Friday, 13 February 2015

complex analysis - inti0nftyfraccos(tx)(x22x+2),mathrmdx for t real




This was a question on an old prelim exam in complex analysis: compute



0cos(tx)x22x+2dx



for t real. I've tried…




  • Residue calculus—it's easy to integrate the similar 0cos(tx)x2+2dx largely because the integrand is even, but this integrand isn't. Similarly 0sin(tx)x2+2dx seems hard. Even if the original integral was from 1 to , so the denominator was even about x=1, we seem to need this latter, hard integral involving sin(tx).

  • Mathematica—even for t=1, it gives the answer in terms of z0sin(t)tdtandz0cos(t)tdt, which is not helpful.

  • Looked through Gamelin's Complex Analysis text for inspiration; everything close used even or odd integrands.


  • Googling/searching here, though it's hard to search for such a specific type of integral.



There's a chance there's just a typo on the old prelim, for what it's worth.


Answer



The integral is even in t, so I will assume t0.
0cos(tx)x22x+2dx=1cos(tx+t)x2+1dx=12i1(1xi1x+i)cos(tx+t)dx=12i1icos(tx+t+it)xdx12i1+icos(tx+tit)xdx=12icos(t+it)1icos(tx)xdx12isin(t+it)1isin(tx)xdx12icos(tit)1+icos(tx)xdx+12isin(tit)1+isin(tx)xdx=12icos(t+it)Ci(tit)12isin(t+it)(π2Si(tit))=+12icos(tit)Ci(t+it)+12isin(tit)(π2Si(t+it))
This matches what Mathematica computes.



If the integral was over the entire real line, the answer would avoid Ci and Si. In fact, using contour integration, we get
cos(tx)x22x+2dx=cos(tx+t)x2+1dx=cos(t)cos(tx)x2+1dxsin(t)sin(tx)x2+1dx=cos(t)Re(eitxx2+1dx)0=cos(t)Re(12iγ(1xi1x+i)eitxdx)=cos(t)Re(12iγeitxxidx12iγeitxx+idx)=cos(t)Re(12i2πiet0)=πcos(t)et
Where γ is the contour along the real axis and circling back counter-clockwise around the upper half-plane.




This may have been the intended question.


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