Sunday, 25 January 2015

complex analysis - Showing intinfty0frac1(x2+1)2(x2+4)=fracpi18 via contour integration



I want to show that:
01(x2+1)2(x2+4)=π18



so considering:
γ1(z2+1)2(z2+4)
where gamma is the curve going from 0 to R along the real axis, from R to R via a semi-circle in the upper plane and then from R to 0 along the real axis.



Using the residue theorem we have that:
γ1(z2+1)2(z2+4)=2πiRes


so re-writing the integrand as 1(z2i)(z+2i)(z+i)2(zi)2



we can see that there is two simple poles at 2i,2i and two poles of order 2 at i,i.
Calculating the residues:
Resz=2i=limz2i1(z+2i)(z+i)2(zi)2=136i




Resz=2i=limz2i1(z2i)(z+i)2(zi)2=136i



Resz=ilimziddz1(z2i)(z+2i)(z+i)2=2i36+224i



Resz=ilimziddz1(z2i)(z+2i)(zi)2=2i36+224i



But now the sum of the residues is 0 and so when I integrate over my curve letting R go to (and the integral over top semi-circle goes to 0) I will just get 0?



Not sure what I've done wrong?

Thanks very much for any help


Answer



Consider the contour C that spans along R to R and around the arc Reiθ for 0θπ.



Letting



f(z):=1(z2+1)2(z2+4)=1(z+i)2(zi)2(z+2i)(z2i)



and we see the poles are located at ±i and ±2i. Letting R, it is very clear that the denominator explodes, causing the integral around the arc to disappear. Then




Cf(z)dz=2πi(Resz=if(z)+Resz=2if(z))



because 2i and i are the only poles in C.
The pole of i is of order 2:



Resz=if(z)=limzi11!ddz(zi)2f(z)=limziddz1(z+i)2(z2+4)=limzi2(2z2+iz+4)(i+z)3(4+z2)2=i36




The pole of 2i is simple:



Resz=2if(z)=limz2i(z2i)f(z)=1(4+1)2(2i+2i)=i36



So finally




0f(x)dx=12f(x)dx=πi(i36i36)=π18


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