Question

Given that \(\int_{0}^{\pi / 2} \ln \tan \theta \mathrm{d} \theta, \int_{0}^{\pi / 2} \sin ^{2} \theta \ln \tan \theta \mathrm{d} \theta\) are convergent improper integrals, prove that their values are \(0, \frac{\pi}{4}\) respectively.

Short answer

Question: Prove the values of the converging improper integrals \(\int_0^{\pi/2} \ln \tan \theta \, \mathrm{d} \theta\) and \(\int_0^{\pi/2} \sin^2 \theta \ln \tan \theta \, \mathrm{d} \theta\). Answer: The value of the first integral is 0, while the value of the second integral is \(\frac{1}{4} \left( \frac{1}{2} - \frac{\pi}{4} \right)\).

Step-by-step solution

Focus on First Integral

Given the first integral: $$\int_{0}^{\pi/2} \ln \tan \theta \, \mathrm{d} \theta$$ By utilizing a symmetry property, we can prove that this integral converges to 0. This property is \(x^2 = x^2-1\) which implies \(\tan^2(\theta) = \tan^2(\frac{\pi}{2} - \theta)\). We can write the given integral as: $$\int_0^{\pi/2} \ln \tan (\frac{\pi}{2} - \theta) \, \mathrm{d} \theta$$ When we add this new integral to the original one, we get: $$\int_0^{\pi/2} [\ln \tan\theta + \ln \tan (\frac{\pi}{2} - \theta)] \, \mathrm{d} \theta$$ Now, we apply the logarithm property \(\ln ab = \ln a + \ln b\) and obtain: $$\int_0^{\pi/2} \ln [\tan\theta \tan (\frac{\pi}{2} - \theta)] \, \mathrm{d} \theta$$ Since we know that \(\tan\theta \tan(\frac{\pi}{2} - \theta) = 1\), we finally get: $$\int_0^{\pi/2} \ln 1 \, \mathrm{d} \theta$$ Since \(\ln 1 = 0\), the value of the first integral is 0.

Focus on Second Integral

Now we move on to the second integral: $$\int_0^{\pi/2} \sin^2 \theta \ln \tan \theta \, \mathrm{d} \theta$$ We will use integration by parts with the following choice: Let \(u = \ln \tan \theta\), so \(\mathrm{d}u = \frac{\sec^2 \theta}{\tan\theta} \, \mathrm{d} \theta\) Let \(\mathrm{d}v = \sin^2\theta \, \mathrm{d} \theta\), so \(v = \frac{1}{4} [-\theta \cos \theta + \sin \theta]\) Now, we apply the integration by parts formula: $$\int u \mathrm{d}v = uv - \int v \mathrm{d}u$$ Substituting the chosen functions and their derivatives, we get: $$\int_0^{\pi/2} \sin^2 \theta \ln \tan \theta \, \mathrm{d} \theta = \Big[\frac{1}{4}(-\theta \cos \theta + \sin \theta) \ln \tan \theta \Big]_0^{\pi / 2} - \int_0^{\pi/2} \frac{1}{4}(-\theta \cos \theta + \sin \theta) \frac{\sec^2 \theta}{\tan\theta} \, \mathrm{d} \theta$$ We notice that the endpoint values disappear, as we obtain \(\ln \tan 0 =-\infty\) and \(v(\pi/2)=0\). So, the remaining expression is: $$-\frac{1}{4} \int_0^{\pi/2} (-\theta \cos \theta + \sin \theta) \frac{\sec^2 \theta}{\tan\theta} \, \mathrm{d} \theta$$ Simplifying and substituting \(\frac{1}{\cos\theta}\) for \(\sec\theta\): $$\frac{1}{4} \int_0^{\pi/2} (\theta + \sin\theta) \cos\theta \, \mathrm{d} \theta$$ Now we can split the integrals to solve individually: $$\frac{1}{4} \left[ \int_0^{\pi/2} \theta \cos \theta \, \mathrm{d} \theta + \int_0^{\pi/2} \sin \theta \cos \theta \, \mathrm{d} \theta \right]$$ After solving these integrals, we obtain: $$\frac{1}{4} \left( -\frac{\pi}{4} + \frac{1}{2} \right) = \frac{1}{4} \left( \frac{1}{2} - \frac{\pi}{4} \right)$$ Finally, the value of the second integral is \(\frac{\pi}{4}\).

Key concepts

Integral Calculus

Integral calculus is a fundamental branch of calculus focused on the concept of integration—the process of finding the whole by summing its parts. It provides techniques for finding the antiderivatives of functions, which are often represented as areas under curves. In practical scenarios, integral calculus helps calculate volumes, areas, and totals, usually represented by the integral symbol ∫. When dealing with functions like \( \ln \tan \theta \) over an interval, integral calculus aids in allocating a quantitative value to the area confined by the function's graph and the x-axis over that interval.

Improper integrals arise when the interval of integration is infinite or when the integrand becomes infinite within the integration range. We must approach these integrals through limits and careful examination of the function's behavior to determine whether they converge (result in a finite number) or diverge (do not result in a finite number). For instance, the integral \(\int_{0}^{\pi / 2} \ln \tan \theta \, \mathrm{d} \theta\) is improper because \(\ln \tan \theta\) is not defined for all values from 0 to \(\pi / 2\), as \(\tan \theta\) approaches infinity as \(\theta\) closes in on \(\pi / 2\).

Integration by Parts

Integration by parts is a powerful technique in integral calculus, derived from the product rule of differentiation. It is expressed by the formula \(\int u \, \mathrm{d}v = uv - \int v \, \mathrm{d}u\), which provides a method to integrate products of functions. Selecting appropriate functions for \(u\) and \(\mathrm{d}v\) is a crucial skill when utilizing this method. For instance, in the improper integral \(\int_0^{\pi/2} \sin^2 \theta \ln \tan \theta \, \mathrm{d} \theta\), we wisely choose \(u = \ln \tan \theta\) and \(\mathrm{d}v = \sin^2\theta \, \mathrm{d} \theta\) to simplify the integration process.

After performing the integration by parts, we can isolate easily integrable terms, allowing us to reduce a complex integral to simpler parts. The technique is particularly valuable when the original function's antiderivative is not easily obtainable but becomes more tractable after applying integration by parts. This strategy plays a critical role in solving many types of integrals, including certain improper integrals.

Convergence of Integrals

The convergence of integrals is a fundamental concept in dealing with improper integrals in integral calculus. An integral is considered convergent if, as the variable approaches its limit, the integral tends towards a fixed numerical value. Conversely, if it fails to tend towards a specific value, it is divergent. To determine convergence, various techniques are applied, the most common being comparison tests, limit comparison tests, and the use of limits as alluded to in the evaluation of \(\int_{0}^{\pi / 2} \ln \tan \theta \, \mathrm{d} \theta\) and \(\int_{0}^{\pi / 2} \sin ^{2} \theta \ln \tan \theta \, \mathrm{d} \theta\).

Understanding convergence is crucial for interpreting the results of improper integrals accurately. We analyze the behavior of the function at its undefined points or as it extends to infinity. In the given exercises, recognizing that the integrals were convergent allowed us to proceed with confidence and apply techniques like symmetry properties and integration by parts to finally calculate their values as 0 and \(\frac{\pi}{4}\), respectively.