Let-I-n-0-pi-4-1-tan-A-1-tan-A-n-dA-what-is-the-Laplace-Transform-and-the-Fourier-Transform-

Question Number 85914 by frc2crc last updated on 26/Mar/20

L e t I_{n} = {\int^{π / 4}}_{0} {(\frac{1 - \tan A}{1 + \tan A})}^{n} d A w h a t i s

t h e L a p l a c e T r a n s f o r m a n d t h e

F o u r i e r T r a n s f o r m

Answered by mind is power last updated on 26/Mar/20

I_{n} = \int_{0}^{\frac{π}{4}} {(\frac{c o s (A) - s i n (A)}{s i n (A) + \cos (A)})}^{n} d A

= \int_{0}^{\frac{π}{4}} {(- 1)}^{n} {\frac{s i n (A - \frac{π}{4})}{c o s (A - \frac{π}{4})}}^{n}

= {(- 1)}^{n} \int_{0}^{\frac{π}{4}} {t g}^{n} (A - \frac{π}{4}) d A

y = A - \frac{π}{4} \Rightarrow d y = d A

= {(- 1)}^{n} \int_{- \frac{π}{4}}^{0} {t g}^{n} (y) d y, t = t g (y) \Rightarrow d y = \frac{d t}{1 + t^{2}}

= {(- 1)}^{n} \int_{- 1}^{0} \frac{t^{n}}{1 + t^{2}} d t

= \int_{0}^{1} \frac{t^{n}}{1 + t^{2}} d t = \int_{0}^{1} t^{n} . (\sum_{k ⩾ 0} {(- t^{2})}^{k}) d t

= \int_{0}^{1} \sum_{k ⩾ 0} {(- 1)}^{k} t^{2 k + n} d t = \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{2 k + n + 1}

= \sum_{k ⩾ 0} \frac{\underset{t = 0}{\prod^{k - 1}} (\frac{n + 1}{2} + t) . \underset{t = 0}{\prod^{k - 1}} (1 + t)}{2 \underset{t = 0}{\prod^{k - 1}} (\frac{n + 3}{2} + t)} \frac{{(- 1)}^{k}}{k!}

= \frac{1}{2} \sum_{k ⩾ 0} \frac{{(1 + t)}_{k} {(\frac{n + 1}{2} + t)}_{k}}{{(\frac{n + 3}{2} + t)}_{k}} . \frac{{(- 1)}^{k}}{k!^{}}

= \frac{1}{2}_{2} F_{1} (1, \frac{n + 1}{2}; \frac{n + 3}{2}; - 1)