# Alignment Problem of Mathematical Equation in Beamer Presentation

\documentclass{beamer}
\newcommand\Fontvi{\fontsize{2}{4}\selectfont}
\usepackage[utf8]{inputenc}
\usepackage[compat=1.1.0]{tikz-feynman}
\usepackage{amsmath, amsthm, amssymb,amsfonts}
\usepackage{graphicx}
\usepackage{tikz}
\usepackage{tcolorbox}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amssymb}
\usepackage{tikz-feynman}
\setcounter{MaxMatrixCols}{20}
%\usepackage{breqn}
\usepackage{tikz-feynman}
\usepackage{amsmath}
\newtheorem{rules}{Rule}
\usetheme{Antibes}
\newcommand{\bracket}[1]{{\left\langle #1 \right\rangle}}
\usecolortheme{}
\date{}
\begin {document}
\begin {frame}
\frametitle{Equivalence with Original Method of Brackets}
\begin{itemize}
\item Single solution of Modified Method of Brackets contains the full solution obtained from Original Method of Bracket
\item If we apply Cauchy's Residue Theorem to the last 1-MB then we get three series for three set of poles to get.
\begin{equation*}
\begin{split}
G_1 = \frac{(-1)^{-D/2}(Q^2)^{D/2-a_1-a_2-a_3}\Gamma(a_1+a_2+a_3-\frac{D}{2})\Gamma(\frac{D}{2}-a_1-a_2)\Gamma(\frac{D}{2}-a_1-a_3)}{\Gamma(a_2)\Gamma(a_3)\Gamma(D-a_1-a_2-a_3)}\\\\
\times _2F_1
\left(
\begin{matrix}
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-\frac{D}{2}\\\\
\hspace{0.5cm}1+a_1+a_3-\frac{D}{2}
\end{matrix}
\Bigg{|} \frac{-M^{2}}{Q^{2}}
\right)
\end{split}
\end{equation*}
For poles: $z_2=-n-a_1-a_2-a_3+d/2$
\end{itemize}
\end{frame}
\end {document}


\documentclass{beamer}
\newcommand\Fontvi{\fontsize{2}{4}\selectfont}
\usepackage[compat=1.1.0]{tikz-feynman}
\usepackage{mathtools}
\usepackage{amsthm, amssymb,amsfonts}
\usepackage{graphicx}
\usepackage{tcolorbox}
\setcounter{MaxMatrixCols}{20}
\newtheorem{rules}{Rule}
\usetheme{Antibes}
\newcommand{\bracket}[1]{{\left\langle #1 \right\rangle}}
\begin{document}
\begin{frame}{Equivalence with Original Method of Brackets}

\begin{itemize}
\item Single solution of Modified Method of Brackets contains the full solution obtained from Original
Method of Bracket
\item If we apply Cauchy's Residue Theorem to the last 1-MB then we get three series for three set of
poles to get.
\begin{multline*}
G_1 = (-1)^{-D/2}(Q^2)^{D/2-a_1-a_2-a_3}\\
\times\frac{\Gamma(a_1+a_2+a_3-\frac{D}{2})
\Gamma(\frac{D}{2}-a_1-a_2)\Gamma(\frac{D}{2}-a_1-a_3)}
{\Gamma(a_2)\Gamma(a_3)\Gamma(D-a_1-a_2-a_3)}\\
\times{} _2F_1
\left(
\begin{matrix}
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-\frac{D}{2}\\\\
\hspace{0.5cm}1+a_1+a_3-\frac{D}{2}
\end{matrix}
\Bigg{|} \frac{-M^{2}}{Q^{2}}
\right)
\end{multline*}
For poles: $z_2=-n-a_1-a_2-a_3+d/2$
\end{itemize}
\end{frame}
\end{document}


An alignment on 3 lines, and various improvements as to the size of fractions, with the medium-size fractions frm nccmath. Also, I simplified the code, deleting the multiply loaded packages, and replacing amsmath with its extension mathtools, for its matrix* environments, which accept an optional argument for the columns alignment.

\documentclass{beamer}
\newcommand\Fontvi{\fontsize{2}{4}\selectfont}
\usepackage[utf8]{inputenc}
\usepackage{mathtools, nccmath}
\usepackage{amsthm, amssymb}
\usepackage{graphicx}
\usepackage{tcolorbox}
\usepackage{tikz}
\usepackage[compat=1.1.0]{tikz-feynman}
\setcounter{MaxMatrixCols}{20}
%\usepackage{breqn}
\newtheorem{rules}{Rule}
\usetheme{Antibes}
\newcommand{\bracket}[1]{{\left\langle #1 \right\rangle}}
\usecolortheme{}
\date{}

\begin {document}
%
\begin {frame}
\setlength{\leftmargini}{12pt}
\frametitle{Equivalence with Original Method of Brackets}
\begin{itemize}
\item Single solution of Modified Method of Brackets contains the full solution obtained from Original Method of Bracket
\item If we apply Cauchy's Residue Theorem to the last 1-MB then we get three series for three set of poles to get.
\begin{align*}
G_1 ={}&(-1)^{-D/2}(Q^2)^{D/2-a_1-a_2-a_3}\times{} \\
& \frac{\Gamma\Bigl(a_1+a_2+a_3-\mfrac{D}{2}\Bigr) \Gamma\Bigl(\mfrac{D}{2}-a_1-a_2\Bigr) \Gamma\Bigl(\mfrac{D}{2}-a_1-a_3\Bigr)}{\Gamma(a_2)\Gamma(a_3)\Gamma(D-a_1-a_2-a_3)} \times {} \\
& _2F_1
\Biggl(
\begin{matrix*}[r]
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-\mfrac{D}{2}\\[0.5ex]
1+a_1+a_3-\mfrac{D}{2}
\end{matrix*}
\Biggm | \mfrac{-M^{2}}{Q^{2}}
\Biggr)
\end{align*}
For poles: $z_2=-n-a_1-a_2-a_3+d/2$
\end{itemize}
\end{frame}

\end {document}


I'd split the thing into three lines.

\documentclass{beamer}
\usetheme{Antibes}

\begin{document}

\begin{frame}
\frametitle{Equivalence with Original Method of Brackets}

\begin{itemize}
\item Single solution of Modified Method of Brackets contains the full
solution obtained from Original Method of Bracket
\item If we apply Cauchy's Residue Theorem to the last 1-MB then we get
three series for three set of poles to get.
\begin{equation*}
\begin{split}
G_1 ={}& (-1)^{-D/2}(Q^2)^{D/2-a_1-a_2-a_3}
\\
&\times
\frac{
\Gamma(a_1+a_2+a_3-\frac{D}{2})
\Gamma(\frac{D}{2}-a_1-a_2)
\Gamma(\frac{D}{2}-a_1-a_3)
}{\Gamma(a_2)\Gamma(a_3)\Gamma(D-a_1-a_2-a_3)}
\\
&\times {}_2F_1
\left(
\begin{smallmatrix}
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-\frac{D}{2}\\
1+a_1+a_3-\frac{D}{2}
\end{smallmatrix}
\;\middle|\; \frac{-M^{2}}{Q^{2}}
\right)
\end{split}
\end{equation*}
For poles: $z_2=-n-a_1-a_2-a_3+d/2$
\end{itemize}

\end{frame}

\end{document}


Note the {} before the symbol for the hypergeometric function, that's meant to avoid the subscript being attached to \times. Note also \middle.

You can divide your equation into three part. Consider the following code:

\begin{align}
G_1 =& (-1)^{-D/2}(Q^2)^{D/2-a_1-a_2-a_3}\nonumber\\
&\times
\frac{\Gamma(a_1+a_2+a_3-\frac{D}{2})\Gamma(\frac{D}{2}-a_1-a_2)\Gamma(\frac{D}{2}-a_1-a_3)}{\Gamma(a_2)\Gamma(a_3)\Gamma(D-a_1-a_2-a_3)}
\nonumber\\
&\times _2F_1
\left(
\begin{matrix}
1+a_1+a_2+a_3-D,  a_1+a_2+a_3-\frac{D}{2}\\
\hspace{0.5cm}1+a_1+a_3-\frac{D}{2}
\end{matrix}
\Bigg{|} \frac{-M^{2}}{Q^{2}}
\right)
\end{align}