How to prove that #int_0^oo##(e^(-alphax)sinx)/x dx=cot^-1alpha# given that #int_0^oo sinx/x dx = pi/2#?

Question

Aurora Alvarado · Accepted Answer

See below

#I(alpha) = int_0^oo \ (e^(-alphax)sinx)/x dx # Liebnitz diff under the integral sign: #(dI)/(d alpha) = int_0^oo \ - x (e^(-alphax)sinx)/x dx # #=- int_0^oo \ e^(-alphax)sinx \ dx # That is very do-able on its own, but is also the Laplace transform of #sin x#: -#implies I(alpha) =- int 1/(1 + alpha^2) \ d alpha= - tan ^(-1) alpha + C# We have an IV: #I(0) = pi/2 = C# #implies I(alpha) = - tan ^(-1) alpha + pi/2# And from a trig identity for #alpha > 0#: #cot^(-1) alpha = = - tan ^(-1) alpha + pi/2# #implies I(alpha) = cot ^(-1) alpha #

Emily Ammerman · Answer

undefined To prove the identity ∫(0 to ∞) (e^(-αx)sinx)/x dx = arccot(α), where ∫(0 to ∞) sinx/x dx = π/2, you can use the Laplace transform method.

1. Start with the Laplace transform of f(x) = sinx:
   L{sinx} = ∫(0 to ∞) e^(-sx)sinx dx = 1/(1+s^2), where s is a complex variable.

2. Multiply both sides by e^(-αx) and integrate from 0 to ∞:
   ∫(0 to ∞) e^(-αx)sinx dx = ∫(0 to ∞) e^(-αx)/(1+s^2) ds

3. Express the left side using the given identity:
   ∫(0 to ∞) e^(-αx)sinx dx = ∫(0 to ∞) e^(-αx)sinx/(x/x) dx = π/2

4. Substitute this value into the equation:
   π/2 = ∫(0 to ∞) e^(-αx)/(1+s^2) ds

5. Apply the inverse Laplace transform to both sides:
   L^(-1){π/2} = L^(-1){∫(0 to ∞) e^(-αx)/(1+s^2) ds}

6. The left side simplifies to π/2.
   For the right side, use the property of inverse Laplace transform:
   L^(-1){F(s)} = f(t) where F(s) = ∫(0 to ∞) e^(-αx)/(1+s^2) ds.
   Thus, the right side becomes f(t) = e^(-αt)sin(t).

7. Take the inverse Laplace transform:
   L^(-1){∫(0 to ∞) e^(-αx)/(1+s^2) ds} = e^(-αt)sin(t) = sin(t)e^(-αt)

8. Recognize the inverse Laplace transform:
   sin(t)e^(-αt) = arccot(α)

9. Therefore, ∫(0 to ∞) (e^(-αx)sinx)/x dx = arccot(α).