application of sine integral at infinity

For finding the value of the improper integral

0sinaxx(1+x2)dx:=f(a)  (a>0) (1)

we first use the partial fraction representation (


Thus we may write


But by the entry sine integral at infinity, the first integral equals π2.  When we check


we see that there is the linear differential equation

f(a)=π2+f′′(a) (2)



satisfied by the sought function af(a).  We have the initial conditionsMathworldPlanetmath

f(0)=00𝑑x= 0,f(0)=0dx1+x2=/0arctanx=π2.

Therefore the general solution


of (2) requires that  C1=0,  C2=π2,  and consequently the sought integral f(a) has the value

0sinaxx(1+x2)𝑑x=π2(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