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Suppose we want to find the asymptotic behavior as $n \rightarrow \infty$ of the integral $$\int_C \frac{dz}{z} \frac{e^z}{z^n}=\int_C \frac{dz}{z} \exp(z-n \ln z)$$ where $C$ is some contour in the complex plane, using the steepest descent method. The saddle point is given by $\frac{d}{dz}(z - n \ln z) = 0$, so $z = n$. We subsitute $z = n w$ in order to make the saddle point a constant ($w=1$), after which we can apply the steepest descent method.

But it is not clear what to do with integrals like $$\int_C \frac{dz}{z} \frac{e^{ e^z } }{z^n}$$

$$\int_C \frac{dz}{z} \frac{z}{e^z-1} \frac{1}{z^n}$$

where the saddle point is not as easily found and it is not clear what substitution makes the method of steepest descent applicable. In the applications I am interested in, $C$ may be taken to be a counterclockwise circle around the origin.

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Assuming that $C$ is a counterclockwise closed curve around the origin, we know that OP's integral is

$$ I ~:=~\oint_0 \frac{dz}{z} \frac{e^z}{z^n}~=~ \frac{2\pi i}{n!} ~\sim~i\sqrt{\frac{2\pi}{n}} \frac{e^n}{n^n} \quad \text{for}\quad n\to \infty , \tag{1}$$

cf. Stirling's formula. Let

$$N~:=~n+1.\tag{2}$$

Define function

$$ S(z)~:=~\ln z - \frac{z}{N}. \tag{3}$$

Then the integral (1) becomes

$$ I ~=~ \oint_0 \!dz ~e^{-N S(z)}. \tag{4}$$

We now apply the method of steepest descent. The stationary point

$$S^{\prime}(z_0)~=~0\tag{5}$$

is

$$z_0~=~N.\tag{6}$$

The Hessian

$$S^{\prime\prime}(z_0)~=~-\frac{1}{N^2}~<~0\tag{7}$$

is negative, so the steepest descent is in the imaginary direction. We choose an integration contour

$$ \mathbb{R}~\ni~y~\mapsto~z~=~z_0+iy ~\in~\mathbb{C} \tag{8}$$

to be a vertical line through the saddle point $z_0$ in the $z$-plane. Next we replace the function $S(z)$ by its quadratic approximation

$$S_2(z)~:=~S(z_0)+ S^{\prime}(z_0)(z-z_0)+\frac{S^{\prime\prime}(z_0)}{2}(z-z_0)^2~\stackrel{(8)}{=}~\ln N -1 +\frac{y^2}{2N^2},\tag{9}$$

so that the integral becomes Gaussian. Then the Gaussian integral reads

$$I_2~:=~ i \int_{\mathbb{R}} \! dy~e^{-NS_2(z)} ~\stackrel{(9)}{=}~ie^{-N\ln N +N}\int_{\mathbb{R}} \! dy ~e^{-\frac{y^2}{2N}} ~=~i\sqrt{2\pi N} \frac{e^N}{N^N}$$ $$~\stackrel{(2)}{\sim}~i\sqrt{\frac{2\pi}{n}} \frac{e^n}{n^n} \quad \text{for}\quad n\to \infty. \tag{10} $$

Now compare eqs. (1) and (10) in the large $n$ limit.

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