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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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Explanation

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Explanation:

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May 19, 2018

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Explanation:

$I \left(\alpha\right) = {\int}_{0}^{\infty} \setminus \frac{{e}^{- \alpha x} \sin x}{x} \mathrm{dx}$

Liebnitz diff under the integral sign:

$\frac{\mathrm{dI}}{d \alpha} = {\int}_{0}^{\infty} \setminus - x \frac{{e}^{- \alpha x} \sin x}{x} \mathrm{dx}$

$= - {\int}_{0}^{\infty} \setminus {e}^{- \alpha x} \sin x \setminus \mathrm{dx}$

That is very do-able on its own, but is also the Laplace transform of $\sin x$:

• $\frac{\mathrm{dI}}{d \alpha} = - m a t h {\boldsymbol{L}}_{\alpha} \left(\sin x\right) = - \frac{1}{1 + {\alpha}^{2}}$

-$\implies I \left(\alpha\right) = - \int \frac{1}{1 + {\alpha}^{2}} \setminus d \alpha = - {\tan}^{- 1} \alpha + C$

We have an IV:

$I \left(0\right) = \frac{\pi}{2} = C$

$\implies I \left(\alpha\right) = - {\tan}^{- 1} \alpha + \frac{\pi}{2}$

And from a trig identity for $\alpha > 0$:

${\cot}^{- 1} \alpha = = - {\tan}^{- 1} \alpha + \frac{\pi}{2}$

$\implies I \left(\alpha\right) = {\cot}^{- 1} \alpha$

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