= Solution
The exponential phase is
$$
\Phi(s)=-s+2\lambda-\frac{\lambda^2}{s}.
$$
Its <saddle points> satisfy
$$
\Phi'(s)=-1+\frac{\lambda^2}{s^2}=0,
\qquad
s=\mathord\pm\lambda.
$$
For positive real $\lambda$, the original positive-real contour passes through $s=\lambda$; the singularity at $s=0$ separates it from the other saddle. At the contributing saddle,
$$
\Phi(\lambda)=0,
\qquad
\Phi''(\lambda)=-\frac2\lambda.
$$
Set $s=\lambda+\lambda^{1/2}t$. Then
$$
\Phi(s)=-t^2+O(\lambda^{-1/2}t^3),
\qquad
ds=\lambda^{1/2}dt.
$$
The <method of steepest descent> therefore gives
$$
I(\lambda)
\sim\lambda^{-1/2}\lambda^{1/2}
\int_{-\infty}^{\infty}e^{-t^2}\,dt
=\boxed{\sqrt\pi}.
$$
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