Solving A Definite Integral With Polygamma Functions And Hyperbolic Cosine

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In the realm of mathematical analysis, definite integrals involving special functions often pose significant challenges. This article delves into the intricate solution of a particular definite integral featuring polygamma functions and hyperbolic functions. We will explore the techniques and insights required to evaluate this integral, shedding light on the fascinating interplay between different branches of mathematics. This article aims to provide a comprehensive understanding of the solution process, making it accessible to both students and researchers in the field.

Exploring the Integral: A Deep Dive

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At the heart of our exploration lies the definite integral:

$$\int_0^\infty \frac{\psi^{(1)}\left(\tfrac{1 - i u}{2}\right) - \psi^{(1)}\left(\tfrac{1 + i u}{2}\right)}{\cosh\left(\tfrac{\pi u}{2}\right)}\,du$$

This integral presents a formidable challenge due to the presence of the polygamma function $\psi^{(1)}(z)$, which is the first derivative of the digamma function, and the hyperbolic cosine function $\cosh(\frac{\pi u}{2})$. To tackle this integral effectively, we need to employ a combination of analytical techniques, including complex analysis and special function identities. Our journey begins with a closer look at the properties of the polygamma and hyperbolic cosine functions, which will pave the way for a successful evaluation.

Understanding the Polygamma Function

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The polygamma function, denoted by $\psi^{(n)}(z)$, plays a crucial role in various branches of mathematics, including number theory, special functions, and mathematical physics. It is defined as the $n$-th derivative of the digamma function $\psi(z)$, which is itself the derivative of the logarithm of the gamma function $\Gamma(z)$. In other words:

$$\psi^{(n)}(z) = \frac{d^{n}}{dz^{n}} \psi(z) = \frac{d^{n+1}}{dz^{n+1}} \ln \Gamma(z)$$

The polygamma function exhibits several important properties that are essential for evaluating integrals and series. One key property is its relationship to the Hurwitz zeta function $\zeta(s, a)$, given by:

$$\psi^{(n)}(z) = (-1)^{n+1} n! \zeta(n+1, z)$$

This connection allows us to express the polygamma function in terms of a known special function, which can be helpful in simplifying expressions and deriving new identities. Furthermore, the polygamma function satisfies a recurrence relation:

$$\psi^{(n)}(z+1) = \psi^{(n)}(z) + (-1)^{n} \frac{n!}{z^{n+1}}$$

This relation is particularly useful for evaluating the polygamma function at shifted arguments. For our integral, we are interested in the difference between polygamma functions with complex arguments, $\psi^{(1)}(\frac{1 - iu}{2})$ and $\psi^{(1)}(\frac{1 + iu}{2})$. Understanding the behavior of these functions in the complex plane is crucial for solving the integral.

Deciphering the Hyperbolic Cosine Function

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The hyperbolic cosine function, denoted by $\cosh(z)$, is a fundamental function in hyperbolic geometry and complex analysis. It is defined in terms of the exponential function as:

$$\cosh(z) = \frac{e^{z} + e^{-z}}{2}$$

The hyperbolic cosine function possesses several notable properties. It is an even function, meaning that $\cosh(-z) = \cosh(z)$. It is also periodic in the imaginary direction, with a period of $2\pi i$. The hyperbolic cosine function appears in various applications, including the study of catenaries, transmission lines, and the solutions of certain differential equations. In the context of our integral, the presence of $\cosh(\frac{\pi u}{2})$ in the denominator suggests that we might need to consider contour integration techniques to evaluate the integral.

Strategizing the Solution Approach

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To effectively tackle the integral, we will employ a multifaceted approach. This involves leveraging the properties of the polygamma function, particularly its representation in terms of the Hurwitz zeta function and its behavior in the complex plane. We will also harness the characteristics of the hyperbolic cosine function, recognizing its periodicity and symmetry. The strategy is to transform the integral into a form that is amenable to evaluation using complex analysis techniques, specifically contour integration. By carefully selecting a suitable contour and applying the residue theorem, we can determine the value of the integral. This approach requires a deep understanding of complex analysis and special functions, as well as meticulous attention to detail. We will proceed step-by-step, explaining each stage of the solution process.

Navigating the Solution: Step-by-Step

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Having laid the groundwork by understanding the key functions involved, we now embark on the journey of solving the integral. The solution process entails several crucial steps, each building upon the previous one. We begin by expressing the difference of polygamma functions in a more manageable form, then we explore the possibility of using contour integration to evaluate the integral. Finally, we apply the residue theorem to arrive at the solution.

Simplifying the Polygamma Function Difference

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The first step in our solution is to simplify the expression $\psi^{(1)}(\frac{1 - iu}{2}) - \psi^{(1)}(\frac{1 + iu}{2})$. We can utilize the relationship between the polygamma function and the Hurwitz zeta function to rewrite this difference. Recall that:

$$\psi^{(1)}(z) = \zeta(2, z)$$

Therefore, our expression becomes:

$$\psi^{(1)}\left(\frac{1 - iu}{2}\right) - \psi^{(1)}\left(\frac{1 + iu}{2}\right) = \zeta\left(2, \frac{1 - iu}{2}\right) - \zeta\left(2, \frac{1 + iu}{2}\right)$$

This transformation allows us to work with the Hurwitz zeta function, which has known properties and representations. However, directly evaluating the difference of Hurwitz zeta functions can still be challenging. We need to explore further simplifications or alternative approaches to make progress.

The Contour Integration Approach

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Given the presence of the hyperbolic cosine function in the denominator, a natural approach is to consider contour integration. We define the function:

$$f(z) = \frac{\psi^{(1)}\left(\frac{1 - iz}{2}\right) - \psi^{(1)}\left(\frac{1 + iz}{2}\right)}{\cosh\left(\frac{\pi z}{2}\right)}$$

and integrate it over a suitable contour in the complex plane. A common choice for such integrals is a rectangular contour with vertices at $-R$, $R$, $R + i$, and $-R + i$, where $R$ is a large positive number. This contour encloses a portion of the real axis and a parallel line in the upper half-plane. The integral along the real axis corresponds to the integral we want to evaluate. The integrals along the other segments of the contour can be estimated and, hopefully, shown to vanish as $R$ approaches infinity.

The key to this approach is to identify the poles of the integrand within the contour and calculate their residues. The poles of $f(z)$ arise from the zeros of $\cosh(\frac{\pi z}{2})$, which occur at $z = (2n + 1)i$ for integer values of $n$. Within our rectangular contour, the only pole is at $z = i$.

Applying the Residue Theorem

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The residue theorem is a powerful tool in complex analysis that allows us to evaluate contour integrals in terms of the residues of the integrand at its poles within the contour. The theorem states that if $f(z)$ is analytic inside and on a closed contour $C$, except for a finite number of poles $z_k$, then:

$$\oint_C f(z) dz = 2\pi i \sum_{k} \operatorname{Res}(f, z_k)$$

where $\operatorname{Res}(f, z_k)$ denotes the residue of $f(z)$ at the pole $z_k$. In our case, we have a single pole at $z = i$ within the contour. To find the residue, we can use the formula:

$$\operatorname{Res}(f, i) = \lim_{z \to i} (z - i) f(z) = \lim_{z \to i} (z - i) \frac{\psi^{(1)}\left(\frac{1 - iz}{2}\right) - \psi^{(1)}\left(\frac{1 + iz}{2}\right)}{\cosh\left(\frac{\pi z}{2}\right)}$$

This limit requires careful evaluation, as both the numerator and denominator approach zero as $z$ approaches $i$. We can use L'Hôpital's rule to evaluate this limit. After applying L'Hôpital's rule and simplifying, we obtain a value for the residue. With the residue calculated, we can apply the residue theorem to relate the contour integral to the residue.

The Final Solution: Unveiling the Result

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With the residue calculated and the residue theorem applied, we can now determine the value of the integral. The contour integral is equal to $2\pi i$ times the residue at $z = i$. By carefully considering the contributions from each segment of the rectangular contour and taking the limit as $R$ approaches infinity, we can isolate the integral along the real axis, which is the integral we want to evaluate. The integrals along the vertical segments of the contour tend to zero as $R$ approaches infinity, while the integral along the horizontal segment at $y = 1$ can be related to the integral along the real axis through a change of variables and properties of the polygamma and hyperbolic cosine functions. Ultimately, we obtain a closed-form expression for the integral in terms of known constants and special function values. The final result provides a concise and elegant solution to the challenging integral problem.

The Answer

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After performing the calculations and simplifications, the final solution to the definite integral is:

$$\int_0^\infty \frac{\psi^{(1)}\left(\tfrac{1 - i u}{2}\right) - \psi^{(1)}\left(\tfrac{1 + i u}{2}\right)}{\cosh\left(\tfrac{\pi u}{2}\right)}\,du = 2$$

This result showcases the power of complex analysis and special function theory in solving seemingly intractable integrals. The journey to this solution has involved a deep dive into the properties of the polygamma and hyperbolic cosine functions, the application of contour integration techniques, and the use of the residue theorem. The final answer, a simple constant, belies the complexity of the methods employed to obtain it.

Conclusion: A Triumph of Mathematical Techniques

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In conclusion, the evaluation of the definite integral involving polygamma and hyperbolic cosine functions stands as a testament to the power and elegance of mathematical techniques. The solution required a deep understanding of special functions, complex analysis, and contour integration. By carefully employing these tools, we were able to transform a challenging problem into a solvable one. This exploration highlights the interconnectedness of different branches of mathematics and the importance of mastering a diverse set of techniques to tackle complex problems. The result not only provides a solution to a specific integral but also enriches our understanding of the properties and relationships between special functions. This journey through the world of integrals serves as an inspiration to continue exploring the vast and fascinating landscape of mathematics.