Perturbation Integral How to show that for $f(x,m)$ = $\frac{mx^2}{2} + \frac{x^4}{4}$,
the integral $Z(\lambda)$ = $\frac{1}{\sqrt \lambda}$$\int_{-\infty}^{\infty} dz e^{-f(z)/\lambda}$   is $\sqrt{\frac{m}{2\lambda}}$$e^{\frac{m^2}{8\lambda}}$
$K_{1/4}(\frac{m^2}{8\lambda})$ for $m>0$?
$m$ here is a mass term, $\lambda$ is a perturbation parameter, the integral is to be asymptotically evaluated near $\lambda = 0$.
I found this reading a paper on the "Power of perturbation series" https://arxiv.org/abs/1702.04148 .
I tried using steepest descent, but not getting this answer.
 A: Start with
\begin{eqnarray}
A &=& \frac{1}{\sqrt{\lambda}} \exp\left[-\frac{1}{\lambda}\left(\frac{m}{2} x^2 +\frac{1}{4}x^4\right)\right]
\\
&=& \frac{1}{\sqrt{\lambda}} \exp\left[\frac{m^2}{8 \lambda}\right]\exp\left\{-\frac{m^2}{8 \lambda}\left[2 \left(\frac{x^2}{m}+1 \right)^2 -1 \right]\right\} \, .
\end{eqnarray}
Then, writing $u=\frac{x}{\sqrt{m}}$ we get
\begin{eqnarray}
\int_{-\infty}^\infty \mathrm{d}x \, A &=&  \sqrt{\frac{m}{\lambda}} \exp\left[\frac{m^2}{8 \lambda}\right] \int_{-\infty}^\infty \mathrm{d}u \,\exp\left\{-\frac{m^2}{8 \lambda}\left[2 \left(u^2+1 \right)^2 -1 \right]\right\}
\\
&=& \sqrt{\frac{m}{2 \lambda}} \exp\left[\frac{m^2}{8 \lambda}\right] \int_{0}^\infty \mathrm{d}u \, 2\sqrt{2} \exp\left\{-\frac{m^2}{8 \lambda}\left[2 \left(u^2+1 \right)^2 -1 \right]\right\} \, .
\end{eqnarray}
Finish with $\cosh t = 2 \left(u^2+1 \right)^2 -1$, which corresponds to $u = \sqrt{\cosh\left[\frac{t}{2}\right]-1}$. Hence $2 \sqrt{2}\mathrm{d}u = \frac{1}{\sqrt{2}} \frac{\sinh\left[\frac{t}{2}\right]}{\sqrt{\cosh\left[\frac{t}{2}\right]-1}} \mathrm{d}t = \cosh\left[\frac{1}{4}t\right] \mathrm{d}t$. Therefore
\begin{eqnarray}
\int_{-\infty}^\infty \mathrm{d}x \, A &=& \sqrt{\frac{m}{2 \lambda}} \exp\left[\frac{m^2}{8 \lambda}\right] \int_{0}^\infty \mathrm{d}t \, \cosh\left[\frac{1}{4}t\right] \exp\left[-\frac{m^2}{8 \lambda} \cosh t\right]
\\
&=& \sqrt{\frac{m}{2 \lambda}} \exp\left[\frac{m^2}{8 \lambda}\right]  K_{\frac{1}{4}} \left[\frac{m^2}{8 \lambda}\right] \,,
\end{eqnarray}
where we used the integral representation in link.
