# Tagged Questions

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156 views

### Computing Reciprocal Gamma

Reciprocal Gamma $1/\Gamma(z)$ is an entire function and so it has a convergent Taylor series expansion which was given in its wikipedia article.
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236 views

### complex contour integral calculation after Möbius transformation

Good day to everyone.
In my scientific research I've got stuck with a contour integration problem.
I would like to evaluate the following integral:
$$I=\int_0^{\infty } \frac{e^{\frac{\alpha -\mathrm ...

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183 views

### An integral with Gamma functions (Part 2)

I was wondering if there is a generalization of the integral discussed here to a case like,
\begin{equation}\int \frac{d^dq}{q^{\nu_1}\vert \vec{q} \pm \vec{k}_1\vert ^{\nu_2}\vert \vec{q} \pm ...

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125 views

### an infinite series expansion in terms of the polylogarithm function

we have the complex valued function :
$$f(z)=\sum_{n=0}^{\infty}a_{n}Li_{-n}(z)$$
we wish to recover the coefficients $a_{n}$ . the only thing i though would work is to try and come up with a function ...

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**1**answer

436 views

### The zeros of the digamma function

I wonder what work have been done on the zeros of the digamma function and on the values of the gamma function at such points (on the negative real axis). Any help please :)

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292 views

### Why is Mellin-inverse of Gamma periodic?

Specific Case
The periodicity is obvious from computation:
$$\cal{M}^{-1}\{\Gamma\}(x) := \frac{1}{2\pi i}\int_{c}\Gamma(s)x^{-s}d s=e^{-x}$$
However, is there a way to see directly from the integral ...

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**3**answers

882 views

### Conformal Mappings for hyperbolic polygon

I am searching for a conformal mapping from the upper halfplane onto a hyperbolic polygon, i.e. the sides of the polygon have to be geodesics.
The classical Schwarz Christoffel theorem does the job ...

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672 views

### Bessel function in polar coordinates

I want to write the Bessel function of the first kind in polar coordinates
$J_\alpha(z)=|J_\alpha(z)|e^{i\varphi_\alpha(z)}$
Is anything known about $\varphi_\alpha(z)$?
In particular, I'm ...