Question Number 149100 by Naser last updated on 02/Aug/21

Answered by Kamel last updated on 02/Aug/21

1/ L(J_0 (t))=(1/π)∫_0 ^π ∫_0 ^(+∞) cos(tsin(θ))e^(−st) dtdθ=(1/π)∫_0 ^π ((sdθ)/(sin^2 (θ)+s^2 ))=((2s)/π)∫_0 ^(π/2) (dθ/(s^2 +cos^2 (θ))) =((2s)/π)∫_0 ^(+∞) (du/(1+s^2 +s^2 u^2 ))=(1/( (√(1+s^2 )))) 2/β(n,n)=((Γ(n)Γ(n))/(Γ(2n)))=Γ((1/2))Γ(n)2^(1−2n) ((Γ(n)2^(2n−1) )/(Γ(2n)Γ((1/2))))=2^(1−2n) ((Γ(n)Γ((1/2)))/(Γ(n+(1/2))))=2^(1−2n) β(n,(1/2)) 3/L^(−1) ((d/ds)L^(−1) (Ln(1+(1/s^2 ))))=L^(−1) (((2s)/(1+s^2 ))−(2/s))=2(cos(t)−1) ∴ L^(−1) (Ln(1+(1/s^2 )))=2((1−cos(t))/t) 4/??? 5/ ∫_0 ^2 x^3 J_0 (x)dx=∫_0 ^2 x^2 (xJ_(1−1) (x))dx=8J_1 (2)−2∫_0 ^2 x^2 J_(2−1) (x)dx=8(J_1 (2)−J_2 (2)) 6−1/∫_0 ^1 x^m Ln^n (x)dx=^(x=e^(−t) ) (−1)^n ∫_0 ^(+∞) t^n e^(−(m+1)t) dt=(((−1)^n n!)/((m+1)^(m+1) )) 6−2/∫_(−∞) ^(+∞) ((e^(2θ) dθ)/(e^(3θ) +1))=^(t=e^(−3θ) ) (1/3)∫_0 ^(+∞) ((t^(−(2/3)) dt)/(1+t))=((2π)/(3(√3))) 7/ L(sin((√t)))=∫_0 ^(+∞) sin((√t))e^(−st) dt=^(u=(√t)) 2∫_0 ^(+∞) usin(u)e^(−su^2 ) du=(1/( s))∫_0 ^(+∞) cos(u)e^(−su^2 ) du I(α)=∫_0 ^(+∞) cos(αx)e^(−sx^2 ) dx=((2s)/α) ∫_0 ^(+∞) xsin(αx)e^(−sx^2 ) dx=−((2s)/α) I′(α) I(α)=Ae^(−(1/(4s))α^2 ) , I(0)=((√π)/(2(√s))) ⇒I(α)=((√π)/(2(√s))) e^(−(1/(4s))α^2 ) ∴ L(sin((√t)))=((√π)/(2s^(3/2) ))e^(−(1/(4s)))

Commented bypuissant last updated on 03/Aug/21

monsieur kamel je pense que vous etes professeur..