Let I_n =∫_(0 ) ^(π/4) (((1−tan A)/(1+tan A)))^n dA what is the Laplace Transform and the Fourier Transform


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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
	$I_n =∫_0 ^(π/4) (((cos(A)−sin(A))/(sin(A)+cos(A))))^n dA =∫_0 ^(π/4) (−1)^n {((sin(A−(π/4)))/(cos(A−(π/4))))}^n =(−1)^n ∫_0 ^(π/4) tg^n (A−(π/4))dA y=A−(π/4)⇒dy=dA =(−1)^n ∫_(−(π/4)) ^0 tg^n (y)dy,t=tg(y)⇒dy=(dt/(1+t^2 )) =(−1)^n ∫_(−1) ^0 (t^n /(1+t^2 ))dt =∫_0 ^1 (t^n /(1+t^2 ))dt=∫_0 ^1 t^n .(Σ_(k≥0) (−t^2 )^k )dt =∫_0 ^1 Σ_(k≥0) (−1)^k t^(2k+n) dt=Σ_(k≥0) (((−1)^k )/(2k+n+1)) =Σ_(k≥0) ((Π_(t=0) ^(k−1) (((n+1)/2)+t). Π_(t=0) ^(k−1) (1+t))/(2Π_(t=0) ^(k−1) (((n+3)/2)+t))) (((−1)^k )/(k!)) =(1/2)Σ_(k≥0) (((1+t)_k (((n+1)/2)+t)_k )/((((n+3)/2)+t)_k )).(((−1)^k )/(k!^ )) =((1 )/2) _2 F_1 (1,((n+1)/2);((n+3)/2);−1)$
	$I_{n} = \int_{0}^{\frac{π}{4}} {(\frac{c o s (A) - s i n (A)}{s i n (A) + \cos (A)})}^{n} d A$ $= \int_{0}^{\frac{π}{4}} {(- 1)}^{n} {\frac{s i n (A - \frac{π}{4})}{c o s (A - \frac{π}{4})}}^{n}$ $= {(- 1)}^{n} \int_{0}^{\frac{π}{4}} {t g}^{n} (A - \frac{π}{4}) d A$ $y = A - \frac{π}{4} \Rightarrow d y = d A$ $= {(- 1)}^{n} \int_{- \frac{π}{4}}^{0} {t g}^{n} (y) d y, t = t g (y) \Rightarrow d y = \frac{d t}{1 + t^{2}}$ $= {(- 1)}^{n} \int_{- 1}^{0} \frac{t^{n}}{1 + t^{2}} d t$ $= \int_{0}^{1} \frac{t^{n}}{1 + t^{2}} d t = \int_{0}^{1} t^{n} . (\sum_{k ⩾ 0} {(- t^{2})}^{k}) d t$ $= \int_{0}^{1} \sum_{k ⩾ 0} {(- 1)}^{k} t^{2 k + n} d t = \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{2 k + n + 1}$ $= \sum_{k ⩾ 0} \frac{\underset{t = 0}{\prod^{k - 1}} (\frac{n + 1}{2} + t) . \underset{t = 0}{\prod^{k - 1}} (1 + t)}{2 \underset{t = 0}{\prod^{k - 1}} (\frac{n + 3}{2} + t)} \frac{{(- 1)}^{k}}{k!}$ $= \frac{1}{2} \sum_{k ⩾ 0} \frac{{(1 + t)}_{k} {(\frac{n + 1}{2} + t)}_{k}}{{(\frac{n + 3}{2} + t)}_{k}} . \frac{{(- 1)}^{k}}{k!^{}}$ $= \frac{1}{2}_{2} F_{1} (1, \frac{n + 1}{2}; \frac{n + 3}{2}; - 1)$
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