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单词 ApplicationOfSineIntegralAtInfinity
释义

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 (http://planetmath.org/PartialFractionsOfExpressions)

1x(1+x2)=1x-x1+x2.

Thus we may write

f(a)=0sinaxx𝑑x-0xsinax1+x2𝑑x.

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

f(a)=0cosax1+x2𝑑x,f′′(a)=-0xsinax1+x2𝑑x,

we see that there is the linear differential equation

f(a)=π2+f′′(a)(2)

i.e.

f′′-f=-π2,

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

f(a)=C1ea+C2e-a+π2

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)
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更新时间:2025/5/4 12:44:30