# application of sine integral at infinity

For finding the value of the improper integral

 $\displaystyle\int_{0}^{\infty}\!\frac{\sin{ax}}{x(1\!+\!x^{2})}\,dx\;:=\;f(a)% \qquad(a>0)$ (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

 $f(a)=\int_{0}^{\infty}\frac{\sin{ax}}{x}\,dx-\int_{0}^{\infty}\frac{x\sin{ax}}% {1+x^{2}}\,dx.$

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

 $f^{\prime}(a)\;=\;\int_{0}^{\infty}\frac{\cos{ax}}{1\!+\!x^{2}}\,dx,\quad f^{% \prime\prime}(a)\;=\;-\!\int_{0}^{\infty}\frac{x\sin{ax}}{1\!+\!x^{2}}\,dx,$

we see that there is the linear differential equation

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

i.e.

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

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

 $f(0)\;=\;\int_{0}^{\infty}{0}\,dx\;=\;0,\quad f^{\prime}(0)\;=\;\int_{0}^{% \infty}\frac{dx}{1\!+\!x^{2}}\;=\;\operatornamewithlimits{\Big{/}}_{\!\!\!0}^{% \,\quad\infty}\!\arctan{x}\;=\;\frac{\pi}{2}.$

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

 $\displaystyle\int_{0}^{\infty}\!\frac{\sin{ax}}{x(1\!+\!x^{2})}\,dx\;=\;\frac{% \pi}{2}(1-e^{-a})$ (3)
Title application of sine integral at infinity ApplicationOfSineIntegralAtInfinity 2013-03-22 18:45:58 2013-03-22 18:45:58 pahio (2872) pahio (2872) 6 pahio (2872) Application msc 34A34 msc 34A12 msc 26A36 msc 26A24 generalisation of sine integral at infinity