Evaluating The Integral Of Polygamma Functions Divided By Hyperbolic Cosine

The integral you've presented, involving the difference of polygamma functions divided by a hyperbolic cosine, is a fascinating problem that touches upon several advanced areas of mathematics. Specifically, it combines the intricacies of improper integrals, the complex behavior of polygamma functions, and the powerful techniques of contour integration. This article aims to dissect this problem, providing a comprehensive approach to understanding and solving it. We will explore the core concepts, the step-by-step methodology, and the underlying mathematical principles that make this integral a compelling challenge.

Understanding the Integral

Let's begin by restating the integral in question:

$$\operatorname{Im}\int_{0}^{\infty}\frac{\psi^{\left(1\right)}\left(\frac{1-iu}{2}\right)-\psi^{\left(1\right)}\left(\frac{iu+1}{2}\right)}{\cosh\left(\frac{\pi u}{2}\right)}\,du$$

At first glance, this integral appears daunting due to the presence of the polygamma function $\psi^{(1)}(z)$ and the hyperbolic cosine function $\cosh(\frac{\pi u}{2})$. However, with a strategic approach and a solid understanding of these functions' properties, we can tackle this problem effectively. The imaginary part operator, $\operatorname{Im}$, suggests that we're dealing with complex-valued functions and that the final result will be the imaginary component of the integral's value. This is a crucial piece of information that will guide our solution process.

To truly grasp the essence of this integral, we need to break it down into its fundamental components. The polygamma function $\psi^{(1)}(z)$ is the first derivative of the digamma function, which itself is the derivative of the logarithm of the gamma function. The gamma function, denoted by $\Gamma(z)$, is a generalization of the factorial function to complex numbers. Its logarithmic derivative, the digamma function $\psi(z)$, and the derivatives of the digamma function, the polygamma functions $\psi^{(n)}(z)$, play significant roles in various areas of mathematics and physics.

The hyperbolic cosine function, $\cosh(x)$, is defined as $\cosh(x) = \frac{e^x + e^{-x}}{2}$. Its presence in the denominator suggests that the integrand might have singularities or specific behaviors that we need to account for when performing the integration. The hyperbolic cosine is an even function, and its behavior is well-understood, which can be advantageous in simplifying the integral.

The limits of integration, from 0 to $\infty$, indicate that we are dealing with an improper integral. Improper integrals require careful treatment because they involve either infinite limits of integration or singularities within the interval of integration. In our case, we need to ensure that the integral converges and that we handle any potential singularities appropriately.

Delving into Polygamma Functions

The polygamma functions, denoted as $\psi^{(n)}(z)$, are defined as the $n$-th derivative of the digamma function $\psi(z)$, which itself is the derivative of the logarithm of the gamma function $\Gamma(z)$. Mathematically, this relationship can be expressed as follows:

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

The polygamma functions possess several key properties that are instrumental in evaluating integrals and solving differential equations. These properties include recurrence relations, reflection formulas, and asymptotic expansions. For instance, a crucial property for our problem is the reflection formula, which relates the polygamma function at $z$ to its value at $1-z$. Specifically, for the first polygamma function (also known as the trigamma function), the reflection formula is given by:

$$\psi^{(1)}(1-z) + \psi^{(1)}(z) = \frac{\pi^2}{\sin^2(\pi z)}$$

This formula is particularly useful when dealing with complex arguments, as it allows us to relate the polygamma function evaluated at a complex number to its value at the complex conjugate. In our integral, we have the terms $\psi^{(1)}(\frac{1-iu}{2})$ and $\psi^{(1)}(\frac{1+iu}{2})$, which are complex conjugates of each other. This suggests that the reflection formula might play a crucial role in simplifying the integrand.

Another important property of polygamma functions is their connection to the Hurwitz zeta function. The Hurwitz zeta function, denoted by $\zeta(s, a)$, is a generalization of the Riemann zeta function and is defined as:

$$\zeta(s, a) = \sum_{k=0}^{\infty} \frac{1}{(k+a)^s}$$

For integer values of $n$, the polygamma function can be expressed in terms of the Hurwitz zeta function as:

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

This relationship provides an alternative way to compute or analyze the polygamma functions, especially when dealing with special values or asymptotic behavior. In the context of our integral, this connection may not be directly applicable for a closed-form solution, but it offers valuable insights into the function's properties and behavior.

Furthermore, understanding the asymptotic behavior of polygamma functions is crucial when dealing with improper integrals. As $|z|$ approaches infinity, the polygamma functions have well-defined asymptotic expansions that allow us to estimate their values. For the first polygamma function, the asymptotic expansion is given by:

$$\psi^{(1)}(z) \sim \frac{1}{z} + \frac{1}{2z^2} + \frac{1}{6z^3} + O\left(\frac{1}{z^5}\right)$$

This expansion is particularly useful for determining the convergence of our integral. By examining the asymptotic behavior of the integrand, we can ascertain whether the integral converges as $u$ approaches infinity. In our case, the difference of the polygamma functions and the presence of the hyperbolic cosine in the denominator will significantly influence the convergence properties.

Leveraging Contour Integration Techniques

Contour integration is a powerful technique in complex analysis for evaluating integrals, especially those that are difficult or impossible to solve using traditional real calculus methods. The fundamental idea behind contour integration is to extend the integration from the real line to a path in the complex plane, known as a contour. By carefully choosing the contour and applying the residue theorem, we can often transform a real integral into a complex integral that can be evaluated more easily.

The residue theorem is the cornerstone of contour integration. It states that the integral of a complex function $f(z)$ around a closed contour $C$ is equal to $2\pi i$ times the sum of the residues of $f(z)$ at the poles enclosed by $C$. Mathematically, the residue theorem can be expressed as:

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

where $\operatorname{Res}(f, z_k)$ denotes the residue of $f(z)$ at the pole $z_k$, and the sum is taken over all poles enclosed by the contour $C$.

To apply contour integration to our problem, we need to consider the complex extension of the integrand. This involves replacing the real variable $u$ with a complex variable $z = x + iy$, where $x$ and $y$ are real numbers. The integral then becomes a contour integral in the complex plane.

The choice of contour is crucial in contour integration. A common strategy is to use a rectangular contour or a semi-circular contour, depending on the properties of the integrand. For our integral, a rectangular contour in the complex plane might be a suitable choice. We can consider a rectangle with vertices at $-R$, $R$, $R + iA$, and $-R + iA$, where $R$ is a large positive number and $A$ is a positive constant. As $R$ approaches infinity, the integral along the real axis segment of the contour will approach our original integral.

The next step is to identify the poles of the integrand within the chosen contour. Poles are the singularities of the complex function, where the function becomes infinite. In our case, the poles will arise from the hyperbolic cosine function in the denominator. The hyperbolic cosine function, $\cosh(\frac{\pi z}{2})$, has zeros when $\frac{\pi z}{2} = i(\frac{\pi}{2} + k\pi)$, where $k$ is an integer. This implies that the poles are located at $z = i(2k+1)$ for integer values of $k$.

Within our rectangular contour, we need to identify the poles that are enclosed. If we choose $A$ such that $0 < A < 2$, then the only pole enclosed by the contour will be at $z = i$. The residue at this pole can be calculated using the formula:

$$\operatorname{Res}(f, i) = \lim_{z \to i} (z-i) f(z)$$

where $f(z)$ is the complex extension of our integrand. After calculating the residue, we can apply the residue theorem to evaluate the contour integral. The contour integral will be equal to $2\pi i$ times the residue at $z = i$.

However, the contour integral is not equal to our original integral directly. We need to consider the contributions from the other segments of the rectangular contour. The integral along the top segment (from $-R + iA$ to $R + iA$) and the vertical segments (from $R$ to $R + iA$ and from $-R + iA$ to $-R$) must be evaluated separately. By carefully choosing the value of $A$ and analyzing the behavior of the integrand along these segments, we can often show that the contributions from these segments vanish as $R$ approaches infinity.

If the contributions from the other segments vanish, then the contour integral will be equal to the integral along the real axis, which is our original integral. By equating the contour integral to $2\pi i$ times the residue and taking the imaginary part, we can obtain the value of the original integral.

Solving the Integral Step-by-Step

Now, let's outline a step-by-step approach to solving the integral using the techniques discussed above:

1. Identify the integrand and its properties:

   • Our integrand is $f(u) = \frac{\psi^{(1)}(\frac{1-iu}{2}) - \psi^{(1)}(\frac{1+iu}{2})}{\cosh(\frac{\pi u}{2})}$.
   • We recognize the presence of polygamma functions and hyperbolic cosine.
   • We note that we are interested in the imaginary part of the integral.

2. Apply the reflection formula for polygamma functions:

   • Use the reflection formula $\psi^{(1)}(1-z) + \psi^{(1)}(z) = \frac{\pi^2}{\sin^2(\pi z)}$.
   • Rewrite the difference of polygamma functions in terms of trigonometric functions.

3. Extend the integrand to the complex plane:

   • Replace the real variable $u$ with a complex variable $z = x + iy$.
   • Consider the complex function $f(z) = \frac{\psi^{(1)}(\frac{1-iz}{2}) - \psi^{(1)}(\frac{1+iz}{2})}{\cosh(\frac{\pi z}{2})}$.

4. Choose a suitable contour:

   • Select a rectangular contour with vertices at $-R$, $R$, $R + iA$, and $-R + iA$.
   • Choose $A$ such that $0 < A < 2$ to enclose only the pole at $z = i$.

5. Identify the poles and calculate the residues:

   • Find the poles of the integrand by solving $\cosh(\frac{\pi z}{2}) = 0$.
   • The poles are at $z = i(2k+1)$ for integer values of $k$.
   • Calculate the residue at $z = i$ using the formula $\operatorname{Res}(f, i) = \lim_{z \to i} (z-i) f(z)$.

6. Apply the residue theorem:

   • Evaluate the contour integral using the residue theorem: $\oint_C f(z) dz = 2\pi i \operatorname{Res}(f, i)$.

7. Evaluate the integrals along the other segments of the contour:

   • Show that the integrals along the top segment and the vertical segments vanish as $R$ approaches infinity.
   • This step often involves careful estimation and bounding techniques.

8. Equate the contour integral to the original integral:

   • If the contributions from the other segments vanish, then the contour integral is equal to the integral along the real axis.

9. Take the imaginary part:

   • Extract the imaginary part of the result to obtain the value of the original integral.

Conclusion

Evaluating the integral $\operatorname{Im}\int_{0}^{\infty}\frac{\psi^{\left(1\right)}\left(\frac{1-iu}{2}\right)-\psi^{\left(1\right)}\left(\frac{iu+1}{2}\right)}{\cosh\left(\frac{\pi u}{2}\right)}\,du$ is a challenging yet rewarding mathematical endeavor. It requires a deep understanding of polygamma functions, their properties, and their connection to other special functions like the Hurwitz zeta function. Moreover, it necessitates the application of advanced techniques such as contour integration and the residue theorem. By carefully analyzing the integrand, choosing an appropriate contour, and evaluating the residues, we can transform this seemingly intractable integral into a solvable problem.

This exploration highlights the power and elegance of complex analysis in solving real-world problems. The techniques discussed here are not only applicable to this specific integral but also to a wide range of other integrals and problems in mathematics, physics, and engineering. The journey through polygamma functions, contour integration, and complex analysis underscores the interconnectedness of mathematical concepts and the beauty of mathematical problem-solving.