text: We have the following complex integral : $$\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}e^{-\frac{\pi}{2}\cot\left(\frac{\pi}{s}\right)}\frac{x^{s}}{s}ds$$ Where $x\in\mathbb{R}:x>1$. i tried closing the contour to the left, and computing the residues at the essential singularities of the integrand $\left( s=-\frac{1}{n}\right)$ in the following manner : Using the partial fraction expansion of $\cot(\pi z)$, we have: $$\frac{\pi}{2}\cot\left(\frac{\pi}{s} \right )=\frac{s}{2}-\sum_{k=1}^{\infty}\frac{1}{2k(ks-1)}+\frac{1}{2k(ks+1)}$$ And around the reciprocal of each negative integer $n$ we have the Taylor expansion: $$e^{-\frac{\pi}{2}\cot\left(\frac{\pi}{s}\right)}\frac{e^{\frac{1}{2n(ns+1)}}}{s}=\sum_{m=0}^{\infty}c_{n,m}\left(s+\frac{1}{n}\right)^{m}$$ Furthermore, using the definition of the Bessel function of the first kind, we have: $$x^{s}e^{-\frac{1}{2n(ns+1)}}=x^{-\frac{1}{n}}\exp\left[\frac{\sqrt{\ln x}}{n\sqrt{2}}\left(\sqrt{\ln x^{2}}(ns+1)-\frac{1}{\sqrt{\ln x^{2}}(ns+1)}\right)\right]$$ $$=x^{-\frac{1}{n}}\sum_{i=-\infty}^{\infty}J_{i}\left(\frac{\sqrt{\ln x^{2}}}{n}\right)\left(\sqrt{\ln x^{2}}(ns+1)\right)^{i}$$ From which we obtain: $$\underset{s=-n^{-1}}{\text{Res}}e^{-\frac{\pi}{2}\cot\left(\frac{\pi}{s}\right)}\frac{x^{s}}{s}=x^{-\frac{1}{n}}\sum_{k=0}^{\infty}c_{n,k}\frac{J_{k+1}\left(\frac{\sqrt{\ln x^{2}}}{n}\right)}{n^{k+1}(\ln x^{2})^{(k+1)/2}}$$ And i claim that: $$\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}e^{-\frac{\pi}{2}\cot\left(\frac{\pi}{s}\right)}\frac{x^{s}}{s}ds=\sum_{n=1}^{\infty}\sum_{k=0}^{\infty}c_{n,k}x^{-\frac{1}{n}}\frac{J_{k+1}\left(\frac{\sqrt{\ln x^{2}}}{n}\right)}{n^{k+1}(\ln x^{2})^{(k+1)/2}}$$ However, when i posted this question on MS, someone pointed out that this scheme is meaningless, since the original integral doesn't exist !! i find myself not content with the answer!! hence the thread.
  We have the following complex integral : $$\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}e^{-\frac{\pi}{2}\cot\left(\frac{\pi}{s}\right)}\frac{x^{s}}{s}ds$$ Where $x\in\mathbb{R}:x>1$. i tried closing the contour to the left, and computing the residues at the essential singularities of the integrand $\left( s=-\frac{1}{n}\right)$ in the following manner : Using the partial fraction expansion of $\cot(\pi z)$, we have: $$\frac{\pi}{2}\cot\left(\frac{\pi}{s} \right )=\frac{s}{2}-\sum_{k=1}^{\infty}\frac{1}{2k(ks-1)}+\frac{1}{2k(ks+1)}$$ And around the reciprocal of each negative integer $n$ we have the Taylor expansion: $$e^{-\frac{\pi}{2}\cot\left(\frac{\pi}{s}\right)}\frac{e^{\frac{1}{2n(ns+1)}}}{s}=\sum_{m=0}^{\infty}c_{n,m}\left(s+\frac{1}{n}\right)^{m}$$ Furthermore, using the definition of the Bessel function of the first kind, we have: $$x^{s}e^{-\frac{1}{2n(ns+1)}}=x^{-\frac{1}{n}}\exp\left[\frac{\sqrt{\ln x}}{n\sqrt{2}}\left(\sqrt{\ln x^{2}}(ns+1)-\frac{1}{\sqrt{\ln x^{2}}(ns+1)}\right)\right]$$ $$=x^{-\frac{1}{n}}\sum_{i=-\infty}^{\infty}J_{i}\left(\frac{\sqrt{\ln x^{2}}}{n}\right)\left(\sqrt{\ln x^{2}}(ns+1)\right)^{i}$$ From which we obtain: $$\underset{s=-n^{-1}}{\text{Res}}e^{-\frac{\pi}{2}\cot\left(\frac{\pi}{s}\right)}\frac{x^{s}}{s}=x^{-\frac{1}{n}}\sum_{k=0}^{\infty}c_{n,k}\frac{J_{-k-1}\left(\frac{\sqrt{\ln \underset{s=-n^{-1}}{\text{Res}}e^{-\frac{\pi}{2}\cot\left(\frac{\pi}{s}\right)}\frac{x^{s}}{s}=x^{-\frac{1}{n}}\sum_{k=0}^{\infty}c_{n,k}\frac{J_{k+1}\left(\frac{\sqrt{\ln x^{2}}}{n}\right)}{n^{k+1}(\ln x^{2})^{(k+1)/2}}$$ However, when i posted this question on MS, someone pointed out that this scheme is meaningless, since the original integral doesn't exist !! i find myself not content with the answer!! hence the thread.
explanation: The document describes a complex integral and the author's attempt to evaluate it using contour integration and residue calculus. The author presents a detailed derivation involving the partial fraction expansion of the cotangent function, Taylor expansions, and the use of Bessel functions. The document also mentions that the author posted this question on Math StackExchange (MS) and received feedback that the original integral does not exist, which the author finds unsatisfactory.
  
  The document contains a question about the validity of the author's approach and the existence of the integral.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: We have the following complex integral : $$\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}e^{-\frac{\pi}{2}\cot\left(\frac{\pi}{s}\right)}\frac{x^{s}}{s}ds$$ Where $x\in\mathbb{R}:x>1$. i tried closing the contour to the left, and computing the residues at the essential singularities of the integrand $\left( s=-\frac{1}{n}\right)$ in the following manner : Using the partial fraction expansion of $\cot(\pi z)$, we have: $$\frac{\pi}{2}\cot\left(\frac{\pi}{s} \right )=\frac{s}{2}-\sum_{k=1}^{\infty}\frac{1}{2k(ks-1)}+\frac{1}{2k(ks+1)}$$ And around the reciprocal of each negative integer $n$ we have the Taylor expansion: $$e^{-\frac{\pi}{2}\cot\left(\frac{\pi}{s}\right)}\frac{e^{\frac{1}{2n(ns+1)}}}{s}=\sum_{m=0}^{\infty}c_{n,m}\left(s+\frac{1}{n}\right)^{m}$$ Furthermore, using the definition of the Bessel function of the first kind, we have: $$x^{s}e^{-\frac{1}{2n(ns+1)}}=x^{-\frac{1}{n}}\exp\left[\frac{\sqrt{\ln x}}{n\sqrt{2}}\left(\sqrt{\ln x^{2}}(ns+1)-\frac{1}{\sqrt{\ln x^{2}}(ns+1)}\right)\right]$$ $$=x^{-\frac{1}{n}}\sum_{i=-\infty}^{\infty}J_{i}\left(\frac{\sqrt{\ln x^{2}}}{n}\right)\left(\sqrt{\ln x^{2}}(ns+1)\right)^{i}$$ From which we obtain: $$\underset{s=-n^{-1}}{\text{Res}}e^{-\frac{\pi}{2}\cot\left(\frac{\pi}{s}\right)}\frac{x^{s}}{s}=x^{-\frac{1}{n}}\sum_{k=0}^{\infty}c_{n,k}\frac{J_{k+1}\left(\frac{\sqrt{\ln x^{2}}}{n}\right)}{n^{k+1}(\ln x^{2})^{(k+1)/2}}$$ And i claim that: $$\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}e^{-\frac{\pi}{2}\cot\left(\frac{\pi}{s}\right)}\frac{x^{s}}{s}ds=\sum_{n=1}^{\infty}\sum_{k=0}^{\infty}c_{n,k}x^{-\frac{1}{n}}\frac{J_{k+1}\left(\frac{\sqrt{\ln x^{2}}}{n}\right)}{n^{k+1}(\ln x^{2})^{(k+1)/2}}$$ However, when i posted this question on MS, someone pointed out that this scheme is meaningless, since the original integral doesn't exist !! i find myself not content with the answer!! hence the thread.
  We have the following complex integral : $$\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}e^{-\frac{\pi}{2}\cot\left(\frac{\pi}{s}\right)}\frac{x^{s}}{s}ds$$ Where $x\in\mathbb{R}:x>1$. i tried closing the contour to the left, and computing the residues at the essential singularities of the integrand $\left( s=-\frac{1}{n}\right)$ in the following manner : Using the partial fraction expansion of $\cot(\pi z)$, we have: $$\frac{\pi}{2}\cot\left(\frac{\pi}{s} \right )=\frac{s}{2}-\sum_{k=1}^{\infty}\frac{1}{2k(ks-1)}+\frac{1}{2k(ks+1)}$$ And around the reciprocal of each negative integer $n$ we have the Taylor expansion: $$e^{-\frac{\pi}{2}\cot\left(\frac{\pi}{s}\right)}\frac{e^{\frac{1}{2n(ns+1)}}}{s}=\sum_{m=0}^{\infty}c_{n,m}\left(s+\frac{1}{n}\right)^{m}$$ Furthermore, using the definition of the Bessel function of the first kind, we have: $$x^{s}e^{-\frac{1}{2n(ns+1)}}=x^{-\frac{1}{n}}\exp\left[\frac{\sqrt{\ln x}}{n\sqrt{2}}\left(\sqrt{\ln x^{2}}(ns+1)-\frac{1}{\sqrt{\ln x^{2}}(ns+1)}\right)\right]$$ $$=x^{-\frac{1}{n}}\sum_{i=-\infty}^{\infty}J_{i}\left(\frac{\sqrt{\ln x^{2}}}{n}\right)\left(\sqrt{\ln x^{2}}(ns+1)\right)^{i}$$ From which we obtain: $$\underset{s=-n^{-1}}{\text{Res}}e^{-\frac{\pi}{2}\cot\left(\frac{\pi}{s}\right)}\frac{x^{s}}{s}=x^{-\frac{1}{n}}\sum_{k=0}^{\infty}c_{n,k}\frac{J_{-k-1}\left(\frac{\sqrt{\ln \underset{s=-n^{-1}}{\text{Res}}e^{-\frac{\pi}{2}\cot\left(\frac{\pi}{s}\right)}\frac{x^{s}}{s}=x^{-\frac{1}{n}}\sum_{k=0}^{\infty}c_{n,k}\frac{J_{k+1}\left(\frac{\sqrt{\ln x^{2}}}{n}\right)}{n^{k+1}(\ln x^{2})^{(k+1)/2}}$$ However, when i posted this question on MS, someone pointed out that this scheme is meaningless, since the original integral doesn't exist !! i find myself not content with the answer!! hence the thread.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document describes a complex integral and the author's attempt to evaluate it using contour integration and residue calculus. The author presents a detailed derivation involving the partial fraction expansion of the cotangent function, Taylor expansions, and the use of Bessel functions. The document also mentions that the author posted this question on Math StackExchange (MS) and received feedback that the original integral does not exist, which the author finds unsatisfactory.
  
  The document contains a question about the validity of the author's approach and the existence of the integral.
  
  Label: question