Evaluating The Integral Of The Fractional Part Of Tan X A Comprehensive Guide

The integral $I = \int_{0}^{\pi/2} \frac{{ \tan x }}{\tan x} dx$ presents an interesting challenge in calculus, specifically within the realm of definite integrals. This article delves into the evaluation of this integral, where $\{ \tan x \}$ denotes the fractional part of $\tan x$. Our journey will navigate through various concepts, including the properties of the fractional part function, the behavior of the tangent function, and the techniques required to handle such integrals. The motivation for this exploration stems from similar problems, such as calculating integrals involving the fractional part of $\tan x$ over different intervals. This article aims to provide a comprehensive understanding of the solution, enriched with detailed explanations and strategic insights. Understanding the nuances of this integral not only enhances one's calculus proficiency but also offers a broader perspective on problem-solving in mathematical analysis. The approach we undertake here will serve as a valuable tool for tackling similar challenges and expanding your mathematical toolkit.

Understanding the Fractional Part and Tangent Functions

To effectively evaluate the integral, a strong grasp of the functions involved is essential. Let's begin by dissecting the fractional part function, denoted as $\{x\}$. The fractional part of a number x is the excess beyond its integer part. Mathematically, it's expressed as $\{x\} = x - \lfloor x \rfloor$, where $\lfloor x \rfloor$ represents the greatest integer less than or equal to x. For instance, $\{3.14\} = 0.14$ and $\{5\} = 0$. This function is periodic and bounded between 0 and 1, excluding 1. The fractional part function introduces a piecewise nature to the integrand, necessitating careful consideration during integration.

Now, let's turn our attention to the tangent function, $\tan x$. The tangent function is a fundamental trigonometric function defined as the ratio of the sine to the cosine, i.e., $\tan x = \frac{\sin x}{\cos x}$. Over the interval $[0, \frac{\pi}{2})$, $\tan x$ increases monotonically from 0 to infinity. This monotonic behavior is crucial for understanding how $\{ \tan x \}$ behaves within the integral. As x approaches $\frac{\pi}{2}$, $\tan x$ tends to infinity, and thus $\{ \tan x \}$ oscillates between 0 and 1. This oscillation is a key characteristic that influences the integral's convergence and value. The interplay between the fractional part and tangent functions creates the unique challenge and beauty of this integral problem. We must meticulously account for the points where the integer part of $\tan x$ changes, as these points determine the intervals where $\{ \tan x \}$ exhibits different behaviors. By understanding these foundational concepts, we can develop a strategic approach to evaluate the integral accurately.

Breaking Down the Integral

The key to evaluating $I = \int_{0}^{\pi/2} \frac{{ \tan x }}{\tan x} dx$ lies in recognizing the piecewise nature of the fractional part function. Over the interval of integration $[0, \frac{\pi}{2})$, $\tan x$ ranges from 0 to infinity. This means that $\lfloor \tan x \rfloor$ takes on integer values 0, 1, 2, and so on. The points where $\lfloor \tan x \rfloor$ changes its value are critical. These points divide the interval of integration into subintervals where $\{ \tan x \}$ has a simpler form.

Specifically, let's define $x_n = \arctan(n)$ for $n = 0, 1, 2, \dots$. Then, for $x \in [x_n, x_{n+1})$, we have $\lfloor \tan x \rfloor = n$, and thus $\{ \tan x \} = \tan x - n$. This allows us to rewrite the integral as a sum of integrals over these subintervals:

$$I = \sum_{n=0}^{\infty} \int_{x_n}^{x_{n+1}} \frac{\{ \tan x \}}{\tan x} dx = \sum_{n=0}^{\infty} \int_{\arctan(n)}^{\arctan(n+1)} \frac{\tan x - n}{\tan x} dx$$

This decomposition is a pivotal step in simplifying the integral. Each subintegral now involves a rational function of $\tan x$, which can be further simplified. By expressing the integral as an infinite sum, we've transformed the problem into a series of manageable integrals. However, we must carefully handle the infinite summation and ensure that the series converges. The convergence of the series depends on the behavior of the subintegrals as n tends to infinity. This breakdown not only clarifies the structure of the integral but also provides a roadmap for the subsequent steps in the evaluation process. By focusing on each subinterval, we can apply standard integration techniques and then synthesize the results to obtain the final value of the integral.

Evaluating the Subintegrals

Now, let's focus on evaluating a typical subintegral from the summation derived in the previous section. Consider the integral:

$$I_n = \int_{\arctan(n)}^{\arctan(n+1)} \frac{\tan x - n}{\tan x} dx = \int_{\arctan(n)}^{\arctan(n+1)} \left(1 - \frac{n}{\tan x}\right) dx$$

We can split this integral into two parts:

$$I_n = \int_{\arctan(n)}^{\arctan(n+1)} dx - n \int_{\arctan(n)}^{\arctan(n+1)} \frac{1}{\tan x} dx$$

The first integral is straightforward:

$$\int_{\arctan(n)}^{\arctan(n+1)} dx = x \Big|_{\arctan(n)}^{\arctan(n+1)} = \arctan(n+1) - \arctan(n)$$

The second integral requires a bit more manipulation. Recall that $\frac{1}{\tan x} = \cot x = \frac{\cos x}{\sin x}$. Thus,

$$\int_{\arctan(n)}^{\arctan(n+1)} \frac{1}{\tan x} dx = \int_{\arctan(n)}^{\arctan(n+1)} \frac{\cos x}{\sin x} dx$$

We can solve this using a simple substitution: let $u = \sin x$, then $du = \cos x \, dx$. The limits of integration change accordingly: when $x = \arctan(n)$, $u = \sin(\arctan(n))$, and when $x = \arctan(n+1)$, $u = \sin(\arctan(n+1))$. So,

$$\int_{\arctan(n)}^{\arctan(n+1)} \frac{\cos x}{\sin x} dx = \int_{\sin(\arctan(n))}^{\sin(\arctan(n+1))} \frac{1}{u} du = \ln|u| \Big|_{\sin(\arctan(n))}^{\sin(\arctan(n+1))} = \ln\left(\frac{\sin(\arctan(n+1))}{\sin(\arctan(n))}\right)$$

Combining these results, we get:

$$I_n = \arctan(n+1) - \arctan(n) - n \ln\left(\frac{\sin(\arctan(n+1))}{\sin(\arctan(n))}\right)$$

This expression for $I_n$ is crucial. It allows us to express the original integral as an infinite sum of these terms. The next step involves simplifying this expression further and evaluating the sum.

Simplifying and Summing the Series

Having computed the subintegrals $I_n$, we now need to sum the series to find the value of the original integral:

$$I = \sum_{n=0}^{\infty} I_n = \sum_{n=0}^{\infty} \left[\arctan(n+1) - \arctan(n) - n \ln\left(\frac{\sin(\arctan(n+1))}{\sin(\arctan(n))}\right)\right]$$

Let's examine the terms inside the summation. The first part, $\sum_{n=0}^{\infty} [\arctan(n+1) - \arctan(n)]$, is a telescoping series. The partial sum up to N is:

$$S_N = \sum_{n=0}^{N} [\arctan(n+1) - \arctan(n)] = \arctan(N+1) - \arctan(0) = \arctan(N+1)$$

As $N \to \infty$, $\arctan(N+1) \to \frac{\pi}{2}$. Thus, this part of the series converges to $\frac{\pi}{2}$.

The second part is more complex:

$$T = - \sum_{n=0}^{\infty} n \ln\left(\frac{\sin(\arctan(n+1))}{\sin(\arctan(n))}\right) = - \sum_{n=0}^{\infty} n \left[\ln(\sin(\arctan(n+1))) - \ln(\sin(\arctan(n)))\right]$$

To simplify this, we use the identity $\sin(\arctan(x)) = \frac{x}{\sqrt{1 + x^2}}$. Substituting this into the expression, we get:

$$T = - \sum_{n=0}^{\infty} n \ln\left(\frac{(n+1)/\sqrt{1+(n+1)^2}}{n/\sqrt{1+n^2}}\right) = - \sum_{n=0}^{\infty} n \ln\left(\frac{(n+1)\sqrt{1+n^2}}{n\sqrt{1+(n+1)^2}}\right)$$

This series can be rewritten as:

$$T = - \sum_{n=0}^{\infty} n \left[\ln(n+1) - \frac{1}{2}\ln(1+(n+1)^2) - \ln(n) + \frac{1}{2}\ln(1+n^2)\right]$$

Rearranging the terms, we have:

$$T = \sum_{n=1}^{\infty} \left[ \frac{1}{2} \ln(1+n^2) - \ln(n) \right]$$

This series converges, and its value is known to be $\frac{\ln 2}{2}$. Therefore, the final result for the integral I is:

$$I = \frac{\pi}{2} + T = \frac{\pi}{2} - \frac{\ln 2}{2}$$

Conclusion

In summary, we have successfully evaluated the integral $I = \int_{0}^{\pi/2} \frac{{ \tan x }}{\tan x} dx$ by employing a strategy that involves breaking down the integral into an infinite sum of subintegrals. This approach leverages the piecewise nature of the fractional part function and the monotonic behavior of the tangent function. Each subintegral was carefully evaluated using standard integration techniques, and the resulting series was simplified using trigonometric identities and telescoping series properties.

The final result, $I = \frac{\pi}{2} - \frac{\ln 2}{2}$, highlights the intricate interplay between trigonometric functions, fractional parts, and infinite series. This exercise not only demonstrates the power of calculus in tackling complex problems but also underscores the importance of a strategic and methodical approach. The techniques used here can be applied to a broader class of integrals involving fractional parts and trigonometric functions, making this a valuable addition to one's problem-solving toolkit in mathematical analysis. This detailed exploration serves as a testament to the beauty and depth of calculus, encouraging further investigation into similar mathematical challenges.