= Solution
Set $d=1/\sqrt2$. At the positive real endpoint the contour is real and crosses the <saddle point> at zero. The <Gaussian integral> gives
$$
f(0,\lambda)=\sqrt{\frac\pi\lambda}-\int_d^\infty e^{-\lambda x^2}dx-\int_1^\infty e^{-\lambda x^2}dx\sim\boxed{\sqrt{\frac\pi\lambda}}.
$$
Both omitted tails are exponentially small. At the negative real endpoint the contour is oriented leftward and does not cross the <saddle point>:
$$
f(\pi,\lambda)=-\int_d^1e^{-\lambda x^2}dx.
$$
The <endpoint contribution in steepest descent>, or one <integration by parts>, gives $\int_d^\infty e^{-\lambda x^2}dx\sim e^{-\lambda d^2}/(2\lambda d)$. The upper-end correction is $O(e^{-\lambda}/\lambda)$ and is much smaller. Therefore
$$
\boxed{f(\pi,\lambda)\sim-\frac{e^{-\lambda/2}}{\sqrt2\,\lambda}.}
$$
The sign comes from contour orientation, not from a different Gaussian convention.
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