Glaisher-Kinkelin Constant: A New Series?

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the fascinating world of mathematical constants, specifically focusing on the Glaisher–Kinkelin constant, often denoted by the letter $A$. You know, the one that pops up in number theory and analysis, chilling with the Bernoulli numbers and the Riemann zeta function. We've got this intriguing identity here:

3${}$\ln A - \frac{1}{4} - \frac{7}{12}${}$\ln 2 = \int_0^1 \frac{1}{\ln z} \left( \frac{2-z}{4} - \frac{1}{(1+z)^2} \right) dz

This identity is super cool because it connects the Glaisher–Kinkelin constant ($A$) to a definite integral. Now, the big question on our minds, and likely yours too, is: can we actually derive a useful series expansion for $\ln(A)$ from this? That's what we're going to explore today. We'll be breaking down the identity, thinking about how integral representations can lead to series, and whether this particular integral is a golden ticket to a new way of calculating $A$. So, grab your thinking caps, and let's get nerdy!

Unpacking the Glaisher–Kinkelin Constant and the Integral Identity

Alright, let's first get acquainted with our star, the Glaisher–Kinkelin constant ($A$). It's a pretty significant number in mathematics, defined by the limit:

$$A = \lim_{n\to\infty} \frac{n! n^{1/n}}{e^{n - \frac{1}{12} + \sum_{k=1}^n \frac{1}{k}} }$$

Whoa, that looks complex, right? But basically, it relates factorials and powers. Its approximate value is $A \approx 1.282427129...$. It shows up in places like the asymptotic behavior of the gamma function and in certain sums involving Bernoulli numbers. Now, let's look at the identity that sparked this discussion:

3${}$\ln A - \frac{1}{4} - \frac{7}{12}${}$\ln 2 = \int_0^1 \frac{1}{\ln z} \left( \frac{2-z}{4} - \frac{1}{(1+z)^2} \right) dz

This isn't just some random equation; it's a clever way to express a relationship involving $\ln A$. The integral on the right-hand side is key. The integrand, $\frac{1}{\ln z} \left( \frac{2-z}{4} - \frac{1}{(1+z)^2} \right)$, has a singularity at $z=1$ (because $\ln 1 = 0$) and also at $z=0$ due to the $\ln z$ in the denominator. Dealing with such integrals can be tricky, but they often hold the secrets to series expansions. The structure of the integrand, particularly the term $\frac{1}{(1+z)^2}$, hints at geometric series or related expansions. The presence of $\ln z$ in the denominator is a bit unusual and might require some special handling, possibly through techniques like integration by parts or by relating it to known special functions. The limits of integration, from 0 to 1, are also typical for generating series involving powers of variables less than 1. We're essentially trying to see if this integral can be transformed into a sum of terms that directly give us $\ln A$. The constants $1/4$ and $7/12 \ln 2$ on the left side are also important clues, suggesting that any series we find might need these terms adjusted to isolate just $\ln A$.

Strategies for Deriving a Series from the Integral

So, how do we get a series from this integral, you ask? That's where the real mathematical magic happens, guys. The primary strategy involves manipulating the integrand into a form that can be integrated term-by-term. Let's focus on the part inside the parenthesis:

$$\frac{2-z}{4} - \frac{1}{(1+z)^2}$$

We know that for $|z| < 1$, the geometric series expansion is $\frac{1}{1+z} = \sum_{n=0}^\infty (-1)^n z^n$. Differentiating this gives us $\frac{-1}{(1+z)^2} = \sum_{n=1}^\infty (-1)^n n z^{n-1}$. Multiplying by -1, we get $\frac{1}{(1+z)^2} = \sum_{n=1}^\infty (-1)^{n-1} n z^{n-1}$.

We can also expand the term $\frac{2-z}{4}$ as $\frac{1}{2} - \frac{z}{4}$.

Substituting these into the expression gives us:

$$\left( \frac{1}{2} - \frac{z}{4} \right) - \sum_{n=1}^\infty (-1)^{n-1} n z^{n-1}$$

This looks a bit messy because we have individual terms and a series. Let's try to combine them. A common tactic is to express everything as a power series. We can write $\frac{1}{2}$ as $\frac{1}{2} z^0$ and $-\frac{z}{4}$ as $-\frac{1}{4} z^1$. So we have:

$$\frac{1}{2} z^0 - \frac{1}{4} z^1 - \sum_{n=1}^\infty (-1)^{n-1} n z^{n-1}$$

Now, let's re-index the sum to have powers of $z^n$. Let $k = n-1$, so $n = k+1$. When $n=1$, $k=0$. The sum becomes:

$$\sum_{k=0}^\infty (-1)^{k} (k+1) z^k$$

So the expression inside the parenthesis is:

$$\frac{1}{2} - \frac{1}{4}z - \sum_{k=0}^\infty (-1)^{k} (k+1) z^k$$

This still doesn't look super clean for term-by-term integration with the $\frac{1}{\ln z}$ factor. The presence of $\frac{1}{\ln z}$ makes direct integration tricky. Usually, we integrate functions of the form $z^n$. The $\frac{1}{\ln z}$ term suggests we might need to use a different approach, perhaps involving the Mellin transform or relating the integral to polylogarithms or other special functions. However, if we ignore the $\frac{1}{\ln z}$ for a moment and focus on integrating the power series part, we'd get:

$$\int_0^1 z^k dz = \left[ \frac{z^{k+1}}{k+1} \right]_0^1 = \frac{1}{k+1}$$

This would lead to a series of terms like $\frac{(-1)^k}{(k+1)^2}$. But we still have that $\frac{1}{\ln z}$ staring us down. One potential avenue is to use the integral representation of certain functions. For instance, $\frac{1}{\ln z} = \int_0^\infty z^{-t} dt$ (though this is a bit informal and requires careful justification, especially near $z=0, 1$). If we could swap the order of integration, we might get somewhere. Another path is to recognize that integrals involving $\frac{1}{\ln z}$ can sometimes be related to the exponential integral function or related integrals. The specific form of the integrand might also allow for integration by parts, potentially simplifying the $\frac{1}{\ln z}$ term. Let's consider the term $\int_0^1 \frac{z^k}{\ln z} dz$. This integral is known to be related to the Logarithmic Integral function $Li(x)$, but the $z^k$ factor complicates things. The identity provided is likely derived using more advanced techniques, perhaps involving contour integration or specific properties of zeta functions. The challenge here is to reverse-engineer that process or find a simpler path.

The Challenge of the $\frac{1}{\ln z}$ Term

Guys, the elephant in the room, or rather, the singularity in the integrand, is that $\frac{1}{\ln z}$ term. This isn't your standard polynomial or exponential function that integrates nicely. When $z$ approaches 1, $\ln z$ approaches 0, making the integrand blow up. And when $z$ approaches 0, $\ln z$ approaches negative infinity, which also poses a problem. This means the integral is an improper integral, and we need to be careful about its convergence and how we handle it, especially if we want to break it down into a series.

One common way to deal with $\frac{1}{\ln z}$ is to use a substitution. Let $z = e^{-t}$. Then $\ln z = -t$, and $dz = -e^{-t} dt$. When $z=0$, $t=\infty$. When $z=1$, $t=0$. The integral becomes:

$$\int_\infty^0 \frac{1}{-t} \left( \frac{2-e^{-t}}{4} - \frac{1}{(1+e^{-t})^2} \right) (-e^{-t} dt)$$

$$= \int_0^\infty \frac{e^{-t}}{t} \left( \frac{2-e^{-t}}{4} - \frac{1}{(1+e^{-t})^2} \right) dt$$

This looks slightly more manageable. We have $e^{-t}$ terms and powers of $e^{-t}$. We can expand $\frac{1}{(1+e^{-t})^2}$ using the geometric series for $\frac{1}{1+x}$ where $x=e^{-t}$. However, $e^{-t}$ is always positive, so we need to be careful. Also, the term $\frac{e^{-t}}{t}$ is problematic at $t=0$. $\frac{e^{-t}}{t}$ behaves like $\frac{1}{t}$ as $t \to 0$. This integral still has a singularity at the lower limit.

Let's consider the structure $\frac{e^{-t}}{t}$. This is related to the exponential integral function $E_1(x) = \int_x^\infty \frac{e^{-t}}{t} dt$. Our integral is $\int_0^\infty \frac{e^{-t}}{t} f(t) dt$, where $f(t) = \frac{2-e^{-t}}{4} - \frac{1}{(1+e^{-t})^2}$. If $f(0) \neq 0$, then the integral $\int_0^\infty \frac{e^{-t}}{t} f(t) dt$ will diverge at $t=0$. Let's check $f(0)$: $f(0) = \frac{2-1}{4} - \frac{1}{(1+1)^2} = \frac{1}{4} - \frac{1}{4} = 0$. Ah, this is good! Since $f(0)=0$, the singularity at $t=0$ might be removable. We need to examine the behavior of $f(t)$ near $t=0$. Using Taylor series for $e^{-t} \approx 1-t+\frac{t^2}{2}$:

$$\frac{2-e^{-t}}{4} \approx \frac{2-(1-t)}{4} = \frac{1+t}{4} = \frac{1}{4} + \frac{t}{4}$$

$$\frac{1}{(1+e^{-t})^2} \approx \frac{1}{(1+(1-t))^2} = \frac{1}{(2-t)^2} = \frac{1}{4(1-t/2)^2} \approx \frac{1}{4} (1-t/2)^{-2} \approx \frac{1}{4} (1 + 2(t/2) + ...) = \frac{1}{4} (1+t) = \frac{1}{4} + \frac{t}{4}$$

This is interesting! It seems $f(t)$ might be zero not just at $t=0$ but up to the first order in $t$. Let's be more precise. We need the Taylor expansion of $f(t)$ around $t=0$. Let's check $f'(t)$:

$f'(t) = \frac{-e^{-t}}{4} - \frac{-2(1+e^{-t})^{-3} (-e^{-t})}{(1+e^{-t})^4} = \frac{-e^{-t}}{4} - \frac{2e^{-t}}{(1+e^{-t})^3}$.

$f'(0) = \frac{-1}{4} - \frac{2(1)}{(1+1)^3} = \frac{-1}{4} - \frac{2}{8} = \frac{-1}{4} - \frac{1}{4} = -\frac{1}{2}$.

So, $f(t) \approx f(0) + f'(0)t = 0 - \frac{1}{2}t = -\frac{t}{2}$ for small $t$.

Then, $\frac{e^{-t}}{t} f(t) \approx \frac{e^{-t}}{t} (-\frac{t}{2}) = -\frac{1}{2} e^{-t}$. As $t o 0$, this approaches $-\frac{1}{2}$. This means the singularity at $t=0$ is indeed integrable.

Now, we need to expand $f(t)$ as a series in $e^{-t}$. Let $x = e^{-t}$. Then $t = -\ln x$. The integral becomes:

$$\int_1^0 \frac{1}{-\ln x} \left( \frac{2-x}{4} - \frac{1}{(1+x)^2} \right) (-dx) = \int_0^1 \frac{1}{\ln x} \left( \frac{2-x}{4} - \frac{1}{(1+x)^2} \right) dx$$

We are back where we started! This substitution didn't immediately yield a series. The key must be in how we handle the $\frac{1}{\ln z}$ term within the integration process, perhaps by relating it to known series representations of functions involving logarithms.

Potential Series Expansions and Connections

Given the challenges, is it still viable to get a series for $\ln(A)$? Absolutely, but it might require some more sophisticated machinery than simple term-by-term integration of basic power series. The identity itself might be a result of applying techniques like the Binet's second formula for the logarithm of the gamma function, or using properties of the Hurwitz zeta function $\zeta(s, q)$.

Let's think about related series. We know that $\ln A = \frac{1}{12} - \int_0^1 \frac{1}{\ln z} \frac{1}{e^z-1} dz$. This is a known representation. The integral we are given is:

$$I = \int_0^1 \frac{1}{\ln z} \left( \frac{2-z}{4} - \frac{1}{(1+z)^2} \right) dz$$

We have $3${}$\ln A = \frac{1}{4} + \frac{7}{12}${}$\ln 2 + I$. So, $\ln A = \frac{1}{12} + \frac{7}{36}${}$\ln 2 + \frac{I}{3}$.

This means if we can evaluate $I$ as a series, we've got our answer. The integrand $\frac{1}{\ln z} \left( \frac{2-z}{4} - \frac{1}{(1+z)^2} \right)$ can be viewed as a product of $\frac{1}{\ln z}$ and a rational function of $z$. Sometimes, functions of the form $\int_0^1 \frac{f(z)}{\ln z} dz$ can be related to $\sum_{n=1}^\infty \frac{f(1/n)}{n o ext{something}}$.

Another approach is to consider the Lerch transcendent $\Phi(z, s, a) = \sum_{n=0}^\infty \frac{z^n}{(n+a)^s}$. Integrals involving $\frac{1}{\ln z}$ can sometimes be connected to these functions or related polylogarithms. The term $\frac{1}{(1+z)^2}$ is strongly suggestive of series involving $\sum \frac{(-1)^n n}{...}$.

Let's try to break down the integrand further. Define $g(z) = \frac{2-z}{4} - \frac{1}{(1+z)^2}$. We saw that $g(z) = \frac{1}{2} - \frac{z}{4} - \sum_{n=1}^\infty (-1)^{n-1} n z^{n-1}$.

Consider the integral $\int_0^1 \frac{z^k}{\ln z} dz$. This integral is related to the logarithmic derivative of the gamma function. More specifically, $\int_0^1 \frac{z^s}{\ln z} dz = -\ln \Gamma(s+1)$ for $Re(s)>-1$. This is not quite what we have.

However, there are known series expansions for $\ln A$. One is:

$$\ln A = \frac{1}{12} - \sum_{n=1}^\infty \frac{\zeta'(n+1)}{n+1}$$

where $\zeta'(s)$ is the derivative of the Riemann zeta function. This series arises from analyzing the derivative of the zeta function itself.

Could our integral be related to this? Let's look at the integrand again:

$$\frac{1}{\ln z} \left( \frac{2-z}{4} - \frac{1}{(1+z)^2} \right)$$

If we could represent $\frac{1}{(1+z)^2}$ and $\frac{2-z}{4}$ as series in a way that combines nicely with $\frac{1}{\ln z}$ after integration, we might get something like $\sum \frac{\zeta'(k)}{k}$.

For example, using the identity $\int_0^1 \frac{x^s - x^t}{\ln x} dx = \ln \frac{s+1}{t+1}$. This is useful for differences of powers. Our function isn't a simple difference of powers.

What if we consider the integral $\int_0^1 \frac{z^n}{\ln z} dz$? This can be related to $\int_0^1 \int_0^\infty n e^{-nt} dt / (-t) dz = \int_0^1 \int_0^\infty -n/t e^{-nt} dt dz$? This seems complicated.

Let's revisit the substitution $z = e^{-t}$:

$$I = \int_0^\infty \frac{e^{-t}}{t} \left( \frac{2-e^{-t}}{4} - \frac{1}{(1+e^{-t})^2} \right) dt$$

We can expand $\frac{1}{(1+e^{-t})^2}$ using the generalized binomial theorem or series. Let $x = e^{-t}$. For $0 < t < \infty$, we have $0 < x < 1$.

$$\frac{1}{(1+x)^2} = \sum_{n=0}^\infty (-1)^n (n+1) x^n = \sum_{n=0}^\infty (-1)^n (n+1) e^{-nt}$$

And $\frac{2-e^{-t}}{4} = \frac{2}{4} - \frac{e^{-t}}{4} = \frac{1}{2} - \frac{1}{4} e^{-t}$.

So the integrand becomes:

$$\frac{e^{-t}}{t} \left( \frac{1}{2} - \frac{1}{4} e^{-t} - \sum_{n=0}^\infty (-1)^n (n+1) e^{-nt} \right)$$

$$= \frac{e^{-t}}{t} \left( \frac{1}{2} - \frac{1}{4} e^{-t} - \left( 1 - 2e^{-t} + 3e^{-2t} - 4e^{-3t} + ... \right) \right)$$

$$= \frac{e^{-t}}{t} \left( (\frac{1}{2}-1) + (-\frac{1}{4} + 2)e^{-t} - 3e^{-2t} + 4e^{-3t} - ... \right)$$

$$= \frac{e^{-t}}{t} \left( -\frac{1}{2} + \frac{7}{4}e^{-t} - 3e^{-2t} + 4e^{-3t} - ... \right)$$

Now we need to integrate $\frac{e^{-t}}{t} \times (\text{series})$. This is $\int_0^\infty \frac{1}{t} \sum_{n=0}^\infty c_n e^{-(n+1)t} dt$ where $c_0 = -1/2$, $c_1 = 7/4$, $c_2 = -3$, etc.

We can potentially swap the integral and the sum: $\sum_{n=0}^\infty c_n \int_0^\infty \frac{e^{-(n+1)t}}{t} dt$.

Each integral $\int_0^\infty \frac{e^{-at}}{t} dt$ diverges at $t=0$. This is where our earlier check $f(0)=0$ becomes crucial. The term inside the parenthesis, let's call it $h(t) = \frac{2-e^{-t}}{4} - \frac{1}{(1+e^{-t})^2}$, must have a zero at $t=0$ that cancels the $1/t$ singularity. We found $h(t) \approx -t/2$ near $t=0$.

So, $\frac{e^{-t}}{t} h(t) \approx \frac{1}{t} (1-t) (-t/2) = (-1/2)(1-t)$. The integral is $\int_0^\infty (e^{-t}/t) h(t) dt$. If $h(t) = \sum_{k=1}^\infty a_k t^k$, then $\frac{e^{-t}}{t} h(t) = \frac{1-t+...}{t} (a_1 t + a_2 t^2 + ...) = (1/t - 1 + ...)(a_1 t + a_2 t^2 + ...) = a_1 + (a_2 - a_1) t + ...$

This implies that $\int_0^1 \frac{1}{\ln z} g(z) dz$ is likely related to \sum_{n=1}^ o rac{a_n}{n}, where $a_n$ are coefficients of a series expansion of $g(z)$.

Conclusion: Yes, it appears viable to obtain a series expansion for $\ln(A)$ from this identity. The key is the careful expansion of the rational function part and the integration of terms involving $\frac{e^{-at}}{t}$. The convergence at $t=0$ is handled because the rational function part has a zero of sufficient order. The integrals $\int_0^\infty \frac{e^{-at}}{t} dt$ are related to the exponential integral $E_1(x)$, but after the cancellation, we should end up with a convergent series. The exact form of the series might be complex, potentially involving zeta function derivatives or other special constants, but the structure suggests a way forward. It's a testament to the power of integral representations in uncovering the structure of fundamental mathematical constants!