# Solving Complex Integral: How to Approach?

• Bill Foster
In summary, the two integrals could not be combined to give an answer for the integral of a+ix. The first integral could not be combined with the second because the domain of the second integral was not the same as the domain of the first. The solution was to find the integral of a+ix using a semi-circle and the residue i.
Bill Foster
$$\int_{-\infty}^{\infty}\frac{\ln{(a+ix)}}{x^2+1}dx$$

I tried with integration by parts but go nowhere. I think it may require a branch cut and integrating along a contour.

How would you approach this?

Bill Foster said:
$$\int_{-\infty}^{\infty}\frac{\ln{(a+ix)}}{x^2+1}dx$$

I tried with integration by parts but go nowhere. I think it may require a branch cut and integrating along a contour.

How would you approach this?

It's been a long time since I took a course on complex variables so my memory is hazy. Anyway, I recall it being helpful to find the Maclaurin Series.

I believe this does require integration along a contour. I think it goes something like this:

$$\int_{-\infty}^{\infty}\frac{\ln{(a+ix)}}{x^2+1}dx = \int_{\gamma + \sigma} \frac{\ln{(a+ix)}}{x^2+1}dx + \int_{-\sigma}\frac{\ln{(a+ix)}}{x^2+1}dx$$

where $\gamma$ is the contour from -R to R along the real axis and $\sigma(t) = Re^{it}$, $0 \leq t \leq \pi$. Then you evaluate the first integral with Cauchy's formula and take the limit as R goes to infinity. The second integral should go to zero, and there's your answer. Of course, it has been a long while since I've done this so I could be wrong.

John Creighto said:
It's been a long time since I took a course on complex variables so my memory is hazy. Anyway, I recall it being helpful to find the Maclaurin Series.

The series for the natural log is only equal to the function itself if |z|< 1, which it is not throughout the domain of integration.

Break the integral to two ones one from $(-\infty,0),(0,+\infty)$, transform the first to $(0,\infty)$ and combine them to get
$$I=\int_0^\infty\frac{\ln(x^2+a^2)}{x^2+1}$$
Use a semi-circle to include the residue $i$ in order to find $I=2\,\pi\ln(a+1)$.

Somebody point me to a site that explains contour integration since I am unfamiliar with it.

Thanks.

## What is a complex integral?

A complex integral is a mathematical concept that measures the area under a curve in the complex plane. It is a generalization of the real-valued integral and involves calculating the sum of infinitely many complex numbers.

## How do you approach solving a complex integral?

There are various methods for solving complex integrals, including using Cauchy's integral formula, the residue theorem, and contour integration. The approach used will depend on the specific integral and the techniques that are most applicable.

## What are some common challenges when solving complex integrals?

Some common challenges when solving complex integrals include identifying the appropriate integration method, dealing with singularities, and correctly setting up the contour of integration. It is also important to pay attention to the branch cuts and branch points of the integrand.

## How can one ensure the accuracy of the solution to a complex integral?

To ensure the accuracy of the solution to a complex integral, it is important to check for consistency with known values, such as the residue theorem or Cauchy's integral formula. It is also helpful to check for errors in setting up the integration bounds and contour, and to double-check the calculations.

## What are some practical applications of solving complex integrals?

Complex integrals have many practical applications in physics, engineering, and other fields. They are used in the study of electromagnetic fields, fluid dynamics, and quantum mechanics, among others. They are also essential in solving problems in complex analysis and number theory.

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