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According to Wolfram website http://functions.wolfram.com/ElementaryFunctions/Csc/introductions/Csc/05/,

There exists a "well-known" integral representation for the cosecant function, i.e. $$\csc(z):=\frac{1}{\sin(z)} = \frac{1}{\pi}\int_0^{\infty} \frac{1}{t^2+t}t^{z/\pi}\,\mathrm d t$$ for complex $z$ such that $0< \Re(z)<\pi$.

I am searching for a demonstration of this formula. I can't find any reference book.

Ideally, I would like to find a similar integral representation on the whole complex plane except $\pi\mathbb{Z}$, or at least on the half plane with positive real part. It may be an indefinite integral.

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1 Answer

up vote 3 down vote accepted

$$\frac{1}{\pi}\int_0^{\infty}\frac{t^{\frac{z}{\pi}}}{t^2+t}dt=\frac{1}{\pi}\int_0^{\infty}\frac{t^{\frac{z}{\pi}-1}}{t+1}dt$$

Let $w=\dfrac{1}{t+1}:$

$$\begin{align*}\frac{1}{\pi}\int_0^{\infty}\frac{t^{\frac{z}{\pi}-1}}{t+1}dt&=\frac{1}{\pi}\int_0^{1}\frac{1}{w}\left(\frac{1}{w}-1\right)^{\frac{z}{\pi}-1}dw\\&=\frac{1}{\pi}\int_0^{1}w^{-\frac{z}{\pi}}\left(1-w\right)^{\frac{z}{\pi}-1}dw\\&=\frac{1}{\pi}\text{B}\left(1-\frac{z}{\pi},\frac{z}{\pi}\right)\end{align*}$$

Where $\text{B}(x,y)$ is the Beta function, defined for $\Re (x),\Re(y)>0$ (hence why we must have $0<\Re (z)<\pi$)

By the powerful Beta-Gamma relation:

$$\frac{1}{\pi}\text{B}\left(1-\frac{z}{\pi},\frac{z}{\pi}\right)=\frac{1}{\pi}\Gamma\left(1-\frac{z}{\pi}\right)\Gamma\left(\frac{z}{\pi}\right)$$

Now, thanks to Euler's reflection formula:

$$\begin{align*}\frac{1}{\pi}\Gamma\left(1-\frac{z}{\pi}\right)\Gamma\left(\frac{z}{\pi}\right)&=\frac{1}{\pi}\cdot \pi\csc\left(\pi\cdot\frac{z}{\pi}\right)\\&=\csc (z)\end{align*}$$

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Thank you very much for your answer. Would you know by chance how to build an integral representation $\csc = \int_0^{\infty}...$ valid on the half plane $\lbrace z\in \mathbb{C} ; \Re(z)>0\rbrace\setminus \pi\mathbb{Z}$ ? –  saposcat Feb 4 '13 at 21:21
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