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Improper Integrals of Approach Functions 📂Complex Anaylsis

Improper Integrals of Approach Functions

Buildup 1

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.

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Considering the representative multivalued function, log 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, Rr0 R \to \infty \\ r \to 0 while L1L_{1} and L2L_{2}, 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 1

As an example, let’s calculate the value of 0Logxx2+a2dx\displaystyle \int_{0}^{ \infty } {{\text{Log} x} \over {x^2 + a^2}} dx.

Solution

Let’s set the integration path C\mathscr{C} as shown in the figure below.

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Then, for the integrand f(z):=logzz2+a2\displaystyle f(z) := {{\log z } \over {z^2 + a^2}}, Cf(z)dz=Γf(z)dz+L0f(z)dz+γf(z)dz+rRf(x)dx \int_{\mathscr{C}} f(z) dz = {\color{red} \int_{\Gamma} f(z) dz } + \int_{L_{0}} f(z) dz + \color{blue} { \int_{\gamma} f(z) dz } + \int_{r}^{R} f(x) dx substituting L0L_{0} with z=ρeiθz = \rho e^{ i \theta } on Cf(z)dz=Rr(Logρ+iθ)eiθρ2e2iθ+a2dρ+Γf(z)dz+γf(z)dz+rRLogxx2+a2dx \int_{\mathscr{C}} f(z) dz = \int_{ R }^{r} { {(\text{Log} \rho + i \theta ) e^{i \theta } } \over {\rho^2 e^{2 i \theta } +a^2} } d \rho + {\color{red} \int_{\Gamma} f(z) dz } + \color{blue} { \int_{\gamma} f(z) dz } + \int_{r}^{R} {{\text{Log} x } \over {x^2 + a^2}} dx As θ\theta gets closer to π\pi, and L0L_{0} approaches the negative real axis, by applying θπ\theta \to \pi we get Cf(z)dz=Rr(Logρ+iπ)ρ2+a2dρ+Γf(z)dz+γf(z)dz+rRLogxx2+a2dx \int_{\mathscr{C}} f(z) dz = \int_{ R }^{r} { { -( \text{Log} \rho + i \pi ) } \over {\rho^2 +a^2} } d \rho + {\color{red} \int_{\Gamma} f(z) dz } + \color{blue} { \int_{\gamma} f(z) dz } + \int_{r}^{R} {{\text{Log} x } \over {x^2 + a^2}} dx

Diverging Semicircular Complex Path Integral: If function ff is continuous on a semicircle Γ \Gamma with radius RR and centered at 00 and if limzzf(z)=0\displaystyle \lim_{z \to \infty } z f(z) = 0, limRΓf(z)dz=0\lim_{R \to \infty} \int_{\Gamma} f(z) dz = 0

Meanwhile, limzzf(z)=limz01zf(1/z)=limz0Logz1/z+a2z=limz0za2z21=0 \lim_{z \to \infty} z f(z) = \lim_{z \to 0} {{1} \over {z}} f(1/z) = \lim_{z \to 0} {{- \text{Log} z } \over { 1/z + a^2 z }} = \lim_{z \to 0} {{ - z } \over {a^2 z^2 -1 }} = 0 thus, limRΓf(z)dz=0 \lim_{R \to \infty} {\color{red} \int_{\Gamma} f(z) dz } = 0

Contracting Semicircular Complex Path Integral: If function ff is continuous on a semicircle γ \gamma with radius rr and centered at 00 and if limz0zf(z)=0\displaystyle \lim_{z \to 0 } z f(z) = 0, limr0γf(z)dz=0\lim_{r \to 0} \int_{\gamma} f(z) dz = 0

And, limz0zf(z)=limz0Logzz+a2/z=limz01/z1a2/z2=0 \lim_{z \to 0} z f(z) = \lim_{z \to 0} {{ \text{Log} z } \over { z + a^2 / z }} = \lim_{z \to 0} {{ 1 / z } \over { 1 - a^2 / z^2 }} = 0 thus,

limr0γf(z)dz=0 \lim_{r \to 0 } \color{blue} { \int_{\gamma} f(z) dz } = 0 Therefore, by taking r0r \to 0 and RR \to \infty we get Cf(z)dz=0(Logρ+iπ)ρ2+a2dρ+0Logxx2+a2dx=0Logρρ2+a2dρ+0iπρ2+a2dρ+0Logxx2+a2dx=20Logxx2+a2dx+i0πρ2+a2dρ \begin{align*} \int_{\mathscr{C}} f(z) dz =& \int_{ \infty }^{0} { { -( \text{Log} \rho + i \pi ) } \over {\rho^2 +a^2} } d \rho + \int_{0}^{\infty} {{\text{Log} x } \over {x^2 + a^2}} dx \\ =& \int_{ 0}^{ \infty } { { \text{Log} \rho } \over {\rho^2 +a^2} } d \rho + \int_{ 0}^{ \infty } { { i \pi } \over {\rho^2 +a^2} } d \rho + \int_{0}^{\infty} {{\text{Log} x } \over {x^2 + a^2}} dx \\ =& 2 \int_{0}^{\infty} {{\text{Log} x } \over {x^2 + a^2}} dx + i \int_{ 0}^{ \infty } { { \pi } \over {\rho^2 +a^2} } d \rho \end{align*}

Residue Theorem: Let the analytic function f:ACCf: A \subset \mathbb{C} \to \mathbb{C} have a finite number of singularities z1,z2,,zmz_{1} , z_{2} , \cdots , z_{m} within a simple closed path C\mathscr{C}. Then, Cf(z)dz=2πik=1mReszkf(z)\int_{\mathscr{C}} f(z) dz = 2 \pi i \sum_{k=1}^{m} \text{Res}_{z_{k}} f(z)

By the Residue Theorem, Cf(z)dz=2πiResaif(z)=2πiLogai2ai=πLoga+iπ2a=πLogaa+iπ22a \int_{\mathscr{C}} f(z) dz = 2 \pi i \text{Res}_{a i} f(z) = 2 \pi i {{\text{Log} a i } \over {2 a i}} = \pi {{\text{Log} a + i {{\pi} \over {2}} } \over {a}} = {{ \pi \text{Log} a } \over {a}} + { i {{\pi^2} \over {2a} }} Summarizing, 20Logxx2+a2dx+i0πρ2+a2dρ=πLogaa+iπ22a 2 \int_{0}^{\infty} {{\text{Log} x } \over {x^2 + a^2}} dx + i \int_{ 0}^{ \infty } { { \pi } \over {\rho^2 +a^2} } d \rho = {{ \pi \text{Log} a } \over {a}} + { i {{\pi^2} \over {2a} }} Taking only the real part from the above equation gives, 0Logxx2+a2dx=πLoga2a \int_{0}^{\infty} {{\text{Log} x } \over {x^2 + a^2}} dx = {{ \pi \text{Log} a } \over {2a}}


  1. Osborne (1999). Complex variables and their applications: p176. ↩︎ ↩︎