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Raabe's integral is usually stated in the form

$$ I(\alpha):=\int_0^1 \log\left(\,\Gamma\left(x+\alpha\right)\,\right)\,{\rm d}x =\frac{\log\left(2\pi\right)}{2}+\alpha \log\left(\alpha\right) -\alpha\,,\qquad \alpha \geq 0. $$

(see for example this question.) I was wondering whether the condition $\alpha\geq0$ is necessary. Since $\Gamma(x+\alpha)$ is a meromorphic function, so is $I(\alpha)$, and therefore differentiation yields

$$I'(\alpha)=\int_0^1\psi(x+\alpha)\,dx =\log\Gamma(1+\alpha)-\log\Gamma(\alpha)$$

from where we deduce that

$$I(\alpha)=\int_0^1(\log\Gamma(1+\alpha)-\log\Gamma(\alpha))\,d\alpha+C$$ for some appropriate constant $C$. If $\alpha$ were real $\log\Gamma(1+\alpha)-\log\Gamma(\alpha)=\log\alpha$ trivially, but when $\alpha$ is complex some care should be taken. We can write $\log\Gamma(1+\alpha)=\log\Gamma(\alpha)+\log\alpha+2\pi i k(\alpha)$ with $k(\alpha)$ an integral-valued function, and hence

$$I(\alpha)=C'+\alpha\log\alpha-\alpha+2\pi i\int k(\alpha)\,d\alpha.$$

How one can show that $k(\alpha)=0$ (if this is really true), or at least, by any other method, compute $I(\alpha)$ for (some) complex arguments?

user246336
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1 Answers1

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The identity theorem says that the formula is valid on $U = \mathbb{C} \setminus \{ t \in \mathbb{R} : t \leqslant 0\}$, provided we use appropriate branches of the logarithm.

$U$ is a simply connected domain, and $\Gamma$ is a holomorphic function without zeros on $U$. Hence there are branches of $\log \Gamma$ on $U$ (note however that a branch $\lambda$ of $\log f$, where $f$ is a zero-free holomorphic function on a simply connected domain can usually not be obtained by composing some branch of the logarithm with $f$; for points $z \neq w$ with $f(z) = f(w)$, one often has $\lambda(z) \neq \lambda(w)$). For every branch $h$ of $\log \Gamma$ on $U$, the function

$$I_h(z) = \int_0^1 h(x+z)\,dx$$

is holomorphic on $U$. Since $U$ is connected, two different branches of $\log \Gamma$ differ by a multiple of $2\pi i$, and hence the integrals differ by the same multiple of $2\pi i$. If $h$ is the branch of $\log \Gamma$ that takes real values on the positive half of the real axis, then Raabe's result says

$$I_h(z) = \log \sqrt{2\pi} + z\log z - z \tag{$\ast$}$$

for $z > 0$ if $\log$ denotes the principal branch of the logarithm on the right hand side. Since the domain of the principal branch of the logarithm is $U$, the identity theorem says that $(\ast)$ holds on all of $U$.

Daniel Fischer
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