The zip file works. I could compile the fractional_integral.tex file using PdfLaTeX (x2) -> BibTeX (x2) -> PdfLaTeX (x2) and the pdf file compiled successfully.
All the files are given below:
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\begin{document}
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\ShortArticleName{Fractional Integrals}
\ArticleName{Evaluation of a Class $n$-fold Integrals by Means of Fractional Integration}
% Names of the authors for the title of the paper
\Author{Ben Orin\,$^{\ast}\!\!\ $
%Roberto S.~Costas-Santos\,$^\S\!\!\ $,
%Mourad E.~H.~Ismail\,$^\ast\!\!\ $
%and Lisa Ritter\,$^\ast\!\!$
}
\AuthorNameForHeading{Ben Orin, %R.S.~Costas-Santos,
%M.~E.~H.~Ismail,
%L.~Ritter
}%\Address{$^\ast$~Applied and Computational Mathematics Division,
%National Institute of Standards and Technology,
%Gaithersburg, MD 20899-8910, USA
%Address of First Author, Country
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%} % Address of First Author
%\EmailD{howard.cohl@nist.gov}
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%\href{http://www.nist.gov/itl/math/msg/howard-s-cohl.cfm}
%{http://www.nist.gov/itl/math/msg/howard-s-cohl.cfm}
%}
\EmailD{benorin@gmail.com} % E-mail address of First Author
%\Address{}
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%} % Address of First Author
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%\Address{$^\S$~Departamento de F\'isica y Matem\'{a}ticas,
%Universidad de Alcal\'{a},
%c.p. 28871, Alcal\'{a} de Henares, Spain
%Address of First Author, Country
%} % Address of First Author
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%\href{http://www.rscosan.com}
%{http://www.rscosan.com}
%}
%\EmailD{rscosa@gmail.com} % E-mail address of 1st Author
%\Address{$^\ast$~Department of Mathematics, University of Central Florida, Orlando, FL 32816, USA
%Address of First Author, Country
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%\href{https://sciences.ucf.edu/math/person/mourad-ismail/}
%{https://sciences.ucf.edu/math/person/mourad-ismail/}
%}
%} % Address of First Author
%\EmailD{mourad.eh.ismail@gmail.com} % E-mail address of First Author%\ArticleDates{Received ?? March 2018 in final form ?; Published online ?}
\ArticleDates{Received ?, in final form ?; Published online ?}
\Abstract{Abstract here.}
%``Symmetry, Integrability and Geometry: Methods and Applications''.}
%\Keywords{
%Generalized hypergeometric series;
%Generalized hypergeometric orthogonal polynomials;
%Linearization coefficient; Connection coefficients;
%Eigenfunction expansions}
%Please type here List of Keywords for your article separated by semicolon.
% Keywords required only for MST, PB, PMB, PM, JOA, JOB?
% Keywords:%\Classification{33C45, 05A15, 33C20, 34L10, 30E20}
%{?} % e.g. 35A30; 81Q05
%For 2010 Mathematics Subject Classification see
%http://www.ams.org/mathscinet/msc/msc2010.html
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
In this note we will be evaluating a certain class of $n$-fold integrals over hypercubes via interpolation of the left Hadamard fractional integral operator. We won't be doing any fractional calculus other than a single interpolation theorem which may be used as a basis for fractional integration of the Hadamard type, note that this is not the more common Riemann nor Reiz types of fractional integral interpolation.
\section{Main Body}
\subsection{Fractional Calculus}
Fractional integrals are a generalization of $n$-fold iterated integrals to arbitrary order $\alpha\in\C$ (see theorem \ref{interpolation}), there at least a few ways to do that, each giving rise to its own fractional calculus.
\cite[p.~1]{Jarad}
\begin{defn}
The left Hadamard fractional integral operator will be denoted by $_aI_x^\alpha $, for $0 < a < x < \infty , \Re\left[ \alpha\right] > 0$ is defined as
\[
_a^HI_x^\alpha g\left( x \right) = \tfrac{1}{{\Gamma \left( \alpha \right)}}\int_{a}^x {{{\log }^{\alpha - 1}}\left( {\tfrac{x}{t}} \right) g\left( t \right)\tfrac{dt}{t}}
\]
\label{defn51}
\noindent assuming the integral is convergent and where $\log$ denotes the natural logarithm, $\Gamma$ is the usual gamma function, and $\Re$ is the real part.
\end{defn}
\cite[p.~2]{Jarad}
\begin{thm}
\label{interpolation}
Interpolation of this n-fold integral by the left Hadamard fractional integral operator.
\[
\int_a^x {\int_a^{{x_1}} { \cdots \int_a^{{x_{n - 1}}} {f\left( {{x_n}} \right)\tfrac{{d{x_n} \ldots d{x_1}}}{{{x_n} \cdots {x_1}}}} } } = \,_a^HI_x^nf\left( x \right) = \tfrac{1}{{\left( {n - 1} \right)!}}\int_a^x {{{\log }^{n - 1}}\left( {\tfrac{x}{t}} \right)f\left( t \right)dt}
\]
\end{thm}
\begin{proof}
The proof is by induction on n:
(i) base case of $n = 1$ is obvious.
(ii) Let $P(n)$ be the statement of theorem \ref{interpolation}. Assume that $P(n)$ holds for some fixed positive integer $n$. Then,
\begin{eqnarray*}P(n+1) &=& \int_a^x \int_a^{x_1} \cdots \int_a^{x_n} f\left( x_{n + 1} \right)\tfrac{d{x_{n + 1}} \ldots {d{x_1}}}{{x_{n + 1}} \cdots {x_1}} \\ &=& \int_a^x {\left[ {\int_a^{{x_1}} { \cdots \int_a^{{x_n}} {f\left( {{x_{n + 1}}} \right)\tfrac{{d{x_{n + 1}} \ldots d{x_2}}}{{{x_{n + 1}} \cdots {x_2}}}} } } \right]\tfrac{{d{x_1}}}{{{x_1}}}} \\ &=& \int_a^x {\left[ {\tfrac{1}{{\left( {n - 1} \right)!}}\int_a^{{x_1}} {{{\log }^{n - 1}}\left( {\tfrac{{{x_1}}}{t}} \right)f\left( t \right)\tfrac{{dt}}{t}} } \right]\tfrac{{d{x_1}}}{{{x_1}}}} \\ &=& \tfrac{1}{{\left( {n - 1} \right)!}}\int_a^x {\int_t^x {{{\log }^{n - 1}}\left( {\tfrac{{{x_1}}}{t}} \right)f\left( t \right)\tfrac{{d{x_1}dt}}{{{x_1}t}}} } \\ &=& \tfrac{1}{{\left( {n - 1} \right)!}}\sum\limits_{k = 0}^{n - 1} {{{\left( { - 1} \right)}^k}\left( {\begin{array}{*{20}{c}}{n - 1} \\ k \end{array}} \right)\int_a^x {\int_t^x {{{\log }^k}\left( {{x_1}} \right){{\log }^{n - k - 1}}\left( t \right) f\left( t \right)\tfrac{{d{x_1}dt}}{{{x_1}t}}} } } \\ &=& \tfrac{1}{\left( {n - 1} \right) ! }\sum\limits_{k = 0}^{n - 1} \left( -1 \right)^k \left( {\begin{array}{*{20}{c}}{n - 1} \\ k \end{array}} \right)\int_a^x {{\log }^{n - k - 1}}\left( t \right) f\left( t \right)\tfrac{1}{t}\left[ \int_t^x {\log }^k\left( x_1 \right) \tfrac{d{x_1}}{x_1}\right] dt \\ &=& \tfrac{1}{{\left( {n - 1} \right) !}}\sum\limits_{k = 0}^{n - 1} {{{\left( { - 1} \right) }^k}\left( {\begin{array}{*{20}{c}}{n - 1} \\ k \end{array}} \right) \tfrac{1}{{k + 1}}\int_a^x {{{\log }^{n - k - 1}}\left( t \right)f\left( t \right)\tfrac{1}{t}\left( {{{\log }^{k + 1}}x - {{\log }^{k + 1}}t} \right)dt} } \\ &=& \tfrac{1}{{n!}}\sum\limits_{k = 0}^n {{\left( - 1 \right) }^k}\left( {\begin{array}{*{20}{c}} n \\ k \end{array}} \right)\int_a^x {{{\log }^{n - k}}\left( t \right){{\log }^k}\left( x \right)f\left( t \right)\tfrac{{dt}}{t}} \\ &=& \tfrac{1}{{n!}}\int_a^x {{{\log }^n}\left( {\tfrac{x}{t}} \right) f\left( t \right)\tfrac{{dt}}{t}} \\ \end{eqnarray*}
and the proof is complete.
\end{proof}
The following non-standard definition we will adopt throughout the rest of this note because it is easier to work with than the standard definition \ref{defn51} for our purposes.
\begin{defn}
\label{defn52}
The modified left Hadamard fractional operator
For $0 < a < x < \infty ,\Re \left[ \alpha \right] > 0$
\[
_a^{eH}I_x^\alpha g\left( x \right) = \tfrac{1}{\Gamma \left( \alpha \right)}\int_{0}^{\log \left( \tfrac{x}{a} \right) } u^{\alpha -1} g\left( x e^{-u}\right)\, du
\]
where we have substituted $u = \log \left( \tfrac{x}{t} \right)$ in the integral of definition \ref{defn51} and $u^{\alpha -1}$ is taken as it’s principle value.
\end{defn}
\subsection{Evaluations of $n$-fold integrals}
Here we reduce the problem of evaluating certain $n$-fold integrals to that of solving a single fractional integral (just think of these as integrals transforms for our purposes).
\begin{thm}
\label{AC}
Analytic continuation of certain $n$-fold integrals over unit hypercubes.
Let $z$ and $\alpha$ be a complex-valued parameters, let $t$ denote a real variable, let $n$ be a positive integer, and for fixed $z=z_0$ let $f(z_0,t)$ be a continuous function of $t$ on $\left[ 0, 1 \right]$. Then for suitable functions $f(z, t)$ (for which the integral converges) define
\[
F_n(z):= \int_0^1\int_0^1\cdots \int_0^1 f\left( z,\prod\limits_{k=1}^n\lambda _k \right)\, d\lambda
\]
where $d\lambda := d\lambda _{n}\ldots d\lambda _1$ (and likewise for other dummy variables as well). Then
\[
G(z, \alpha ) := \tfrac{1}{\Gamma (\alpha )}\int_0^\infty {u^{\alpha -1}e^{-u} f(z,e^{-u})\, d{u}}\text{ and } G(z,n)=F_n(z)
\]
is the Hadamard fractional integral of order $\alpha$ which is the analytic continuation of $F_n(z)$ from integer $n$ to complex-valued $\alpha$ restricted to values for which the integral converges.
\end{thm}
\begin{proof}
\cite{PhysicsForums}
We will use the change of variables ${y_k} = \prod\limits_{i = 1}^k {\lambda _i} ,k = 1,2, \ldots ,n$ on the integral $F_n(z)$ to formulate an integral that represents the function for complex values of the argument via theorem \ref{interpolation} .
Note that the for given change of variables we have ${\lambda _1} = {y_1},{\lambda _k} = \tfrac{y_k}{y_{k - 1}},k = 2,3, \ldots ,n$, hence
\[
\tfrac{{\partial {\lambda _i}}}{{\partial {y_j}}} = \left\{ {\begin{array}{*{20}{c}}{1,}&{i = j = 1} \\ {\tfrac{1}{{{y_{i - 1}}}},}&{i = j \ne 1} \\ { - \tfrac{{{y_i}}}{{y_{i - 1}^2}},}&{i = j - 1} \\ {0,}&{{\text{otherwise}}} \end{array}} \right.
\]
hence the Jacobian determinant is the product along the diagonal, $\left| {\tfrac{{\partial \left( {{\lambda _1}, \ldots ,{\lambda _n}} \right)}}{{\partial \left( {{y_1}, \ldots ,{y_n}} \right)}}} \right| = \tfrac{{d{y_n} \ldots d{y_1}}}{{{y_{n - 1}} \cdots {y_1}}}$.
\noindent Notice that this change of variables maps the unit hypercube ${\left[ {0,1} \right]^n}$ to the simplex
\[
\left\{ \vec y \in {\mathbb{R}^n}|0 \leq {y_1} \leq 1,0 \leq {y_i} \leq {y_{i - 1}},\text{ for } i = 2,3, \ldots ,n \right\} .
\]
We replace the upper bound of ${y_1}$ with $x$ so that
\begin{eqnarray*} F_n(z) &=& \lim\limits_{a \to 0^{+}}\lim\limits_{x \to 1^{-}}\int_a^x \int_a^{y_1} \cdots \int_a^{y_{n-1}} y_n f(z, y_n)\, \tfrac{d{y_n} \ldots d{y_1}}{{y_n} \ldots {y_1}} \\ &=& \lim\limits_{a \to 0^{+}}\lim\limits_{x \to 1^{-}} \, _a^{eH}I_x^n\left( x f(z, x) \right) \\ &=& \tfrac{1}{\left( n - 1 \right) !}\int_0^\infty u^{n-1}e^{-t} f\left( z, e^{-t}\right) \\ \end{eqnarray*}
by theorem \ref{interpolation} which we analytically continue to
\begin{eqnarray*}G(z, \alpha ) = \lim\limits_{a \to 0^{+}}\lim\limits_{x \to 1^{-}} \,_a^{eH}I_x^\alpha \left[ xf(z,x) \right] &=& \tfrac{1}{\Gamma (\alpha )}\int_0^\infty {u^{\alpha -1}e^{-u} f(z,e^{-u})\, du} \\ \end{eqnarray*}
\end{proof}
The next few computer and table-assisted examples will illustrate the use of theorem \ref{AC}.
\begin{ex}
\label{ex1}
\cite[p.~193]{BorosMoll}
Let $G( z, \alpha ) =\tfrac{1}{\Gamma (\alpha )} \int_0^\infty t^{\alpha -1} e^{-t}\cdot\tfrac{e^t}{(1+t)^z}dt$. We see that this is a beta integral upon canceling $e^{-t}e^t=1$ giving the value $G( z, \alpha ) = \tfrac{\Gamma (z- \alpha )}{\Gamma (z)}$. We then determine what $f(z, t)$ is by comparing the integrand of the integral defining $G(z, \alpha )$ in this example to the corresponding integrand in theorem \ref{AC}, to see that $f(z, e^{-t}) = \tfrac{e^t}{(1+t)^z}$ which implies that $f(z, t) = \tfrac{t^{-1}}{(1-\log (t))^z}$ and hence the evaluation of the $n$-fold integral of theorem \ref{AC} is
\[
F_n (z) = \int_0^1\int_0^1\cdots\int_0^1 \left( \prod\limits_{k=1}^n \lambda _k\right) ^{-1} \left( 1- \log \prod\limits_{k=1}^n \lambda _k \right) ^{-z}\, d\lambda = \tfrac{\Gamma (z-n)}{\Gamma (z)} =G(z, n).
\]
Note that other integrals may be deduced from this by differentiation under the integral sign w.r.t. $z$, such as
\[
F^{\prime }_n (z) = \int_0^1\int_0^1\cdots\int_0^1 \tfrac{-\log\left( 1-\log\prod\limits_{j=1}^n \lambda _j \right) }{\left( \prod\limits_{m=1}^n \lambda _m\right) \left( 1- \log \prod\limits_{k=1}^n \lambda _k \right) ^z} \, d\lambda = \tfrac{\Gamma (z-n)\left( \psi ^{(0)}(z-n)-\psi ^{(0)}(z) \right)}{\Gamma (z)}
\]
where $\psi ^{(m)}$ is the $m^{th}$ derivative of the diagamma function.
\end{ex}
\begin{ex}
\cite{FunctionsWolfram}
\label{ex2}
Let $G(z, \alpha , y) = \tfrac{1}{\Gamma ( \alpha )}\int_0^\infty t^{\alpha -1}e^{-t}\cdot\tfrac{e^{-(y-1)t}}{1-z e^{-t}}dt$. Wolfram.functions.com gives the value $G(z, \alpha ,y) = \sum\limits_{k= 0}^\infty \frac{z^k}{(k+y) ^\alpha } = \Phi ( z,\alpha ,y)$ where $\Phi$ is the Lerch Transcendent. We see that $f(z,t,y)= \tfrac{t^{y-1}}{1-zt}$ and hence the evaluation we seek is
\[
F_n (z,y) = \int_0^1 \int_0^1 \cdots \int_0^1 \left( 1 - z\prod\limits_{k = 1}^n \lambda _k \right) ^{-1}\prod\limits_{j = 1}^n \lambda _j^{y-1} \, d{\lambda} =\sum\limits_{k = 0}^\infty \frac{z^k}{(k+y)^n} = \Phi (z,n,y).
\]
More integrals maybe calculated by differentiation under the integral sign w.r.t. $y$,
\begin{eqnarray*}\tfrac{\partial F_n}{\partial y} (z,y) &=& \int_0^1 \int_0^1 \cdots \int_0^1 \left( 1 - z\prod\limits_{k = 1}^n \lambda _k \right) ^{-1}\prod\limits_{j = 1}^n \lambda _j^{y-1}\cdot \log\prod\limits_{\ell = 1}^n\lambda _\ell d{\lambda} \\ &=& -n\sum\limits_{k = 0}^\infty \frac{z^k}{(k+y)^{n+1}} = -n\Phi (z,n+1,y) \\ \end{eqnarray*}
Differentiating $m$ times w.r.t. $y$, we get
\begin{eqnarray*} \tfrac{\partial ^m F_n}{\partial y^m} (z,y) &=& \int_0^1 \int_0^1 \cdots \int_0^1 \left( 1 - z\prod\limits_{k = 1}^n \lambda _k \right) ^{-1}\prod\limits_{j=1}^n\lambda _j^{y-1}\cdot \log^{m}\prod\limits_{\ell = 1}^n\lambda _\ell d{\lambda} \\ &=& (-1)^{m}\tfrac{(n+m-1)!}{(n-1)!}\sum\limits_{k = 0}^\infty \frac{z^k}{(k+y)^{n+m}} = (-1)^{m}\tfrac{(n+m-1)!}{(n-1)!}\Phi (z,n+m,y) \\ \end{eqnarray*}
\end{ex}
%\begin{defn}
%\label{MellinT}
%The Mellin transform. For $s\in\mathbb{C},$
%\[
%\mathfrak{M}\left[ g(t)\right] = \int_0^\infty t^{s-1}g(t)dt.
%\]
%\end{defn}
\subsection*{Acknowledgements}
What does a man have that God has not given him? Nothing.
\noindent Thanks to Howard Cohl for helping me.
\bibliographystyle{plain}
\bibliography{refbiborin}
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@article{Jarad,
AUTHOR = {Jarad F.~Abdeljawad T.~Baleanu, D.},
TITLE = {Caputo-type modification of the Hadamard
fractional derivatives},
JOURNAL = {Advances in Difference Equations},
VOLUME = {2012},
NUMBER = {142},
YEAR = {2012},
PAGES = {142-150},
URL = {\newline\url{https://advancesindifferenceequations.springeropen.com/track/pdf/10.1186/1687-1847-2012-142.pdf}}
}
@book{BorosMoll,
AUTHOR = {Boros, G.~ Moll, V.},
TITLE = {Irresistible Integrals},
PUBLISHER = {Cambridge University Press, New York},
YEAR = {2004},
PAGES = {xvi+306},
}
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