COSECANT INFINITE SERIES REPRESENTATION VIA THE DIGAMMA FUNCTION

In this previous post we have proved the series representation for the cosecant and hyperbolic cosecant based on a Fourier series approach. Today we will prove it again but this time relying on the Digamma function.

$\pi \mathrm{csc}( x)=\frac{1}{x}+2x\sum_{k=1}^{\infty}\frac{(-1)^k}{x^2-k^2}$ (1)

$\pi \mathrm{csch}(\pi x)=\frac{ 1}{x}+2x\sum_{k=1}^{\infty}\frac{(-1)^{k} }{x^2+k^2}$ (2)

As an application of (2) we will compute also the following integrals, a variation of Binet´s second representation of the Digamma function

$2\int_0^{\infty} \frac{x}{(x^2+s^2) (e^{2\pi x}+1)} dx=\psi\left( s+\frac{1}{2}\right)-\ln s$ (3)

$4\int_0^{\infty} \frac{sx}{(x^2+s^2)^2 (e^{2\pi x}+1)} dx=\frac{1}{s}-\psi^{\prime}\left( s+\frac{1}{2}\right)$ (4)

$\begin{aligned} &S= 2x\sum_{k=1}^{\infty}\frac{(-1)^k}{k^2-x^2}\\ &=2x\sum_{k=1}^{\infty}\frac{(-1)^k}{2x}\left(\frac{1}{k-x}-\frac{1}{k+x} \right)\\ &=\sum_{k=1}^{\infty}(-1)^k\left(\int_0^1t^{k-x-1}dt-\int_0^1t^{k+x-1}dt\right)\\ &=-\int_0^1 t^{-x}\left(\sum_{k=1}^{\infty}(-1)^{k-1}t^{k-1}\right)dt+\int_0^1 t^x\sum_{k=1}^{\infty}(-1)^{k-1}t^{k-1}\right)dt\\ &=\int_0^1\frac{t^x}{1+t}dt-\int_0^1\frac{t^{-x}}{1+t}dt\\ &=\beta(x+1)-\beta(1-x)\\ &=\frac{1}{x}-\frac{\pi}{\sin\pi x} \end{aligned}$

$\boxed{\pi \mathrm{csc}( x)=\frac{1}{x}+2x\sum_{k=1}^{\infty}\frac{(-1)^k}{x^2-k^2}}$ (5)

Next, we let $x \longmapsto ix$ in (5)

$\frac{\pi}{\sin(i\pi x)}=\frac{ 1}{ix}+2ix\sum_{k=1}^{\infty}\frac{(-1)^{k} }{(ix)^2-k^2}$

$-\frac{2\pi i}{e^{\pi x}-e^{-\pi x}}=-\frac{ i}{x}-2ix\sum_{k=1}^{\infty}\frac{(-1)^{k} }{x^2+k^2}$

$\frac{\pi }{\sinh(\pi x)}=\frac{ 1}{x}+2x\sum_{k=1}^{\infty}\frac{(-1)^{k} }{x^2+k^2}$

And we obtain:

$\boxed{\pi \mathrm{csch}(\pi x)=\frac{ 1}{x}+2x\sum_{k=1}^{\infty}\frac{(-1)^{k} }{x^2+k^2}}$ (6)

To compute (3), We first recall the Laplace transform of the sine function

$\int_0^{\infty}e^{-st}\sin(xt)dt=\frac{x}{x^2+s^2}$

The we can write our integral as

$\int_0^{\infty} \frac{x}{(x^2+s^2) (e^{2\pi x}+1)} dx=\int_0^{\infty} \frac{1}{(e^{2\pi x}+1)}\int_0^{\infty}e^{-st}\sin(xt)dtdx$

Switching the order of integration

$=\int_0^{\infty}e^{-st}\int_0^{\infty} \frac{\sin(xt)}{e^{2\pi x}+1}dx\,\,dt$ (7)

Lets now focus only in the inner integral of (7)

$\begin{aligned} &J=\int_0^{\infty} \frac{\sin(xt)}{e^{2\pi x}+1}dx\\ &=\int_0^{\infty} \frac{e^{-2\pi x}\sin(xt)}{1+e^{-2\pi x}}dx\\ &=\int_0^{\infty}\sin(xt)\sum_{k=0}^{\infty}(-1)^k e^{-2\pi x(k+1)}dx\\ &=\int_0^{\infty}\sin(xt)\sum_{k=1}^{\infty}(-1)^{k-1} e^{-2\pi xk}dx\\ &=-\sum_{k=1}^{\infty}(-1)^k\int_0^{\infty}\sin(xt) e^{-2\pi xk}dx\\ &=-\sum_{k=1}^{\infty}(-1)^k\frac{t}{t^2+(2 \pi k)^2}\\ &=-\frac{2t}{2}\sum_{k=1}^{\infty}\frac{(-1)^k}{t^2+(2 \pi k)^2}\\ &=-\frac{2t}{2(2 \pi)^2}\sum_{k=1}^{\infty}\frac{(-1)^k}{\left(\frac{t}{2\pi}\right)^2+k^2}\\ &=-\frac{1}{4 \pi}\cdot\frac{2t}{2 \pi}\sum_{k=1}^{\infty}\frac{(-1)^k}{\left(\frac{t}{2\pi}\right)^2+k^2}\\ &=-\frac{1}{4 \pi}\left(\pi\mathrm{csch}\left(\pi\frac{t}{2 \pi} \right)-\frac{2 \pi}{t} \right) \end{aligned}$

$\boxed{\int_0^{\infty} \frac{\sin(xt)}{e^{2\pi x}+1}dx=-\frac{1}{4 }\left(\mathrm{csch}\left(\frac{t}{2 } \right)-\frac{2}{t} \right)}$ (8)

Plugging (8) in (7) we get

$\int_0^{\infty} \frac{x}{(x^2+s^2) (e^{2\pi x}+1)} dx=-\frac{1}{4 }\int_0^{\infty}e^{-st}\left(\mathrm{csch}\left(\frac{t}{2 } \right)-\frac{2}{t} \right)dt$

$=\frac{1}{4 }\int_0^{\infty}e^{-st}\left(\frac{2}{t}-\mathrm{csch}\left(\frac{t}{2 } \right) \right)dt$

Let st=w then

$\begin{aligned} &=\frac{1}{4s }\int_0^{\infty}e^{-w}\left(\frac{2s}{w}-\mathrm{csch}\left(\frac{w}{2s } \right) \right)dw\\ &=\int_0^{\infty}\left(\frac{e^{-w}}{2w}-\frac{e^{-w}\mathrm{csch}\left(\frac{w}{2s } \right)}{4s} \right)\,dw\\ &=\int_0^{\infty}\left(\frac{e^{-w}}{2w}-\frac{e^{-w}}{4s}\cdot\frac{2}{e^{\frac{w}{2s}}-e^{-\frac{w}{2s}}} \right)\,dw\\ &=\int_0^{\infty}\left(\frac{e^{-w}}{2w}-\frac{e^{-w}}{2s}\cdot\frac{e^{-\frac{w}{2s}}}{1-e^{-\frac{w}{s}}} \right)\,dw \end{aligned}$

let $e^{-w}=u \,\, \Rightarrow \,\, w=-\ln u \,\, \Rightarrow \,\, dw=-\frac{du}{u}$ , then

$=\frac{1}{2}\int_1^{0}\left(\frac{u}{-\ln u}-\frac{u}{s}\cdot\frac{u^{\frac{1}{2s}}}{1-u^{\frac{1}{s}}} \right)\,\frac{(-du)}{u}$

$=-\frac{1}{2}\int_0^{1}\left(\frac{1}{\ln u}+\frac{1}{s}\cdot\frac{u^{\frac{1}{2s}}}{\left(1-u^{\frac{1}{s}}\right)} \right)\,du$

Let $u=y^s \,\, \Rightarrow \,\, du= sy^{s-1}dy$

$=-\frac{1}{2}\int_0^{1}\left(\frac{1}{\ln y^s}+\frac{1}{s}\cdot\frac{y^{\frac{1}{2}}}{1-y} \right)\,(sy^{s-1}dy)$

$=-\frac{1}{2}\int_0^{1}\left(\frac{y^{s-1}}{s \ln y}+\frac{1}{s}\cdot\frac{y^{s-\frac{1}{2}}}{1-y} \right)\,sdy$

$=-\frac{1}{2}\int_0^{1}\left(\frac{y^{s-1}}{ \ln y}+\frac{y^{s-\frac{1}{2}}}{1-y} \right)\,dy$ (9)

Recall the following integral representation of the Digamma function

$\int_0^1\left(\frac{x^{z-1}}{\ln x}+\frac{x^{w-1}}{1-x} \right) dx=\ln z-\psi(w)$

where z=s and $w=s+\frac{1}{2}$ , then we obtain

$\int_0^{\infty} \frac{x}{(x^2+s^2) (e^{2\pi x}+1)} dx=\frac{\psi\left( s+\frac{1}{2}\right)}{2}-\frac{\ln s}{2}$

$\boxed{2\int_0^{\infty} \frac{x}{(x^2+s^2) (e^{2\pi x}+1)} dx=\psi\left( s+\frac{1}{2}\right)-\ln s}$ (10)

Differentiating (10) w.r. to we obtain

$-4\int_0^{\infty} \frac{sx}{(x^2+s^2)^2 (e^{2\pi x}+1)} dx=\psi^{\prime}\left( s+\frac{1}{2}\right)-\frac{1}{s}$

$\boxed{4\int_0^{\infty} \frac{sx}{(x^2+s^2)^2 (e^{2\pi x}+1)} dx=\frac{1}{s}-\psi^{\prime}\left( s+\frac{1}{2}\right)}$ (11)

Appendix I

Recall the properties of the digamma function:

$\psi(x+1)=\frac{1}{x}+\psi(x)$ (A.1)

$\psi\left(x\right)-\psi\left(1-x\right)=-\pi \cot \pi x$ (A.2)

$\psi\left(x+\frac{1}{2}\right)-\psi\left(\frac{1}{2}-x\right)=\pi \tan \pi x$ (A.3)

Recall the result

$\int_0^1\frac{t^{m}}{1+t^n}dt=\frac{1}{2n}\left(\psi\left( \frac{m+n+1}{2n}\right)-\psi\left(\frac{m+1}{2n} \right)\right)$ (A.4)

Setting $n=1\,\,\text{and}\,\,m=x-1$ in (A.4) we obtain

$\int_0^1\frac{t^{x-1}}{1+t}dt=\frac{1}{2}\left(\psi\left( \frac{x+1}{2}\right)-\psi\left(\frac{x}{2} \right)\right)$ (A.5)

We now define the function $\beta(x)$ by:

$\begin{aligned} &\beta(x)=\int_0^1\frac{t^{x-1}}{1+t}dt\\ &\beta(x)=\frac{1}{2}\left(\psi\left( \frac{x+1}{2}\right)-\psi\left(\frac{x}{2} \right)\right) \end{aligned}$ (A.6)

By this definition we can proof some useful functional relation, namely:

$\beta(x+1)=\frac{1}{x}-\beta(x)$ (A.7)

$\beta(1-x)=\frac{\pi}{\sin \pi x}-\beta(x)$ (A.8)

$\beta(x+1)-\beta(1-x)=\frac{1}{x}-\frac{\pi}{\sin \pi x}$ (A.9)

By the definition (A.6) of $\beta(x)$ we have that

$\begin{aligned} \beta(x+1)&=\frac{1}{2}\left(\psi\left( \frac{x+2}{2}\right)-\psi\left(\frac{x+1}{2} \right)\right)\\ &=\frac{1}{2}\left(\psi\left( \frac{x}{2}+1\right)-\psi\left(\frac{x+1}{2} \right)\right)\\ &=\frac{1}{2}\left(\frac{2}{x}+\psi\left( \frac{x}{2}\right)-\psi\left(\frac{x+1}{2} \right)\right)\\ &=\frac{1}{x}-\beta(x)\\ \end{aligned}$

And (A.7) is proved. Now for (A.8)

$\begin{aligned} \beta(x)+\beta(1-x)&=\frac{1}{2}\left(\psi\left( \frac{1}{2}+\frac{x}{2}\right)-\psi\left(\frac{x}{2} \right)+\psi\left(1- \frac{x}{2}\right)-\psi\left(\frac{1}{2}-\frac{x}{2} \right)\right)\\ &=\frac{1}{2}\left(\psi\left( \frac{1}{2}+\frac{x}{2}\right)-\psi\left(\frac{1}{2}-\frac{x}{2} \right)\right)+\frac{1}{2}\left(\psi\left(1- \frac{x}{2}\right)-\psi\left(\frac{x}{2} \right)\right)\\ \end{aligned}$

From (A.2) and (A.3) we obtain

$\begin{aligned} &=\frac{\pi}{2} \tan\left( \frac{\pi x}{2}\right)+\frac{\pi}{2}\cot\left(\frac{\pi x}{2}\right)\\ &=\frac{\pi}{2} \left(\frac{1-\cos\left( \pi x\right)}{\sin(\pi x)}+\frac{1+\cos\left( \pi x\right)}{\sin(\pi x)}\right)\\ &=\frac{\pi}{\sin \pi x} \end{aligned}$

And (A.8) is proved. To estabilish (A.9) we just need to subtract (A.8) from (A.7).

Appendix II

$\tan \left(\frac{x}{2}\right)=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{2 \sin \frac{x}{2} \sin \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}=\frac{2 \sin ^{2} \frac{x}{2}}{\sin x}=\frac{1-\cos x}{\sin x}$

$\boxed{\tan \left(\frac{x}{2}\right)=\frac{1-\cos x}{\sin x}}$ (A.10)

$\cot \left(\frac{x}{2}\right)=\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}=\frac{2\cos \frac{x}{2}\cos \frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}}=\frac{2\cos^2 \frac{x}{2}}{\sin x}=\frac{1+\cos x}{\sin x}$

$\boxed{\cot \left(\frac{x}{2}\right)=\frac{1+\cos x}{\sin x}}$ (A.11)

Reference

The integrals in Gradshteyn and Ryzhik. Part 11: The incomplete beta function Khristo Boyadzhiev, Luis Medina, Victor Moll, arXiv:0808.2751v1

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