application of sine integral at infinity

For finding the value of the improper integral

${\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle \frac{\sin ax}{x(1+x^2)}}dx:=f(a)\mathit{\hspace{1em}\hspace{1em}}(a>0)$ (1)

we first use the partial fraction representation (http://planetmath.org/PartialFractionsOfExpressions)

$$\frac{1}{x(1+x^2)}=\frac{1}{x}-\frac{x}{1+x^2}.$$

Thus we may write

$$f(a)={\int }_0^{\mathrm{\infty }}\frac{\sin ax}{x}𝑑x-{\int }_0^{\mathrm{\infty }}\frac{x\sin ax}{1+x^2}𝑑x.$$

But by the entry sine integral at infinity, the first integral equals ${\displaystyle \frac{\pi }{2}}$.  When we check

$$f^{\prime }(a)={\int }_0^{\mathrm{\infty }}\frac{\cos ax}{1+x^2}𝑑x,f^{\prime \prime }(a)=-{\int }_0^{\mathrm{\infty }}\frac{x\sin ax}{1+x^2}𝑑x,$$

we see that there is the linear differential equation

$f(a)={\displaystyle \frac{\pi }{2}}+f^{\prime \prime }(a)$ (2)

i.e.

$$f^{\prime \prime }-f=-\frac{\pi }{2},$$

satisfied by the sought function $a\mapsto f(a)$.  We have the initial conditions

$$f(0)={\int }_0^{\mathrm{\infty }}0𝑑x=\mathrm{\hspace{0.33em}0},f^{\prime }(0)={\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^2}=\underset{0}{\overset{\mathrm{\infty }}{/}}\arctan x=\frac{\pi }{2}.$$

Therefore the general solution

$$f(a)=C_1{e}^a+C_2{e}^{-a}+\frac{\pi }{2}$$

of (2) requires that  $C_1=0$,  $C_2=\frac{\pi }{2}$,  and consequently the sought integral $f(a)$ has the value

${\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle \frac{\sin ax}{x(1+x^2)}}𝑑x={\displaystyle \frac{\pi }{2}}(1-e^{-a})$ (3)

Title	application of sine integral at infinity
Canonical name	ApplicationOfSineIntegralAtInfinity
Date of creation	2013-03-22 18:45:58
Last modified on	2013-03-22 18:45:58
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	6
Author	pahio (2872)
Entry type	Application
Classification	msc 34A34
Classification	msc 34A12
Classification	msc 26A36
Classification	msc 26A24
Synonym	generalisation of sine integral at infinity