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Question Number 63662 by mathmax by abdo last updated on 06/Jul/19

l e t A_{n} = \int_{0}^{\infty} \frac{x^{a - 1}}{1 + x^{n}} d x w i t h n i n t e g r a n d n ⩾ 2 a n d 0 < a < 1

1) c a l c u l a t e A_{n}

2) f i n d t h e v a l u e s o f \int_{0}^{\infty} \frac{x^{a - 1}}{1 + x^{2}} d x a n d \int_{0}^{\infty} \frac{x^{a - 1}}{1 + x^{3}} d x

3) c a l c u l a t e \int_{0}^{\infty} \frac{d x}{\sqrt{x} (1 + x^{4})} a n d \int_{0}^{\infty} \frac{d x}{(^{3} \sqrt{x^{2}}) (1 + x^{4})}

Commented by mathmax by abdo last updated on 10/Jul/19

1) w e h a v e A_{n} = \int_{0}^{\infty} \frac{x^{a - 1}}{1 + x^{n}} d x c h a n g e m e n t x^{n} = t g i v e x = t^{\frac{1}{n}} \Rightarrow

A_{n} = \int_{0}^{\infty} \frac{{(t^{\frac{1}{n}})}^{a - 1}}{1 + t} \frac{1}{n} t^{\frac{1}{n} - 1} d t = \frac{1}{n} \int_{0}^{\infty} \frac{t^{\frac{a - 1}{n} + \frac{1}{n} - 1}}{1 + t} d t

= \frac{1}{n} \int_{0}^{\infty} \frac{t^{\frac{a}{n} - 1}}{1 + t} d t = \frac{1}{n} \frac{π}{s i n (\frac{π a}{n})} b y u s e o f r e s u l t \int_{0}^{\infty} \frac{t^{α - 1}}{1 + t} d t = \frac{π}{s i n (π α)}

2) \int_{0}^{\infty} \frac{x^{a - 1}}{1 + x^{2}} d x = A_{2} = \frac{π}{2 s i n (\frac{π a}{2})}

\int_{0}^{\infty} \frac{x^{a - 1}}{1 + x^{3}} d x = A_{3} = \frac{π}{3 s i n (\frac{π a}{3})}

3) \int_{0}^{\infty} \frac{d x}{\sqrt{x} (1 + x^{4})} = \int_{0}^{\infty} \frac{x^{- \frac{1}{2}}}{1 + x^{4}} d x = \int_{0}^{\infty} \frac{x^{\frac{1}{2} - 1}}{1 + x^{4}} (a = \frac{1}{2} a n d n = 4)

= \frac{π}{4 s i n (\frac{π}{4})} = \frac{π}{4 \frac{\sqrt{2}}{2}} = \frac{π}{2 \sqrt{2}} .

\int_{0}^{\infty} \frac{d x}{(^{3} \sqrt{x^{2}}) (1 + x^{4})} = \int_{0}^{\infty} \frac{x^{- \frac{2}{3}}}{1 + x^{4}} d x = \int_{0}^{\infty} \frac{x^{\frac{1}{3} - 1}}{1 + x^{4}} (\to n = 4 a n d a = \frac{1}{3})

= \frac{π}{4 s i n (\frac{π}{6})} = \frac{π}{4 . \frac{1}{2}} = \frac{π}{2}