let-f-n-t-0-dx-x-2-t-2-n-with-n-from-N-and-n-1-1-find-a-explicit-form-of-f-n-t-2-what-is-the-value-of-g-n-t-0-t-dx-x-2-t-2-n-1-3-calculate-0-d

Question Number 56932 by turbo msup by abdo last updated on 27/Mar/19

l e t f_{n} (t) = \int_{0}^{\infty} \frac{d x}{{(x^{2} + t^{2})}^{n}}

w i t h n f r o m N a n d n ⩾ 1

1 . f i n d a e x p l i c i t f o r m o f f_{n} (t)

2 . w h a t i s t h e v a l u e o f

g_{n} (t) = \int_{0}^{\infty} \frac{t d x}{{(x^{2} + t^{2})}^{n + 1}} ?

3 . c a l c u l a t e \int_{0}^{\infty} \frac{d x}{{(x^{2} + 3)}^{4}}

a n d \int_{0}^{\infty} \frac{d x}{{(x^{2} + 16)}^{3}}

Answered by Smail last updated on 27/Mar/19

l e t x = t \times t a n θ \Rightarrow d x = \frac{t d θ}{{c o s}^{2} θ}

f_{n} (t) = \int_{0}^{π / 2} \frac{t}{t^{2 n} {c o s}^{2} θ {(1 + {t a n}^{2} θ)}^{n}} d θ

= t^{1 - 2 n} \int_{0}^{π / 2} {c o s}^{2 (n - 1)} θ d θ

l e t A_{n - 1} = \int_{0}^{π / 2} {c o s}^{2 (n - 1)} θ d θ

= \int_{0}^{π / 2} ({c o s}^{2 (n - 2)} θ - {s i n}^{2} θ {c o s}^{2 (n - 2)} θ) d θ

B y p a r t s

u = s i n θ \Rightarrow u^{'} = c o s θ

v^{'} = s i n θ {c o s}^{2 n - 4} θ \Rightarrow v = - \frac{1}{2 n - 3} {c o s}^{2 n - 3}

A_{n - 1} = A_{n - 2} - \frac{1}{2 n - 3} A_{n - 1}

A_{n - 1} = \frac{2 n - 3}{2 n - 2} A_{n - 2} = \frac{(2 n - 3) (2 n - 5)}{(2 n - 2) (2 n - 4)} A_{n - 3}

= \frac{(2 (n - 1) - 1) (2 (n - 2) - 1) (2 (n - 3) - 1) \dots (2 (n - (n - 1)) - 1)}{(2 (n - 1)) (2 (n - 2)) \dots (2 (n - (n - 1))} A_{n - (n - 1)}

= \frac{(2 n - 3) (2 n - 5) (2 n - 7) \dots 1}{2^{n - 1} (n - 1)!} \int_{0}^{π / 2} {c o s}^{2} θ d θ

= \frac{(2 n - 2) (2 n - 3) (2 n - 4) (2 n - 5) (2 n - 6) \dots 1}{2^{n - 1} (n - 1)! \times 2^{n - 1} (n - 1)!} \times \frac{1}{2} {[θ + \frac{1}{2} s i n 2 θ]}_{0}^{π / 2}

= \frac{(2 n - 2)!}{2^{2 (n - 1)} {((n - 1)!)}^{2}} \times \frac{π}{2 \times 2}

A_{n - 1} = \frac{π (2 n - 2)!}{2^{2 n} {((n - 1)!)}^{2}}

f_{n} (t) = t^{1 - 2 n} \times \frac{π (2 n - 2)!}{2^{2 n} {((n - 1)!)}^{2}}

f_{n} (t) = \frac{π (2 n - 2)!}{2^{2 n} {((n - 1)!)}^{2}} t^{1 - 2 n}

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