改变指数的大小

Question

当我写指数函数时

与其他数学方程相比，指数函数的大小变得很小。我该如何匹配大小？

Answer 1

\documentclass{article}

\usepackage{amsmath}

\begin{document}

\noindent Bad: \verb`e^{...}`
\begin{align*}
Z(\lambda)&=\sum_{N_+=0}^{\infty}\sum_{N_-=0}^{\infty}\frac{\lambda_c^{N_+}\lambda_a^{N_-}}{N_+!N_-!}Z\\
&=\int\mathcal{D}\phi_e(r)e^{
-\frac{\beta_e}{2}\int dr(\nabla\phi_e(r))^2-\mathrm{i}\beta\int dr\phi_e(r)\rho_f(r)
}
\end{align*}
Better: \verb`\exp\Bigl(...\Bigr)`
\begin{align*}
Z(\lambda)&=\sum_{N_+=0}^{\infty}\sum_{N_-=0}^{\infty}\frac{\lambda_c^{N_+}\lambda_a^{N_-}}{N_+!N_-!}Z\\
&=\int\mathcal{D}\phi_e(r)\exp\Bigl(
-\frac{\beta_e}{2}\int dr(\nabla\phi_e(r))^2-\mathrm{i}\beta\int dr\phi_e(r)\rho_f(r)
\Bigr)
\end{align*}
Maybe: \verb`e^{\displaystyle{...}}`
\begin{align*}
Z(\lambda)&=\sum_{N_+=0}^{\infty}\sum_{N_-=0}^{\infty}\frac{\lambda_c^{N_+}\lambda_a^{N_-}}{N_+!N_-!}Z\\
&=\int\mathcal{D}\phi_e(r)e^{\displaystyle{
-\frac{\beta_e}{2}\int dr(\nabla\phi_e(r))^2-\mathrm{i}\beta\int dr\phi_e(r)\rho_f(r)
}}
\end{align*}
Another option:
\begin{align*}
Z(\lambda)&=\sum_{N_+=0}^{\infty}\sum_{N_-=0}^{\infty}\frac{\lambda_c^{N_+}\lambda_a^{N_-}}{N_+!N_-!}Z\\
&=\int\mathcal{D}\phi_e(r)e^K,
\end{align*}
where $K=-\frac{\beta_e}{2}\int dr(\nabla\phi_e(r))^2-\mathrm{i}\beta\int dr\phi_e(r)\rho_f(r)$.

\end{document}