I am aware that there is a theorem which states that for $0
My attempt is the following: Let's take f(z)=1z1/2(z2+1) as the complexification of our integrand. Define γR=γ1R+γ2R+γ3R, where γ1R(t)=t for t from 1/R to R; γ2R(t)=1Reit, where t goes from π/4 to 0 ; and γ3R(t)=eπi/4t, where t goes from R to 1/R (see drawing).
The poles of the integrand are at 0,±i, but those are not contained in the contour, so by the residue theorem ∫γRf(z)dz=0. On the other hand, ∫γR=∫γ1R+∫γ2R+∫γ3R. As R→∞, ∫γ1Rf(z)dz→∫∞01x1/2(x2+1)dx. Also, = |∫γ2Rf(z)dz|≤π4R⋅1R2−1 and the lattest expression tends to 0 as R→∞. However, ∫γ3Rf(z)=i∫1/RRtdt=i/R2−iR22, which is unbounded in absolute value for large R.
Is there a better contour to choose? If so, what is the intuition for finding a good contour in this case?
Answer
For this, you want the keyhole contour γ=γ1∪γ2∪γ3∪γ4, which passes along the positive real axis (γ1), circles the origin at a large radius R (γ2), and then passes back along the positive real axis (γ3), then encircles the origin again in the opposite direction along a small radius-ϵ circle (γ4). Picture (borrowed from this answer):
γ1 is green, γ2 black, γ3 red, and γ4 blue.
It is easy to see that the integrals over the large and small circles tend to 0 as R→∞, ϵ→0, since the integrand times the length is O(R−3/2) and O(ϵ1/2) respectively. The remaining integral tends to
∫γ1∪γ3=∫∞0x−1/21+x2dx+∫0∞(xe2πi)−1/21+(xe2πi)2dx,
because we have walked around the origin once, and chosen the branch of the square root based on this. This simplifies to
(1−e−πi)∫∞0x−1/21+x2dx=2I.
Now you need to compute the residues of the function at the two poles, using the same branch of the square root. The residues of 1/(1+z2)=1(z+i)(z−i) are at z=eπi/2,e3πi/2, so you find
2I=2πi((eπi/2)−1/22i+(e3πi/2)−1/2−2i)=2πsin14π=2π√2
However, I do recommend that you don't attempt to use contour integration for all such problems: imagine trying to do
∫∞0xs−1(a+bxn)mdx,
for general a,b,s,m,n such that it converges, using that method! No, the useful thing to know is that
1An=1Γ(n)∫∞0αn−1e−αxdx,
which enables you to do more general integrals of this type. Contour integration's often a quick and cheap way of doing simple integrals, but becomes impractical in some general cases.
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