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

Answer 1

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 #
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Answer 2

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).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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