How to prove that $\int_{0}^{\infty}$ $int_0^oo$ $\frac{e^{- α x} sin x}{x} d x = {cot}^{- 1} α$ $(e^(-alphax)sinx)/x dx=cot^-1alpha$ given that $\int_{0}^{\infty} \frac{sin x}{x} d x = \frac{π}{2}$ $int_0^oo sinx/x dx = pi/2$ ?

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Answer 1 · 2018-05-19T18:58:50

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 $alpha > 0$ :

$cot^(-1) alpha = = - tan ^(-1) alpha + pi/2$

$implies I(alpha) = cot ^(-1) alpha$

How to prove that int_0^oo∫∞0int_0^oo(e^(-alphax)sinx)/x dx=cot^-1alphae−αxsinxxdx=cot−1α(e^(-alphax)sinx)/x dx=cot^-1alpha given that int_0^oo sinx/x dx = pi/2∫∞0sinxxdx=π2int_0^oo sinx/x dx = pi/2?