Integral of sine integral times exponential: ∫₀^∞ Si(ax)e^(-x)dx = arctan(a)

[hmiamid]

Hello PF,

I just found a curious integral. I wondered if it comes from a bigger group of integral definitions:
$$\int_0^\infty \mathrm{Si}(ax)e^{-x}\mathrm{d}x=\mathrm{atan}(a)$$
Where Si(x) is the sine integral function $\mathrm{Si}(x)=\int_0^x \frac{\mathrm{sin}x}{x}\mathrm{d}x$
I proved the equation by developing Si(x) in Taylor series and there is a nice simplification between Si Taylor coefficients and the Gamma function.

 

[lurflurf]

Sure something like

$$\int_0^\infty \! \! \! \left\{ \int_0^x \mathrm{f}( \tau ) \, \mathrm{d} \tau \right\} e^{-s \, x}\mathrm{d}x=\dfrac{1}{s} \int_0^\infty \mathrm{f}( x ) \, e^{-s \, x}\mathrm{d}x$$