z)}^{(n - 1)}

= {l i m}_{z \to i} \frac{1}{(n - 1)!} {{(z + i)}^{- n}}^{(n - 1)} l e t f i r s t f i n d {{(z + i)}^{- n}}^{(p)}

w e h a v e {{(z + i)}^{- n}}^{(1)} = - n {(z + i)}^{- n - 1} = - n {(z + i)}^{- (n + 1)}

{(z + i)}^{(2)} = {(- 1)}^{2} n (n + 1) {(z + i)}^{- (n + 2)}

\dots \Rightarrow {{(z + i)}^{- n}}^{(p)} = {(- 1)}^{p} n (n + 1) \dots (n + p - 1) {(z + i)}^{- (n + p)}

p = n - 1 \Rightarrow {{(z + i)}^{- n}}^{(n - 1)} = {(- 1)}^{n - 1} n (n + 1) \dots (2 n - 2) {(z + i)}^{- (2 n - 1)}

R e s (W, i) = {l i m}_{z \to i} \frac{1}{(n - 1)!} {(- 1)}^{n - 1} n (n + 1) \dots . (2 n - 2) {(z + i)}^{- 2 n + 1}

= \frac{{(- 1)}^{n - 1} n (n + 1) (n + 2) \dots (2 n - 2)}{(n - 1)!} \times \frac{1}{{(2 i)}^{2 n - 1}}

= \frac{{(- 1)}^{n - 1} n (n + 1) \dots . (2 n - 2)}{2^{2 n - 1} {(- 1)}^{n}} \times i = - i \frac{n (n + 1) (n + 2) \dots (2 n - 2)}{2^{2 n - 1}} \Rightarrow

\int_{- \infty}^{+ \infty} W (z) d z = 2 i π \times - i \frac{n (n + 1) (n + 2) \dots . (2 n - 2)}{2^{2 n - 1}}

= \frac{π}{4^{n - 1}} \times n (n + 1) \dots . (2 n - 2) = 2 A_{n} \Rightarrow

A_{n} = \frac{π n (n + 1) (n + 2) \dots (2 n - 2)}{2 \times 4^{n - 1}} w i t h n > 1 a n d

A_{1} = \frac{π}{2}

Answered by mind is power last updated on 06/Oct/19

f (z) = \frac{1}{{(1 + z^{2})}^{n}}

l i m z f (z) = 0

\int_{0}^{\infty} f (z) d z = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{d z}{{(1 + z^{2})}^{n}} = 2 i π R e s (f, i)

R e s (f, i) = \frac{1}{(n - 1)!} \frac{d^{n - 1}}{{d z}^{n - 1}} {(z - i)}^{n} . \frac{1}{{(z - i)}^{n} {(z + i)}^{n}} ∣_{z = i}

= \frac{1}{(n - 1)!} . \frac{d^{n - 1}}{{d z}^{n - 1}} . \frac{1}{{(z + i)}^{n}}

{(\frac{1}{{(z - i)}^{n}})}^{(k)} = \frac{{(- 1)}^{k} n \dots (n + k - 1)}{{(z + i)}^{n + k}}

\Rightarrow \frac{d^{n - 1}}{{d z}^{n - 1}} . \frac{1}{{(z + i)}^{n}} = \frac{{(- 1)}^{n} . n \dots . . (2 n - 2)}{{(z + i)}^{2 n - 1}}

R e s (f, i) = \frac{1}{(n - 1)!} . \frac{{(- 1)}^{n - 1} \times n \dots . (2 n - 2)}{{(2 i)}^{2 n - 1}} = \frac{2 i . {(- 1)}^{n - 1} . n \dots (2 n - 2)}{(n - 1)! . {(- 4)}^{n} .}

\int_{- \infty}^{+ \infty} f (z) d z = \frac{4 π . n . . (2 n - 2)}{(n - 1)! . 4^{n}} = \frac{4 π . (2 n - 2)!}{(n - 1)! 4^{n}}

\Rightarrow \int_{0}^{+ \infty} \frac{d x}{{(1 + x^{2})}^{n}} = \frac{2 . 4^{- n} π (2 n - 2)!}{(n - 1)!}

Commented by mathmax by abdo last updated on 06/Oct/19

t h a n k y o u s i r .