Riemann Zeta Integral: Does It Converge? Perron's Formula

Hey everyone! Today, we're diving deep into a fascinating question in analytic number theory: Does the integral $\int_{-\infty}^{\infty}\zeta(c+it)x^{c - 1 + it} \,dt$ even exist? And if it doesn't, what are the implications for using Perron's formula, a powerful tool we often use in this field? This is a question that touches on the heart of how we connect the seemingly discrete world of integers with the continuous realm of complex analysis. So, buckle up, and let's explore this together!

The Integral in Question: A Deep Dive into Convergence

Let's start by dissecting the integral itself. We're looking at $\int_{-\infty}^{\infty}\zeta(c+it)x^{c - 1 + it} \,dt$, where:\

• $\zeta(s)$ is the Riemann zeta function, a cornerstone of number theory. It's defined for complex numbers s with a real part greater than 1 by the infinite series $\sum_{n=1}^{\infty} n^{-s}$, and it has an analytic continuation to the entire complex plane (except for a simple pole at s = 1).
• c is a real number, typically chosen to be greater than 1 to ensure the absolute convergence of the Riemann zeta function's series representation.
• x is a positive real number.
• t is the variable of integration, ranging from negative infinity to positive infinity.

So, the million-dollar question is: does this integral converge? To answer this, we need to understand the behavior of the integrand, particularly the Riemann zeta function, as t gets very large. The Riemann zeta function, while beautifully defined, can be quite tricky to handle. Its behavior in the critical strip (where the real part of s is between 0 and 1) is particularly complex, and that's precisely where the action happens in our integral.

Understanding the Oscillations: The term $x^{it}$ introduces oscillations into the integrand. Remember Euler's formula: $e^{it} = \cos(t) + i \sin(t)$. So, $x^{it} = e^{it \log(x)} = \cos(t \log(x)) + i \sin(t \log(x))$. These trigonometric functions oscillate rapidly as t increases, which can potentially lead to cancellation and convergence. However, this cancellation isn't guaranteed, especially when we multiply by the zeta function, which has its own complex behavior.

The Role of $\zeta(c + it)$: The Riemann zeta function $\zeta(c + it)$ is the real wild card. For a fixed c > 1, we know that $|\zeta(c + it)|$ grows at most polynomially in t. This means there exist constants M and k such that $|\zeta(c + it)| \le M|t|^k$ for large |t|. This polynomial growth is crucial. If $\zeta(c + it)$ grew exponentially, the integral would almost certainly diverge. However, polynomial growth doesn't automatically guarantee convergence either. The oscillations of $x^{it}$ need to be “in sync” with the behavior of $\zeta(c + it)$ for the integral to converge.

Divergence in the Wild: Unfortunately, the integral $\int_{-\infty}^{\infty}\zeta(c+it)x^{c - 1 + it} \,dt$ does not converge in general. The oscillations introduced by $x^{it}$ are not sufficient to tame the polynomial growth of $\zeta(c + it)$ as t goes to infinity. There are rigorous ways to show this, often involving careful estimation of the zeta function and integration by parts, but the core idea is that the integrand doesn't decay fast enough as t goes to infinity. The interplay between the oscillatory term and the growth of the zeta function is a delicate dance, and in this case, divergence wins.

Perron's Formula: A Powerful Tool with Limitations

Now that we know the integral often diverges, let's talk about Perron's formula. Perron's formula is a fantastic tool in analytic number theory. It allows us to express sums of arithmetic functions as contour integrals involving their Dirichlet series. This is a powerful connection, as it lets us use the machinery of complex analysis to study number-theoretic problems. In essence, it states:

$\sum_{n<x} a(n) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} F(s) \frac{x^s}{s} ds$

Where:

• $a(n)$ is an arithmetic function (a function defined on the positive integers).
• $F(s)$ is the Dirichlet series associated with $a(n)$, i.e., $F(s) = \sum_{n=1}^{\infty} \frac{a(n)}{n^s}$.
• x is a real number.
• c is a real number chosen appropriately to ensure convergence (typically c > 1 if F(s) converges absolutely for Re(s) > 1).
• The integral is taken along a vertical line in the complex plane.

The Case of $a(n) = 1$ and the Zeta Function: When we apply Perron's formula to the simple case where $a(n) = 1$, we're essentially counting the number of integers less than or equal to x. The associated Dirichlet series is simply the Riemann zeta function, $\zeta(s)$. So, Perron's formula gives us:

$\sum_{n\le x}’ 1 = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \zeta(s) \frac{x^s}{s} ds$

The prime on the summation indicates that if x is an integer, we only count x with a weight of 1/2. This is a technical detail that arises from the way the contour integral handles discontinuities.

The Problem with Direct Substitution: Now, let's try to make a direct substitution, setting $s = c + it$ in the integral. This gives us:

$\frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \zeta(s) \frac{x^s}{s} ds = \frac{1}{2\pi} \int_{-\infty}^{\infty} \zeta(c + it) \frac{x^{c + it}}{c + it} dt$

This looks very similar to the integral we started with! If we factor out $x^c$ and let's try to simplify a bit, we get:

$\frac{x^{c-1}}{2\pi} \int_{-\infty}^{\infty} \zeta(c + it) \frac{x^{it}}{c + it} dt$

This integral, while related, is subtly different from our original integral $\int_{-\infty}^{\infty}\zeta(c+it)x^{c - 1 + it} \,dt$. The key difference is the presence of the $c + it$ term in the denominator. This term provides additional decay as t goes to infinity, which is crucial for convergence.

Why Perron's Formula Still Works (With a Twist): So, if the original integral diverges, why does Perron's formula seem to work? The answer lies in the fact that Perron's formula doesn't directly involve the divergent integral we initially considered. The factor of $1/s$ in the integrand, which becomes $1/(c + it)$ after the substitution, is the key. This factor provides enough decay to ensure the convergence of the integral in Perron's formula. The integral $\int_{-\infty}^{\infty} \zeta(c + it) \frac{x^{c + it}}{c + it} dt$ does converge, thanks to the $1/(c + it)$ term.

The Takeaway: The divergence of $\int_{-\infty}^{\infty}\zeta(c+it)x^{c - 1 + it} \,dt$ doesn't invalidate Perron's formula. Perron's formula, in its correct form, includes the crucial $1/s$ term, which ensures convergence. The original integral is a related, but distinct, integral that simply doesn't converge in general. This highlights the importance of carefully applying these tools and understanding their limitations.

The Importance of Convergence in Analytic Number Theory

This exploration underscores a vital principle in analytic number theory: convergence matters! We can't just blindly manipulate integrals and series without considering whether they converge. Divergent integrals can lead to nonsensical results, and that's why we need to be meticulous in our analysis. Perron's formula is a powerful tool, but it's not a magic wand. It works because the integrals involved, in their proper form, converge. The seemingly small factor of $1/s$ makes all the difference.

In conclusion, while the integral $\int_{-\infty}^{\infty}\zeta(c+it)x^{c - 1 + it} \,dt$ generally does not exist due to the insufficient decay of the integrand, Perron's formula remains valid because it involves a related integral with a crucial convergence factor. This journey into the intricacies of convergence and divergence reminds us of the beautiful and sometimes subtle nature of mathematics, especially in the realm of analytic number theory. Keep exploring, keep questioning, and never stop diving deep into the fascinating world of numbers!