Question

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

Explanation:

`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`:

• `(dI)/(d alpha) =- mathbb L_alpha ( sin x) = - 1/(1 + alpha^2)`

-`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`