Tuesday, 26 November 2013

integration - Improper integral of sin(x)/x from zero to infinity




I was having trouble with the following integral:

0sin(x)xdx. My question is, how does one go about evaluating this, since its existence seems fairly intuitive, while its solution, at least to me, does not seem particularly obvious.


Answer



Let I(s) be given by



I(s)=0esxsin(x)xdx



for s0. Note that limsI(s)=0 and that I(0)=0sin(x)xdx is the integral of interest.







Differentiating I(s) as given by (1) (this is justified by uniform convergence of the "differentiated" integral for sδ>0) reveals



I(s)=0esxsin(x)dx=11+s2



Integrating (2), we find that I(s)=π/2arctan(s) whence setting s=0 yields the coveted result



0sin(x)xdx=π2



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