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

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May 19, 2018

Answer:

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 #

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