Real Analysis – Closed-Form for Infinite Sum Involving Arccoth

analysiscalculusreal-analysissequences-and-seriesspecial functions

I'm looking for a closed-form for the following sum:

$$\sum_{n=1}^{\infty} \left(m n \, \text{arccoth} \, (m n) – 1\right)$$ for $|m|>1.$

In a previous question of mine, the following similar sum was determined:

$$\sum_{n=1}^{\infty} (-1)^n \left(m n \, \text{arccoth} \, (m n) – 1\right)=\frac{1}{2}+\frac{1}{2} \log \left|\cot\left(\frac{\pi}{2m}\right)\right|+\frac{m}{2\pi} \left(\frac{1}{2}\text{Cl}_2\left(\frac{2\pi}{m}\right)-2\text{Cl}_2\left(\frac{\pi}{m}\right)\right)$$
for $|m|>1,$ where $\text{Cl}_2$ is the Clausen function of order 2.

I determined the following closed-forms for example as mentioned in that question:

$$\sum_{n=1}^{\infty} \left( 4n \, \text{arccoth} \, (4n)-1\right) = \frac{1}{2} – \frac{G}{\pi}- \frac{1}{4} \ln (2)$$
$$\sum_{n=1}^{\infty} \left(6n \, \text{arccoth} \, (6n) – 1\right) = \frac{1}{2} – \frac{3}{2\pi} \, \text{Cl}_2 \left( \frac{\pi}{3}\right)$$
$$\sum_{n=1}^{\infty} \left( 8n \, \text{arccoth} \, (8n) – 1\right) = \frac{1}{2} – \frac{2}{\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{4}\right) – \frac{1}{4} \ln (2-\sqrt{2})$$

I suspect a similar method to that used by @skbmoore involving the Barnes G function might be applicable. Any help would be much appreciated.

Here is my attempt:

$$S(m) := \sum_{n=1}^{\infty} \left(m n \, \text{arccoth} \, (m n) – 1\right) = \frac{m}{2} \log \left( \prod_{n=1}^{\infty} \frac{1}{e} \left(\frac{1+1/(m n)}{1-1/(m n)}\right)^{n}\right)\\ = \frac{1}{2} + \frac{m}{2} \zeta' \left(-1,1-\frac{1}{m}\right)-\frac{m}{2}\zeta' \left(-1,1+\frac{1}{m}\right)+\frac{1}{2} \zeta' \left(0,1-\frac{1}{m}\right)+\frac{1}{2} \zeta' \left(0,1+\frac{1}{m}\right) \\ =\frac{1}{2} + \frac{m}{2} \left(\zeta' \left(-1,1-\frac{1}{m}\right)-\zeta' \left(-1,1+\frac{1}{m}\right)\right) \\+ \frac{1}{2} \ln\left( \Gamma \left(1-\frac{1}{m}\right)\Gamma \left(1+\frac{1}{m}\right)\right) – \frac{1}{2}\ln (2\pi) \\=\frac{1}{2} + \frac{m}{2} \left(\zeta' \left(-1,1-\frac{1}{m}\right)-\zeta' \left(-1,1+\frac{1}{m}\right)\right) + \frac{1}{2} \ln\left( \frac{\pi}{m} \csc \left(\frac{\pi}{m}\right)\right) – \frac{1}{2}\ln (2\pi)$$

However, I would like to write the solution in terms of the Clausen function if possible, like in the alternating sum, but I cannot see how to do that.

Best Answer

EDIT: There was a question asked a long time ago about the related infinite series $$\sum_{n=1}^{\infty} \left( n \operatorname{arccot}(n)-1 \right). $$

The approach used in the accepted answer is similar to the approach I used for this question.

For $|x| >1 $, the inverse hyperbolic cotangent function has the Laurent series representation $$\operatorname{arcoth}(x) = \frac{1}{2} \log \left(\frac{1+ \frac{1}{x}}{1- \frac{1}{x}} \right) = \frac{1}{2} \log \left[ \left(1+ \frac{1}{x} \right) - \log\left(1- \frac{1}{x} \right) \right]= \sum_{n=0}^{\infty}\frac{1}{(2n+1)x^{2n+1}}. $$ And for $0 < |x| < 1$, $\cot (\pi x)$ has the Laurent series representation $$\cot(\pi x) = \frac{1}{\pi x} - \frac{2}{\pi x} \sum_{k=1}^{\infty}\zeta(2k) x^{2k}. $$

Therefore, for $m>1$, we have $$ \begin{align} \sum_{n=1}^{\infty} \left(mn \operatorname{arcoth} (mn)-1 \right) &= \sum_{n=1}^{\infty} \left(mn \sum_{k=0}^{\infty} \frac{1}{(2k+1) (mn)^{2k+1}} -1\right) \\ &= \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{(2k+1)(mn)^{2k}} \\ &= \sum_{k=1}^{\infty}\frac{1}{(2k+1)m^{2k}} \sum_{n=1}^{\infty} \frac{1}{n^{2k}} \\ &= \sum_{k=1}^{\infty}\frac{\zeta(2k)}{(2k+1)m^{2k}} \\ &= \frac{m}{2} \int_{0}^{1/m} \, \mathrm dx - \frac{m \pi}{2} \int_{0}^{1/m} x \cot (\pi x) \, \mathrm dx \\ &= \frac{1}{2} - \frac{m x \log(\sin \pi x)}{2}\Bigg|_{0}^{1/m} + \frac{m}{2} \int_{0}^{1/m} \log(\sin \pi x) \, \mathrm dx \\ &= \frac{1}{2} - \frac{\log \left(\sin \frac{\pi}{m} \right)}{2} + \frac{m}{4 \pi} \int_{0}^{2 \pi /m} \log \left(\sin \frac{ u}{2}\right) \, \mathrm du \\ &= \frac{1}{2} - \frac{\log \left(\sin \frac{\pi}{m} \right)}{2} + \frac{m}{4 \pi}\int_{0}^{2 \pi /m} \log \left(2 \sin \frac{u}{2}\right) \, \mathrm du - \frac{\log 2}{2} \\ &= \frac{1}{2} - \frac{\log \left(\sin \frac{\pi}{m} \right)}{2} - \frac{m \operatorname{Cl}_{2} \left(\frac{2 \pi}{m} \right)}{4 \pi } - \frac{\log 2}{2}. \end{align} $$

Related Solutions

General form for $\sum_{n=1}^{\infty} (-1)^n \left(m n \, \text{arccoth} \, (m n) – 1\right)$

Let

$$S\left( m \right)=\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n}}\left( mn\operatorname{arcoth}\left( mn \right)-1 \right)},\,\,\left| m \right|>1$$

And note the logarithmic representation of the inverse function

$$\operatorname{arcoth}\left( z \right)=\frac{1}{2}\log \left( \frac{z+1}{z-1} \right)$$

The series can therefore be written as

$$S\left( m \right)=\sum\limits_{n=1}^{\infty }{\left( \log \left( \frac{1}{{{e}^{{{\left( -1 \right)}^{n}}}}}{{\left( \frac{nm+1}{nm-1} \right)}^{\frac{1}{2}{{\left( -1 \right)}^{n}}nm}} \right) \right)}=\log \left( \prod\limits_{n=1}^{\infty }{\frac{1}{{{e}^{{{\left( -1 \right)}^{n}}}}}{{\left( \frac{nm+{{\left( -1 \right)}^{n}}}{nm-{{\left( -1 \right)}^{n}}} \right)}^{\frac{1}{2}nm}}} \right)$$

Breaking the product into two convergent products consider then

$${{e}^{S}}=\prod\limits_{n=1}^{\infty }{\frac{1}{e}{{\left( \frac{2nm+1}{2nm-1} \right)}^{nm}}}\prod\limits_{n=1}^{\infty }{e{{\left( \frac{\left( 2n-1 \right)m-1}{\left( 2n-1 \right)m+1} \right)}^{\frac{1}{2}\left( 2n-1 \right)m}}}={{p}_{1}}\left( m \right){{p}_{2}}\left( m \right)$$

The first of these is

$${{p}_{1}}\left( m \right)=\prod\limits_{n=1}^{\infty }{\frac{1}{e}\frac{{{\left( 1+\frac{1}{2nm} \right)}^{nm}}}{{{\left( 1-\frac{1}{2nm} \right)}^{nm}}}}$$

However for the moment note that to ensure the convergence of

$${{p}_{a}}\left( m \right)=\prod\limits_{n=1}^{\infty }{{{\left( 1+\frac{1}{2nm} \right)}^{nm}}a}$$

we must consider the convergence of

$$\log {{p}_{a}}\left( m \right)=\sum\limits_{n=1}^{\infty }{nm\log \left( 1+\frac{1}{2nm} \right)+\log \left( a \right)}$$

For large n the summand behaves

$$nm\log \left( 1+\frac{1}{2nm} \right)+\log \left( a \right)\simeq \frac{1}{2}+\log \left( a \right)-\frac{1}{8mn}+O\left( {{n}^{-2}} \right)$$

so let $a={{e}^{\frac{1}{8mn}-\frac{1}{2}}}$. The product can now be written in the following manner

$${{p}_{1}}\left( m \right)=\prod\limits_{n=1}^{\infty }{\frac{{{\left( 1+\frac{1}{2nm} \right)}^{nm}}{{e}^{-\frac{1}{2}+\frac{1}{8mn}}}}{{{\left( 1-\frac{1}{2nm} \right)}^{nm}}{{e}^{\frac{1}{2}+\frac{1}{8mn}}}}}=\frac{\prod\limits_{n=1}^{\infty }{{{\left( 1+\frac{1}{2nm} \right)}^{nm}}{{e}^{-\frac{1}{2}+\frac{1}{8mn}}}}}{\prod\limits_{n=1}^{\infty }{{{\left( 1-\frac{1}{2nm} \right)}^{nm}}{{e}^{\frac{1}{2}+\frac{1}{8mn}}}}}$$

Therefore consider now

$${{p}_{1\pm }}=\prod\limits_{n=1}^{\infty }{{{\left( 1\pm \frac{1}{2nm} \right)}^{nm}}{{e}^{\mp \frac{1}{2}+\frac{1}{8mn}}}}$$

or alternatively

$$\log {{p}_{1\pm }}=\sum\limits_{n=1}^{\infty }{nm\log \left( 1\pm \frac{1}{2nm} \right)}\mp \frac{1}{2}+\frac{1}{8mn}$$

Using the series representation for the logarithm

$$\log {{p}_{1+}}=\sum\limits_{n=1}^{\infty }{nm\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k+3}}}{\left( k+2 \right){{\left( 2nm \right)}^{k+2}}}}}=-\frac{1}{2}\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}\zeta \left( k+1 \right)}{\left( k+2 \right){{\left( 2m \right)}^{k+1}}}}$$

$$\log {{p}_{1-}}=-\sum\limits_{n=1}^{\infty }{nm\sum\limits_{k=3}^{\infty }{\frac{1}{k{{\left( 2nm \right)}^{k}}}}}=-\frac{1}{2}\sum\limits_{k=1}^{\infty }{\frac{\zeta \left( k+1 \right)}{\left( k+2 \right){{\left( 2m \right)}^{k+1}}}}$$

From Srivastava [pg. 257, (63)] the following series takes the form

$$\begin{align}\sum\limits_{k=2}^{\infty }{\frac{\zeta \left( k,a \right)}{k+n}{{x}^{k+n}}}&=\sum\limits_{k=0}^{n}{\left( \begin{matrix} n \\ k \\ \end{matrix} \right)\zeta '\left( -k,a-x \right){{x}^{n-k}}}-\sum\limits_{k=0}^{n-1}{\frac{\zeta \left( -k,a \right)}{n-k}{{x}^{n-k}}}\\&+\left\{ \psi \left( a \right)-{{H}_{n}} \right\}\frac{{{x}^{n+1}}}{n+1}-\zeta '\left( -n,a \right),\ \ \left| x \right|<\left| a \right|,\,\,n=0,1,2...\end{align}$$

where $\zeta '\left( s,a \right)=\frac{\partial }{\partial s}\zeta \left( s,a \right)$ . Hence

$$\sum\limits_{k=1}^{\infty }{\frac{\zeta \left( k+1 \right)}{k+2}{{x}^{k+1}}}=\frac{1}{x}\left\{ \zeta '\left( 0,1-x \right)x+\zeta '\left( -1,1-x \right)+\frac{1}{2}\left( x-{{x}^{2}} \right)-\tfrac{1}{2}\gamma {{x}^{2}}-\zeta '\left( -1 \right) \right\}$$

We have then

$$\log {{p}_{-}}\left( m \right)=-\frac{1}{2}\zeta '\left( 0,1-\frac{1}{2m} \right)-m\zeta '\left( -1,1-\frac{1}{2m} \right)-\frac{1}{4}\left( 1-\frac{\gamma +1}{2m} \right)+m\zeta '\left( -1 \right)$$

$$\log {{p}_{+}}\left( m \right)=\frac{1}{2}\zeta '\left( 0,1+\frac{1}{2m} \right)-m\zeta '\left( -1,1+\frac{1}{2m} \right)+\frac{1}{4}\left( 1+\frac{1+\gamma }{2m} \right)+m\zeta '\left( -1 \right)$$

Continuing, recall $${{p}_{2}}\left( m \right)=\prod\limits_{n=1}^{\infty }{e\frac{{{\left( 1-\frac{1}{2m\left( n-\tfrac{1}{2} \right)} \right)}^{m\left( n-\tfrac{1}{2} \right)}}}{{{\left( 1+\frac{1}{2m\left( n-\tfrac{1}{2} \right)} \right)}^{m\left( n-\tfrac{1}{2} \right)}}}}$$

and so for a moment consider the convergence of the product

$${{q}_{a}}=\prod\limits_{n=1}^{\infty }{{{\left( 1-\frac{1}{2m\left( n-\tfrac{1}{2} \right)} \right)}^{m\left( n-\tfrac{1}{2} \right)}}a}\Rightarrow a={{e}^{\frac{1}{2}+\frac{1}{8m\left( n-\tfrac{1}{2} \right)}}}$$

Therefore re-write the product

$$\begin{align}{{p}_{2}}\left( m \right)&=\prod\limits_{n=1}^{\infty }{\frac{{{\left( 1-\frac{1}{2m\left( n-\tfrac{1}{2} \right)} \right)}^{m\left( n-\tfrac{1}{2} \right)}}{{e}^{\frac{1}{2}+\frac{1}{8m\left( n-\tfrac{1}{2} \right)}}}}{{{\left( 1+\frac{1}{2m\left( n-\tfrac{1}{2} \right)} \right)}^{m\left( n-\tfrac{1}{2} \right)}}{{e}^{-\frac{1}{2}+\frac{1}{8m\left( n-\tfrac{1}{2} \right)}}}}}\\&=\frac{\prod\limits_{n=1}^{\infty }{{{\left( 1-\frac{1}{2m\left( n-\tfrac{1}{2} \right)} \right)}^{m\left( n-\tfrac{1}{2} \right)}}{{e}^{\frac{1}{2}+\frac{1}{8m\left( n-\tfrac{1}{2} \right)}}}}}{\prod\limits_{n=1}^{\infty }{{{\left( 1+\frac{1}{2m\left( n-\tfrac{1}{2} \right)} \right)}^{m\left( n-\tfrac{1}{2} \right)}}{{e}^{-\frac{1}{2}+\frac{1}{8m\left( n-\tfrac{1}{2} \right)}}}}}\end{align}$$

Consider then the products and their log transforms

$${{p}_{2\pm }}=\prod\limits_{n=1}^{\infty }{{{\left( 1\pm \frac{1}{2m\left( n-\tfrac{1}{2} \right)} \right)}^{m\left( n-\tfrac{1}{2} \right)}}{{e}^{\mp \frac{1}{2}+\frac{1}{8m\left( n-\tfrac{1}{2} \right)}}}}$$ $$\log {{p}_{2\pm }}=\sum\limits_{n=1}^{\infty }{2m\left( n-\tfrac{1}{2} \right)\log \left( 1\pm \frac{1}{2m\left( n-\tfrac{1}{2} \right)} \right)\mp \frac{1}{2}+\frac{1}{8m\left( n-\tfrac{1}{2} \right)}}$$ Taking each case separately and expanding the logarithm in its series $$\log {{p}_{2+}}=-\frac{1}{2}\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}}{\left( k+2 \right){{\left( 2m \right)}^{k+1}}}}\sum\limits_{n=1}^{\infty }{\frac{1}{{{\left( n-\tfrac{1}{2} \right)}^{k+1}}}}=-\frac{1}{2}\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}\left( -1+{{2}^{k+1}} \right)\zeta \left( k+1 \right)}{\left( k+2 \right){{\left( 2m \right)}^{k+1}}}}$$ $$\log {{p}_{2-}}=-\frac{1}{2}\sum\limits_{k=1}^{\infty }{\frac{1}{\left( k+2 \right){{\left( 2m \right)}^{k+1}}}}\sum\limits_{n=1}^{\infty }{\frac{1}{{{\left( n-\tfrac{1}{2} \right)}^{k+1}}}}=-\frac{1}{2}\sum\limits_{k=1}^{\infty }{\frac{\left( -1+{{2}^{k+1}} \right)\zeta \left( k+1 \right)}{\left( k+2 \right){{\left( 2m \right)}^{k+1}}}}$$ So $$\log {{p}_{2+}}=\frac{1}{2}\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}\zeta \left( k+1 \right)}{\left( k+2 \right){{\left( 2m \right)}^{k+1}}}}-\frac{1}{2}\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}\zeta \left( k+1 \right)}{\left( k+2 \right){{m}^{k+1}}}}=-\log {{p}_{1+}}\left( m \right)+\log {{p}_{1+}}\left( \tfrac{1}{2}m \right)$$ $$\log {{p}_{2-}}=\frac{1}{2}\sum\limits_{k=1}^{\infty }{\frac{\zeta \left( k+1 \right)}{\left( k+2 \right){{\left( 2m \right)}^{k+1}}}}-\frac{1}{2}\sum\limits_{k=1}^{\infty }{\frac{\zeta \left( k+1 \right)}{\left( k+2 \right){{m}^{k+1}}}}=-\log {{p}_{1-}}\left( m \right)+\log {{p}_{1-}}\left( \tfrac{1}{2}m \right)$$ Therefore $${{p}_{2}}\left( m \right)=\frac{{{e}^{-\log {{p}_{1-}}\left( m \right)+\log {{p}_{1-}}\left( \tfrac{1}{2}m \right)}}}{{{e}^{-\log {{p}_{1+}}\left( m \right)+\log {{p}_{1+}}\left( \tfrac{1}{2}m \right)}}}$$ The final product is $${{e}^{S}}={{p}_{1}}\left( m \right){{p}_{2}}\left( m \right)={{e}^{2\log {{p}_{1+}}\left( m \right)-2\log {{p}_{1-}}\left( m \right)+\log {{p}_{1-}}\left( \tfrac{1}{2}m \right)-\log {{p}_{1+}}\left( \tfrac{1}{2}m \right)}}$$ and so the series becomes $$S=2\left\{ \log {{p}_{1+}}\left( m \right)-\log {{p}_{1-}}\left( m \right) \right\}+\log {{p}_{1-}}\left( \tfrac{1}{2}m \right)-\log {{p}_{1+}}\left( \tfrac{1}{2}m \right)$$

Substitution of previous results yields

$$\begin{align} S&=\zeta '\left( 0,1+\frac{1}{2m} \right)+\zeta '\left( 0,1-\frac{1}{2m} \right)-\frac{1}{2}\zeta '\left( 0,1-\frac{1}{m} \right)-\frac{1}{2}\zeta '\left( 0,1+\frac{1}{m} \right) \\ & -2m\zeta '\left( -1,1+\frac{1}{2m} \right)+2m\zeta '\left( -1,1-\frac{1}{2m} \right)-\frac{1}{2}m\zeta '\left( -1,1-\frac{1}{m} \right)\\&+\frac{1}{2}m\zeta '\left( -1,1+\frac{1}{m} \right)+\frac{1}{2} \\ \end{align}$$

Note: $\zeta '\left( 0,a \right)=\log \left( \Gamma \left( a \right) \right)-\tfrac{1}{2}\log \left( 2\pi \right)$, to obtain

$$\begin{align} S\left( m \right)&=\log \left( \frac{\Gamma \left( 1+\frac{1}{2m} \right)\Gamma \left( 1-\frac{1}{2m} \right)}{\sqrt{2\pi \Gamma \left( 1+\frac{1}{m} \right)\Gamma \left( 1-\frac{1}{m} \right)}} \right) \\ &+2m\left\{ \zeta '\left( -1,1-\frac{1}{2m} \right)-\zeta '\left( -1,1+\frac{1}{2m} \right) \right\}\\&+\frac{1}{2}m\left\{ \zeta '\left( -1,1+\frac{1}{m} \right)-\zeta '\left( -1,1-\frac{1}{m} \right) \right\}+\frac{1}{2} \\ \end{align}$$ Now fix m as a positive integer greater than one. Note also $\zeta \left( s,a+1 \right)=\zeta \left( s,a \right)-{{a}^{-s}}$ so $\zeta '\left( s,a+1 \right)=\zeta '\left( s,a \right)+{{a}^{-s}}\log \left( a \right)$ and DLMF 25.11.21

$$\begin{align} \zeta '\left( -1,\frac{h}{k} \right)&=\frac{\left( 1-\gamma -\ln \left( 2\pi k \right) \right)\left( \tfrac{1}{6}-\tfrac{h}{k}+{{\left( \tfrac{h}{k} \right)}^{2}} \right)}{2}-\frac{\left( 1-\gamma -\ln \left( 2\pi \right) \right)}{12{{k}^{2}}} \\ &+\frac{1}{4\pi {{k}^{2}}}\sum\limits_{r=1}^{k-1}{\sin \left( \frac{2\pi rh}{k} \right){{\psi }^{\left( 1 \right)}}\left( \frac{r}{k} \right)}+\frac{1}{2{{\pi }^{2}}{{k}^{2}}}\sum\limits_{r=1}^{k-1}{\cos \left( \frac{2\pi rh}{k} \right)\zeta '\left( 2,\frac{r}{k} \right)}\\&+\frac{1}{{{k}^{2}}}\zeta '\left( -1 \right) \\ \end{align}$$ From this, finally: $$\begin{align} S\left( m \right)&=\frac{1}{2}+\log \left( \frac{\sqrt{2m}\Gamma \left( 1+\frac{1}{2m} \right)\Gamma \left( 1-\frac{1}{2m} \right)}{\sqrt{\pi \Gamma \left( 1+\frac{1}{m} \right)\Gamma \left( 1-\frac{1}{m} \right)}} \right)\\&+\frac{1}{4\pi m}\sum\limits_{r=1}^{m-1}{\sin \left( \frac{2\pi r}{m} \right){{\psi }^{\left( 1 \right)}}\left( \frac{r}{m} \right)}-\frac{1}{4\pi m}\sum\limits_{r=1}^{2m-1}{\sin \left( \frac{\pi r}{m} \right){{\psi }^{\left( 1 \right)}}\left( \frac{r}{2m} \right)}\end{align}$$

Simple example: m=2 $$S\left( 2 \right)=\frac{1}{2}+\log \left( \frac{2\sqrt{2}\Gamma \left( \frac{5}{4} \right)\Gamma \left( \frac{3}{4} \right)}{\pi } \right)-\frac{1}{8\pi }\left\{ {{\psi }^{\left( 1 \right)}}\left( \frac{1}{4} \right)-{{\psi }^{\left( 1 \right)}}\left( \frac{3}{4} \right) \right\}$$ Simplifying $$S\left( 2 \right)=\frac{1}{2}-\frac{1}{8\pi }\left\{ {{\psi }^{\left( 1 \right)}}\left( \frac{1}{4} \right)-{{\psi }^{\left( 1 \right)}}\left( \frac{3}{4} \right) \right\}=\frac{1}{2}-\frac{2G}{\pi }$$ where G is Catalan’s constant.

References: Srivastava, H. M., Choi, J. Zeta and q-Zeta Functions and Associated Series and Integrals (Elsevier 2012)

Closed form of $\int e^{i\csc^2(x)}dx=\int \cos\left(\csc^2(x)\right)dx+i\int \sin\left(\csc^2(x)\right)dx$

The Owen T function is used for statistics, but has applications like the following definition:

$$2\pi \cdot\text T(x,a)\mathop=^\text{def} \int_0^a\frac{e^{-\frac{x^2}2\left(t^2+1\right)}}{t^2+1}dt$$

We can easily use a series expansion for the exponential function with index $n$, but that would case a hypergeometric function with a negative argument index of $-n$ which cannot really be converted into $n$ in the function. This problem is due to the positive exponent in the $e^y$, it is better to do the following substitution to shape make it easier to turn into a Kampé de Feriét function. We just need an identity, so keep in mind possible restrictions:

$$\int_0^a\frac{e^{-x^2\left(t^2+1\right)}}{t^2+1}dt\ \mathop=^{t^2+1=\frac1{u^2+1}\implies t=\pm \sqrt{\frac1{u^2+1}-1}}_{dt=\frac i{(u^2+1)^\frac32}du,u>0}\ i\int_0^{\sqrt{-\frac{a^2}{a^2+1}}} (u^2+1)\frac{e^{-\frac{x^2}{u^2+1}}}{(u^2+1)^\frac32}du=i\int_0^{\sqrt{-\frac{a^2}{a^2+1}}} \frac{e^{-\frac{x^2}{u^2+1}}}{(u^2+1)^\frac12}du=i\int_0^{\sqrt{-\frac{a^2}{a^2+1}}} \sum_{m=0}^\infty\frac{(-1)^mx^{2m}}{m!(u^2+1)^m(u^2+1)^\frac12}du=i \sum_{m=0}^\infty\frac{(-1)^mx^{2m}}{m!}\int_0^{b} (u^2+1)^{-m-\frac12}du$$

The upper bound will be called $b$ by definition. Let there be an integration using a Gauss hypergeometric series and Pochhammer symbol:

$$i \sum_{m=0}^\infty\frac{(-1)^mx^{2m}}{m!}\int_0^b (u^2+1)^{-m-\frac12}du=i \sum_{m=0}^\infty\frac{(-1)^mx^{2m}}{m!}b\sum_{n=0}^\infty \frac{\left(\frac12\right)_n \left(m+\frac12\right)_n(-b^2)^n}{\left(\frac32\right)_n n!}$$

Notice the similar coefficients to the original problem. We can change the pochhammer symbol by:

$$\left(m+\frac12\right)_n =\frac{\Gamma\left(m+n+\frac12\right)\Gamma\left(\frac12\right)}{\Gamma\left(m+\frac12\right)\Gamma\left(\frac12\right)}=\frac{\left(\frac12\right)_{m+n}}{\left(\frac12\right)_m}$$

Therefore we can use the question’s Kampé de Feriét function link:

$$i \sum_{m=0}^\infty\frac{(-1)^mx^{2m}}{m!}b\sum_{n=0}^\infty \frac{\left(\frac12\right)_n \left(m+\frac12\right)_n(-b^2)^n}{\left(\frac32\right)_n n!}=ib \sum_{m=0}^\infty\sum_{n=0}^\infty \frac{\left(\frac12\right)_{m+n}\left(\frac12\right)_n \left(-x^2\right)^m\left(-b^2\right)^n}{\left(\frac12\right)_m\left(\frac32\right)_n m!n!}=i b\,\text F^{1,0,1}_{0,1,1}\left(^{\frac12;;\frac12}_{;\frac12;\frac32}\ -x^2,-b^2\right)$$

which we now need to form into:

$$\int e^{i\csc^2(x)}dx=\int \cos\left(\csc^2(x) \right)dx+i\int \sin\left(\csc^2(x) \right)dx= C-\sqrt{\sin^2(x)}\cot(x)\text F^{1,0,1}_{0,1,1}\left(^{\frac12;;\frac12}_{;\frac12;\frac32}\ i\csc^2(x)\sin^2(x),\cos^2(x)\right)=$$

To finish the answer, let’s calculate $b$:

$$b={\sqrt{-\frac{a^2}{a^2+1}}}\implies -b^2=\frac{a^2}{a^2+1}$$

Therefore:

$$\text T(x,a)=-\frac{{\sqrt{\frac{a^2}{a^2+1}}}}{2\pi}\,\text F^{1,0,1}_{0,1,1}\left(^{\frac12;;\frac12}_{;\frac12;\frac32}\ -x^2,\frac{a^2}{a^2+1}\right)$$

Solving for $a,x$ and taking the positive square root branch gives the final answer as:

$$\int e^{i\csc^2(x)}dx=\int \cos\left(\csc^2(x) \right)dx+i\int \sin\left(\csc^2(x) \right)dx= -2\pi \sqrt{\cot^2(x)}\tan(x)\text T\left(1-i,\sqrt{\cot^2(x)}\right)+C\mathop=^{x\in\Bbb R} -2\pi\text{sgn}(\cot(x))\text T\left(1-i,|\cot(x)|\right)+C$$

Here is proof of the result with this computation

Therefore one special case is with the Imaginary Error function and Fresnel Integrals:

$$\int_0^\frac\pi4 e^{i\csc^2(x)}dx= \int_0^\frac\pi2 e^{i\csc^2(x)}dx -\int_\frac\pi4^\frac\pi2 e^{i\csc^2(x)}dx =\int_0^1 \frac{e^{i(x^2+1)}}{x^2+1}dx=2\pi(\text T(1-i,\infty)-\text T(1-i,1))=\frac\pi2+\frac\pi2\left((i-1)\text C\left(\sqrt{\frac2\pi}\right)-(1+i) \text S\left(\sqrt{\frac2\pi}\right)\right)+\frac\pi 4\left(\text{erfi}^2\left(\sqrt[4]{-1}\right)+1\right) $$

Note that the following code also applies with the Complementery Error function:

π/2 + 1/2 ((-1 + i) π C(sqrt(2/π)) - (1 + i) π S(sqrt(2/π))) = π/2 + 1/2 (-(1 + i) π (1/2 + 1/4 (1 + i) (i erfc(1/2 (1 - i) sqrt(π) sqrt(2/π)) - erfc(1/2 (1 + i) sqrt(π) sqrt(2/π)))) + (-1 + i) π (1/2 - 1/4 (1 - i) (i erfc(1/2 (1 - i) sqrt(π) sqrt(2/π)) + erfc(1/2 (1 + i) sqrt(π) sqrt(2/π)))))

Here is a plot of the real, cosine, part:

Here is a plot of the imaginary, sine, part:

Please correct me and give me feedback!

It is hard to separate real and imaginary parts of the OwenT function unless you use Complex Components

Best Answer

Related Solutions

General form for $\sum_{n=1}^{\infty} (-1)^n \left(m n \, \text{arccoth} \, (m n) – 1\right)$

Closed form of $\int e^{i\csc^2(x)}dx=\int \cos\left(\csc^2(x)\right)dx+i\int \sin\left(\csc^2(x)\right)dx$

Related Question