关于含Gamma函数对数的定积分能否用闭形式或特殊函数表示的问询

Great question—you’re absolutely right to connect this integral to Raabe’s work; it’s a natural generalization of the classic Raabe integral, and we can indeed express it using special functions and closed-form integrals. Let’s break this down clearly:

Key Observations & Setup #

First, rewrite your integral as the difference of two convergent integrals (divergent terms from expanding $\ln\Gamma$ will cancel out):
$$
\int_{0}^{1}\log\left(\frac{\Gamma(at+b)}{\Gamma(ct+b)}\right)\frac{dt}{t} = \int_0^1 \frac{\ln\Gamma(at+b)}{t}dt - \int_0^1 \frac{\ln\Gamma(ct+b)}{t}dt
$$

Define a helper function for a single non-negative parameter $p$:
$$
J(p) = \int_0^1 \frac{\ln\Gamma(pt + b) - \ln\Gamma(b)}{t}dt
$$
Your integral simplifies to $J(a) - J(c)$. Note $J(p)$ is convergent: as $t \to 0^+$, $\ln\Gamma(pt+b) - \ln\Gamma(b) \sim pt\psi(b)$ (where $\psi$ is the digamma function), so the integrand behaves like $p\psi(b) + o(1)$.

Closed-Form via Integral of Log-Gamma #

To find $J(p)$, take its derivative with respect to $p$ (we can interchange differentiation and integration via dominated convergence):
$$
J'(p) = \int_0^1 \frac{\psi(pt + b)}{t} \cdot t dt = \int_0^1 \psi(pt + b)dt
$$
Using the antiderivative of the digamma function ($\int \psi(z)dz = \ln\Gamma(z) + C$), we get:
$$
J'(p) = \frac{1}{p}\left[\ln\Gamma(p + b) - \ln\Gamma(b)\right]
$$
Since $J(0) = 0$, integrate this derivative to get a closed-form integral expression:
$$
J(p) = \int_0^p \frac{\ln\Gamma(q + b) - \ln\Gamma(b)}{q}dq
$$
Your original integral is therefore:
$$
\int_{0}^{1}\log\left(\frac{\Gamma(at+b)}{\Gamma(ct+b)}\right)\frac{dt}{t} = \int_c^a \frac{\ln\Gamma(q + b) - \ln\Gamma(b)}{q}dq
$$
(If $c > a$, swap the limits and add a negative sign.)

Series Expansion with Hurwitz Zeta Function #

For an explicit series representation, use the expansion of $\ln\Gamma(z+b) - \ln\Gamma(b)$ in terms of the Hurwitz zeta function $\zeta(k, b)$ (a generalization of the Riemann zeta function to non-integer offsets):
$$
\ln\Gamma(z + b) - \ln\Gamma(b) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \zeta(k, b) z^k
$$
Substitute this into $J(p)$ and integrate term-by-term (valid for $|p| < 1$, with analytic continuation for $p \geq 1$):
$$
J(p) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k2} \zeta(k, b) p^k
$$

Connection to Classic Raabe Integral #

The classic Raabe integral is:
$$
\int_0^1 \frac{\ln\Gamma(t + a) - \ln\Gamma(t)}{t}dt = \ln\left(\frac{\Gamma(a)}{a^a}\right) + a\gamma + \sum_{k=1}^\infty \frac{(-1)^{{k+1}\zeta(k+1)}{k}a}k
$$
Your integral extends this by replacing the linear term $t+a$ with $pt+b$. When $b=0$ (and $a,c > 0$, since $\Gamma(0)$ is undefined), $J(p)$ reduces to an integral of $\ln\Gamma(q)/q$, which can also be linked to the Barnes G function (a two-variable generalization of the gamma function) via its logarithmic derivatives.

内容的提问来源于stack exchange，提问作者Thomas