含(q;q)∞对数的积分求解技术求助

Hey there, let's break down this integral step by step. You've already tried two key expansions for $\log\phi(z^{-s})$, so let's build on those with standard techniques to make progress.

First Approach: Integration by Parts (Most Straightforward) #

Since your integral includes a derivative term $\frac{d}{dz}\log\phi(z^{-s})$, integration by parts is a natural first move. Recall the formula:
$$\int u , dv = uv - \int v , du$$

Let’s set:

• $u = z - 1$ (so $du = dz$)
• $dv = \frac{d}{dz}\log\phi(z^{-s}) dz$ (so $v = \log\phi(z^{-s})$)

Applying this to your integral gives:
$$\left.(z-1)\log\phi(z^{{-s})\right|_{1}}{q} - \int_{1}^{{q}\log\phi(z}{-s}) dz$$

Now we handle each part:

1. Boundary Term at $z=1$:
   When $z \to 1^+$, we get an indeterminate form $0 \cdot -\infty$. Using your first expansion $\log\phi(z^{-s}) = -\sum_{k=1}^{{\infty}\frac{1}{k(z}{ks}-1)}$, approximate $z^{ks} \approx 1 + ks(z-1)$ for $z$ near 1. This simplifies to:
   $$\log\phi(z^{-s}) \approx -\frac{1}{s(z-1)}\sum_{k=1}^\infty \frac{1}{k^2} + O(1) = -\frac{\pi^2}{6s(z-1)} + O(1)$$
   Multiplying by $(z-1)$ gives the limit $\lim_{z\to1^{+}(z-1)\log\phi(z}{-s}) = -\frac{\pi^2}{6s}$.

2. Boundary Term at $z=q$:
   This is straightforward: $(q-1)\log\phi(q^{-s})$.

3. Remaining Integral:
   Use your second expansion $\log\phi(z^{-s}) = -\sum_{k=1}^{{\infty}\frac{\sigma(k)}{k}z}{-ks}$ (valid for $\text{Re}(s) > 0$ and $z > 1$, since the series converges uniformly here). Swap the sum and integral (justified by the Weierstrass M-test):
   $$\int_{1}^{{q}\log\phi(z}{-s}) dz = -\sum_{k=1}^{{\infty}\frac{\sigma(k)}{k}\int_{1}}{q}z^{-ks} dz$$
   Compute the inner integral:
   $$\int_{1}^{q}z{-ks} dz = \frac{q^{1-ks} - 1}{1 - ks}$$

Putting it all together, your integral becomes:
$$(q-1)\log\phi(q^{-s}) + \frac{\pi^2}{6s} + \sum_{k=1}^{\infty}\frac{\sigma(k)}{k} \cdot \frac{q^{1-ks} - 1}{1 - ks}$$

This expression works well for analyzing convergence, performing analytic continuation, or taking limits (e.g., $q \to \infty$).

Second Approach: Variable Substitution #

Let’s substitute $w = z^{-s}$, so $z = w^{-1/s}$ and $dz = -\frac{1}{s}w^{-1/s - 1}dw$. When $z=1$, $w=1$; when $z=q$, $w=q^{-s}$. The integral transforms to:
$$\frac{1}{s}\int_{q^{-s}}{1} (w^{-1/s} - 1) w^{-1/s - 1} \cdot \frac{d}{dw}\log\phi(w) dw$$

The logarithmic derivative of $\phi(w)$ has a well-known series expansion:
$$\frac{d}{dw}\log\phi(w) = -\sum_{n=1}^\infty \frac{n w^{n-1}}{1 - w^n}$$

This substitution shifts the problem to $w \in (0,1)$ (since $\text{Re}(s) > 0$ and $q > 1$ makes $|q^{-s}| < 1$), which might align better with modular form properties of $\phi(w)$ if you're working in number theory contexts.

Third Approach: Direct Series Expansion of the Derivative #

Starting from your first expansion, compute the derivative directly:
$$\frac{d}{dz}\log\phi(z^{-s}) = \sum_{k=1}^{\infty}\frac{s k z^{ks - 1}}{(z^{ks} - 1)^2}$$

Substitute this into the integral and swap sum and integral (valid due to uniform convergence):
$$\int_{1}^{{q}(z-1)\frac{d}{dz}\log\phi(z}{-s}) dz = s\sum_{k=1}^{{\infty}k\int_{1}}{q}\frac{(z-1)z^{ks - 1}}{(z^{ks} - 1)^2} dz$$

Let $t = z^{ks}$, so $dt = s k z^{ks - 1} dz$. The inner integral becomes:
$$\int_{1}^{q{ks}} \frac{t^{1/(ks)} - 1}{(t-1)^2} dt$$

You can expand $\frac{t^a - 1}{(t-1)^2}$ (where $a = 1/(ks)$) as a power series or use integration by parts again, though this path is more tedious than the first approach.

内容的提问来源于stack exchange，提问作者Mohammad Al Jamal