Wednesday, 24 August 2016

integration - Using Complex Analysis to Compute inti0nftyfracdxx1/2(x2+1)




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).enter image description here



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)dz01x1/2(x2+1)dx. Also, = |γ2Rf(z)dz|π4R1R21 and the lattest expression tends to 0 as R. However, γ3Rf(z)=i1/RRtdt=i/R2iR22, 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=γ1γ2γ3γ4, 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):




enter image description here



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, ϵ0, since the integrand times the length is O(R3/2) and O(ϵ1/2) respectively. The remaining integral tends to
γ1γ3=0x1/21+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
(1eπi)0x1/21+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)(zi) are at z=eπi/2,e3πi/2, so you find
2I=2πi((eπi/2)1/22i+(e3πi/2)1/22i)=2πsin14π=2π2







However, I do recommend that you don't attempt to use contour integration for all such problems: imagine trying to do
0xs1(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)0αn1eαxdx,

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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