I want to show that:
∫∞01(x2+1)2(x2+4)=π18
so considering:
∫γ1(z2+1)2(z2+4)
Using the residue theorem we have that:
∫γ1(z2+1)2(z2+4)=2πi∑Res
so re-writing the integrand as 1(z−2i)(z+2i)(z+i)2(z−i)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=limz→2i1(z+2i)(z+i)2(z−i)2=136i
Resz=−2i=limz→2i1(z−2i)(z+i)2(z−i)2=−136i
Resz=ilimz→iddz1(z−2i)(z+2i)(z+i)2=2i36+224i
Resz=−ilimz→−iddz1(z−2i)(z+2i)(z−i)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 0≤θ≤π.
Letting
f(z):=1(z2+1)2(z2+4)=1(z+i)2(z−i)2(z+2i)(z−2i)
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)=limz→i11!ddz(z−i)2f(z)=limz→iddz1(z+i)2(z2+4)=limz→i2(2z2+iz+4)(i+z)3(4+z2)2=−i36
The pole of 2i is simple:
Resz=2if(z)=limz→2i(z−2i)f(z)=1(−4+1)2(2i+2i)=−i36
So finally
∫∞0f(x)dx=12∫∞−∞f(x)dx=πi(−i36−i36)=π18
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