Method of Setting Integral Path of Complex Integral


I described a method of solving complex integrals by theorems before, but some complex integrals can not be solved only applying theorems. However, even with such complex integrals, it becomes possible to solve by setting an integration route yourself.

Furthermore, real integrals can be solved as complex integrals. In this article, I’ll write about some integrals that can be solved by applying this.


The Radius of Convergence


Consider the next power series.

$$\sum_{n=0}^∞ a_nz^n・・・(1)$$

We take the absolute value of \(a_n\) and \(z\) which make up this power series.

$$\sum_{n=0}^∞ |a_n||z|^n・・・(2)$$

This power series (2) is called absolute convergence if it is finite. Then if the power series (2) converges, the power series (1) also converges.

The convergence radius R refers to the boundary between the area the power series (2) converges absolutely and doesn’t. If \(|z|<R\), the power series (1) converges absolutely. Then if \(|z|>R\), it diverges. When \(|z|=R\), some power series converge, others diverge. Therefore, we need to think about converging or diverging for each power series.


Laplace Transform and Algebraic Equation


Laplace transform refers to the following conversion \(F(s)\).

$$F(s)=\int_0^∞ f(t)e^{-st}dt$$

The function before Laplace transform \(f(t)\) depends on \(t\), but the function after Laplace transform \(F(s)\) depends on \(s\).


Cauchy-Riemann Equation and Holomorphic Function


A complex-valued function \(f(z)\) is said to be a holomorphic function if it is differentiable at every point in its domain.

By the way, let \(f(z)\) be defined as


where \(z\) is a complex variable, \(u(x,y)\) and \(v(x,y)\) are real-number functions, and \(x\) and \(y\) are real variables. The following two equations are collectively called Cauchy-Riemann’s equations.



You can determine the complex-valued function \(f(z)\) is a holomorphic function when \(u(x, y)\) and \(v(x, y)\) satisfy the Cauchy-Riemann equations above.


Fourier Transform


I thought about Fourier series expansion before, but this time it is an introduction of Fourier transform.

Fourier transform

$$F(k)=\int_{-∞}^{∞} f(x)e^{-ikx} dx$$

inverse Fourier transform

$$f(x)=\frac{1}{2π}\int_{-∞}^{∞} F(k)e^{ikx} dk$$

Suppose a wave is given by the function \(f(x)\). Using Fourier transform, the function of wave \(f (x)\) is expressed as a function \(F(k)\) of wavenumber \(k\). \(F(k)\) is the function which decomposes \(f(x)\) into the sine waves of all wavenumbers \(k\) and expresses the size of each the sine wave as a function. You can convert the wave \(f (x)\) to \(F(k)\) by Fourier transform and \ (F (k) \) to \(F(k)\) by inverse Fourier transform.


Fourier Series Expansion Method


Fourier series expansion and Fourier transform are completely different. In this article, I’ll write about Fourier series expansion.

Fourier series expansion method is to express the periodic function \(f(x)\) by using the sum of \(sin\) and \(cos\). Using the method, the periodic function \(f(x)\) whose period is \(L\) can be expressed as follows.

$$f(x)=\frac{a_0}{2}+\sum_{n=1}^{∞} \left[ {a_{n}cos(\frac{2πnx}{L})}+b_{n}sin(\frac{2πnx}{L}) \right]$$

$$a_0=\frac{2}{L}\int_{-L/2}^{L/2} f(x) dx$$

$$a_n=\frac{2}{L}\int_{-L/2}^{L/2} f(x)cos(\frac{2πnx}{L}) dx$$

$$b_n=\frac{2}{L}\int_{-L/2}^{L/2} f(x)sin(\frac{2πnx}{L}) dx$$


The Foundation of Determinant


A 2 × 2 determinant is expressed as follows.

$$\left[\begin{array}{cc}a_{11}&a_{12}\\  a_{21}&a_{22}\\  \end{array}\right]=a_{11}a_{22}-a_{21}a_{12}$$

Then a 3 × 3 determinant is expressed as follows.

$$\left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\  a_{21}&a_{22}&a_{23}\\  a_{31}&a_{32}&a_{33}\\ \end{array}\right]\\=a_{11}a_{22}a_{33}+a_{21}a_{32}a_{13}+a_{31}a_{23}a_{12}\\-a_{31}a_{22}a_{13}-a_{21}a_{12}a_{33}-a_{11}a_{23}a_{32}$$

So far you can memorize easily, but it is hard to memorize from 4 × 4. In this article, I’ll show you how to find a 4×4 determinant by looking at the elements of the first row.


Taylor Expansion


Taylor expansion around x = a is to express an arbitrary function f (x) in a form like the expression below.


When a = 0, it can be written as follows. This case is called “Maclaurin’s expansion” specially.


\(f^{(k)}(x)\) means the \(k\)th order derivative of the function \(f(x)\).