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

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

$${Let}\:{I}_{{n}} =\overset{\pi/\mathrm{4}} {\int}_{\mathrm{0}\:} \left(\frac{\mathrm{1}−\mathrm{tan}\:{A}}{\mathrm{1}+\mathrm{tan}\:{A}}\right)^{{n}} {dA}\:\:{what}\:{is} \\ $$$${the}\:{Laplace}\:{Transform}\:{and}\:{the} \\ $$$${Fourier}\:{Transform} \\ $$

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)

$${I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\frac{{cos}\left({A}\right)−{sin}\left({A}\right)}{{sin}\left({A}\right)+\mathrm{cos}\left(\mathrm{A}\right)}\right)^{{n}} {dA} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(−\mathrm{1}\right)^{{n}} \left\{\frac{{sin}\left({A}−\frac{\pi}{\mathrm{4}}\right)}{{cos}\left({A}−\frac{\pi}{\mathrm{4}}\right)}\right\}^{{n}} \\ $$$$=\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {tg}^{{n}} \left({A}−\frac{\pi}{\mathrm{4}}\right){dA} \\ $$$${y}={A}−\frac{\pi}{\mathrm{4}}\Rightarrow{dy}={dA} \\ $$$$=\left(−\mathrm{1}\right)^{{n}} \int_{−\frac{\pi}{\mathrm{4}}} ^{\mathrm{0}} {tg}^{{n}} \left({y}\right){dy},{t}={tg}\left({y}\right)\Rightarrow{dy}=\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\left(−\mathrm{1}\right)^{{n}} \int_{−\mathrm{1}} ^{\mathrm{0}} \frac{{t}^{{n}} }{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{n}} }{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{n}} .\left(\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−{t}^{\mathrm{2}} \right)^{{k}} \right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} {t}^{\mathrm{2}{k}+{n}} {dt}=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+{n}+\mathrm{1}} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\underset{{t}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\prod}}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}+{t}\right).\:\:\underset{{t}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\prod}}\left(\mathrm{1}+{t}\right)}{\mathrm{2}\underset{{t}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\prod}}\left(\frac{{n}+\mathrm{3}}{\mathrm{2}}+{t}\right)}\:\:\:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{1}+{t}\right)_{{k}} \left(\frac{{n}+\mathrm{1}}{\mathrm{2}}+{t}\right)_{{k}} }{\left(\frac{{n}+\mathrm{3}}{\mathrm{2}}+{t}\right)_{{k}} }.\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!^{} } \\ $$$$=\frac{\mathrm{1}\:}{\mathrm{2}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\mathrm{1},\frac{{n}+\mathrm{1}}{\mathrm{2}};\frac{{n}+\mathrm{3}}{\mathrm{2}};−\mathrm{1}\right) \\ $$

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