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I cite this paper by Iaroslav Blagouchine.

As mentioned in the comments under this post, Blagouchine erroneously claims that the Log-Gamma function has no branch cuts (fortunately none of his contours make this error an issue). It does have branch cuts, but allegedly has a much simpler branch structure (one cut along the negative reals) than the logarithm of the gamma function, which in general is a different function with a complicated branch structure. Blagouchine's method relies on the following, where $\Lambda(z)=\mathrm{Log}\Gamma(z)=-\gamma z-\ln z+\sum_{n=1}^\infty\frac{z}{n}-\mathrm{Log}\left(1+\frac{z}{n}\right)$:

$$\Lambda(z+1)-\Lambda(z)=\mathrm{Log}(z)$$

And the niceness of the $\Lambda$ function allows him (apparently) to take simpler contours to yield his impressive results. He emphasises how just using the complex logarithm as it stands is difficult:

"But the evaluation of more complicated logarithmic integrals may become a difficult task, because it may be very hard (or even impossible) to find an appropriate line integral... Thus, in order to evaluate such kinds of integrals, one may be led to consider unusual integration paths and more sophisticated forms of the integrand."

Certainly his results are impressive, and as they do feature the gamma function it stands to reason that the use of $\Lambda$ has indeed helped us in the integration process.

I am uncertain about some things here. The complex logarithm is implicitly used in the log gamma function, and they share the same branch cut along the negative real axis. If some $\zeta$ on the rectangular or semi circular contour is a problematic pole for $\mathrm{Log}$, then so it is also for $\Lambda$. The contours used by Blagouchine do not, to my untrained eye, cause problems for the plain logarithm.

Let me present an (abbreviated) example:

Let $L_\beta$ be the rectangular contour, counter clockwise, with vertices $\{-\beta,\beta,\beta+2\pi i,-\beta+2\pi i\}$, $R$ a real valued rational function (with some growth condition), and $\alpha\gt0$ a real constant. Then one can safely integrate: $$\lim_{\beta\to\infty}\oint_{L_\beta}R(e^z)\Lambda\left(\frac{z}{2\pi i}+\alpha\right)\,\mathrm{d}z$$And find an equation, involving the residue theorem, with: $$\int_{-\infty}^\infty R(e^x)\left[\Lambda\left(\frac{x}{2\pi i}+\alpha\right)-\Lambda\left(\frac{x}{2\pi i}+\alpha+1\right)\right]\,\mathrm{d}z$$And obtain an integral with the usual complex logarithm.

My question is, why is this more powerful than performing:

$$\lim_{\beta\to\infty}\oint_{\gamma_\beta}R(e^z)\mathrm{Log}(g(z))\,\mathrm{d}z$$

Where $g(z)$ is some suitably simple expression in $z$ (e.g. Blagouchine uses $\frac{z}{2\pi i}+\alpha$ for his integrals). I would not be able to obtain:

$$\frac{1}{2}\int_0^\infty\frac{\ln(x)}{\cosh(x)}\,\mathrm{d}x$$

Anyway, since I am not properly trained in this business, but Blagouchine claims a "naive" approach using the usual logarithm would fail here (whereas his log gamma contour integral computes the above surprisingly "easily"). I wouldn't know myself, but what stops one from finding a nice contour over $\mathrm{Log}$ here?

To clarify:

I understand his method. I just don't understand the merits of his method over more basic ones, since I do not understand branch cuts well enough. I am under the impression that the logarithm and the log gamma function share the same cut along the negative reals, and as such any integral with one should be just as easy with the other. In the above integral, he writes that the integrand has infinitely many poles (on a complex semi circular contour). Clearly I am missing something! Yes, the end results involve log-gamma, which would be a mystery using any other method, but on principle what's the problem with $\mathrm{Log}$?

FShrike
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  • I'm not sure if this answers your question, but you can show that $\int_{-\infty}^{\infty} \frac{\ln(a^{2}+x^{2})}{1+x^{2}} , \mathrm dx = 2 \pi \ln(a+1), a \ge 0, $ by integrating the function $\frac{\ln(z+ia)}{1+z^{2}}$ around a large semicircular contour in the upper half-plane. I posted this approach a long time ago here. – Random Variable Dec 09 '21 at 03:08
  • But if you try the same thing with $\frac{\ln(z+ia)}{\cosh z}$, you'll run into a problem because the integral doesn't vanish along the semicircular arc as the radius of the semicircle goes to infinity, even if the radius is increased in a way that avoids the poles on the imaginary axis. – Random Variable Dec 09 '21 at 03:08
  • @RandomVariable Thank you for having helped me with this research on Blagouchine so much over the different posts. LogGamma grows faster at infinity than log (and so I believe would fail also in the integral you talk about), so I find it curious that LogGamma is more convenient and easy to find contours for. I think Log would have also worked with his rectangular contour, given the growth control on the rational function $R$. The problem must lie not in the growth but more on the branching, I feel, but I don’t understand that point well enough. – FShrike Dec 09 '21 at 07:18
  • If you integrate the function $\frac{\ln(z+ia)}{\cosh z }$ around a rectangular contour of height $2 \pi i $, you'll end up showing that $$\frac{1}{2} \int_{-\infty}^{\infty} \frac{\ln(x^{2}+a^{2})}{\cosh x} , \mathrm dx - \frac{1}{2}\int_{-\infty}^{\infty} \frac{\ln(x^{2}+(2 \pi +a)^{2})}{\cosh x} , \mathrm dx= 2 \pi \ln \left(\frac{\pi +2a}{3\pi + 2a} \right). $$ – Random Variable Dec 09 '21 at 17:03
  • @RandomVariable So it is not merely the branching properties, but also the fact that one can exploit the functional equation $\Gamma(z+1)=z\cdot\Gamma(z)$ to simplify such expressions, but the normal logarithm of course does not have that property. That was a good example – FShrike Dec 09 '21 at 18:26

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