Improper Integrals of Approach Functions
📂Complex AnaylsisImproper Integrals of Approach Functions
Buildup
The biggest problem when integrating a multivalued function is that the function value can change unexpectedly when the path encounters a branch cut. When integrating such a function, one uses the same trick as before to make the path circumvent the branch cut.
Considering the representative multivalued function, log log, with the negative real axis as the branch cut and the origin as the branching point, one can imagine a path like the one above. Then,
R→∞r→0
while L1 and L2, as it approaches the negative real axis, it becomes possible to use the Residue Theorem over the entire complex plane. At first glance, it might seem too good to be true, but this powerful idea mobilizes all the methods learned so far.
Example
As an example, let’s calculate the value of ∫0∞x2+a2Logxdx.
Solution
Let’s set the integration path C as shown in the figure below.
Then, for the integrand f(z):=z2+a2logz,
∫Cf(z)dz=∫Γf(z)dz+∫L0f(z)dz+∫γf(z)dz+∫rRf(x)dx
substituting L0 with z=ρeiθ on
∫Cf(z)dz=∫Rrρ2e2iθ+a2(Logρ+iθ)eiθdρ+∫Γf(z)dz+∫γf(z)dz+∫rRx2+a2Logxdx
As θ gets closer to π, and L0 approaches the negative real axis, by applying θ→π we get
∫Cf(z)dz=∫Rrρ2+a2−(Logρ+iπ)dρ+∫Γf(z)dz+∫γf(z)dz+∫rRx2+a2Logxdx
Diverging Semicircular Complex Path Integral: If function f is continuous on a semicircle Γ with radius R and centered at 0 and if z→∞limzf(z)=0,
R→∞lim∫Γf(z)dz=0
Meanwhile,
z→∞limzf(z)=z→0limz1f(1/z)=z→0lim1/z+a2z−Logz=z→0lima2z2−1−z=0
thus,
R→∞lim∫Γf(z)dz=0
Contracting Semicircular Complex Path Integral: If function f is continuous on a semicircle γ with radius r and centered at 0 and if z→0limzf(z)=0,
r→0lim∫γf(z)dz=0
And,
z→0limzf(z)=z→0limz+a2/zLogz=z→0lim1−a2/z21/z=0
thus,
r→0lim∫γf(z)dz=0
Therefore, by taking r→0 and R→∞ we get
∫Cf(z)dz===∫∞0ρ2+a2−(Logρ+iπ)dρ+∫0∞x2+a2Logxdx∫0∞ρ2+a2Logρdρ+∫0∞ρ2+a2iπdρ+∫0∞x2+a2Logxdx2∫0∞x2+a2Logxdx+i∫0∞ρ2+a2πdρ
Residue Theorem: Let the analytic function f:A⊂C→C have a finite number of singularities z1,z2,⋯,zm within a simple closed path C. Then,
∫Cf(z)dz=2πik=1∑mReszkf(z)
By the Residue Theorem,
∫Cf(z)dz=2πiResaif(z)=2πi2aiLogai=πaLoga+i2π=aπLoga+i2aπ2
Summarizing,
2∫0∞x2+a2Logxdx+i∫0∞ρ2+a2πdρ=aπLoga+i2aπ2
Taking only the real part from the above equation gives,
∫0∞x2+a2Logxdx=2aπLoga