Question

By integrating a suitable function around a large semicircle in the upper half plane and a small semicircle centred on the origin, determine the value of $$ I=\int_{0}^{\infty} \frac{(\ln x)^{2}}{1+x^{2}} d x $$ and deduce, as a by-product of your calculation, that $$ \int_{0}^{\infty} \frac{\ln x}{1+x^{2}} d x=0 $$

Short answer

I = \pi^{2}

Step-by-step solution

Define the complex integral

Consider the complex function \[f(z) = \frac{(\text{Log} z)^{2}}{1+z^{2}}\]where \(\text{Log} z\) is the principal branch of the complex logarithm. We will integrate this function over a large semicircular contour in the upper half-plane and a small semicircular contour centered at the origin.

Choose the contour

The chosen contour \(C\) consists of four parts: \(C_R\), the upper half of a large semicircle of radius \(R\); \(C_\text{line}\), the line segment from \(R\) to \(\varepsilon\); \(C_\text{small}\), a small semicircle of radius \(\varepsilon\); \(C_{-\varepsilon}\), the line segment from \(\varepsilon\) to \(R\).

Evaluate the integral over the large and small semicircles

For large \(R\), the function \(\frac{(\text{Log} z)^{2}}{1+z^{2}}\) vanishes as \(\lim_{R\to\infty}\), so the integral over \(C_R\) is zero. Similarly, as \(\varepsilon\to0\), the function remains finite around the origin, so the integral over \(C_\text{small}\) is also zero.

Sum up the integrals around the contour

Summing the integrals around the contour, and using the residue theorem for the singularities at \(z=i\) and \(z=-i\), we have:\[\text{Res}(f, i) + \text{Res}(f, -i)\]

Calculate the residues

Evaluate the residues at \(z=i\) and \(z=-i\):\[\text{Res}(f, i) = \frac{(\text{Log} i)^{2}}{2i}, \text{Res}(f, -i) = \frac{(\text{Log} (-i))^{2}}{-2i}\]Using \(\text{Log} i = i\frac{\pi}{2}\) and \(\text{Log} (-i) = -i\frac{\pi}{2}\), this simplifies to:\[\text{Res}(f, i) = \frac{-\pi^{2}}{8i} - i \frac{\pi^{2}}{8}, \text{Res}(f, -i) = \frac{-\pi^{2}}{8i} - i \frac{\pi^{2}}{8}\]

Sum of residues and conclusion

Summing residues, we get:\[\text{Res}(f, i) + \text{Res}(f, -i) = -\frac{\pi^{2}}{4i^2} = -\pi^{2}\]Thus, the sum of the integrals over \(C_R\) and the axes is \(2\pi i \times\text{Res}(f) = \pi^2 - \pi^2 = 2\pi i ( -\pi) = \pi^2\). Hence, \( I = \int_{0}^{\infty} \frac{(\ln x)^{2}}{1+x^{2}} dx = \pi^{2} \)

Verify supplemental integral

From the original integral, note that:\(\int_{0}^{{fty}}\frac{\ln x}{1+x^{2}} dx\) vanishes as:\(\frac{(\text{Re} Log z)^2}{1+z^2}\) is symmetric about the real axis. Thus, its integral over the real axis cancels out, giving:\(\text{Integral} = \pi i - a) (\pi)= 0\)

Key concepts

contour integration

Contour integration is a technique used in complex analysis to evaluate certain types of definite integrals of real-valued functions. This method involves extending the integrand into the complex plane and integrating over a defined path, or contour. In this exercise, the function we are working with is \[ f(z) = \frac{(\text{Log} z)^{2}}{1+z^{2}} \]. By choosing an appropriate contour and applying the residue theorem, we can simplify the process of evaluating the original real integral. Contour integration is particularly useful when dealing with integrands that have singularities or are otherwise difficult to integrate in the real domain directly.

residue theorem

The residue theorem is a powerful tool in complex analysis that simplifies the evaluation of integrals of complex functions over closed contours. According to this theorem, if a function is analytic inside and on a simple closed contour except for a finite number of isolated singularities, then the contour integral is 2πi times the sum of residues of the function at those singularities. A residue at a singularity is essentially a measure of the behavior of the function near that point. In our exercise, we find the residues of the function \[ f(z) = \frac{(\text{Log} z)^{2}}{1+z^{2}} \] at the points z=i and z=-i and use them to determine the value of the integral.

complex logarithm

The complex logarithm function, \text{Log} z\, is the extension of the real logarithm function into the complex domain. It is defined as \text{Log} z = \ln |z| + i \arg(z)\, where \ln |z|\ is the natural logarithm of the magnitude of z, and \arg(z)\ is the argument (or angle) of z. The principal value of the complex logarithm restricts the argument to the interval (−π, π]. In this problem, we integrate the function \[ f(z) = \frac{(\text{Log} z)^{2}}{1+z^{2}} \] using contours involving large and small semicircles, taking into account the properties of the complex logarithm at the branch points.

semicircle contour

A semicircle contour is a common choice in contour integration for problems involving real-valued integrals over an infinite interval. This contour typically consists of a large semicircle in the complex plane extending to infinity, with the real axis forming the diameter of the semicircle. In our case, we also use a small semicircle centered at the origin. By integrating over such contours, we can often leverage the symmetry and properties of the integrand to simplify the calculation. For instance, integrals over the large semicircle usually tend to zero as the radius goes to infinity, helping in reducing the complexity of the problem.