Evaluating the closed contour integral $\int_\gamma\frac{e^{-\sqrt{z}}}{(z-b)^{k+1}}dz$

Question

I'm hoping to evaluate the contour integral

$$\int_\gamma\frac{e^{-\sqrt{z}}}{(z-b)^{k+1}}dz$$

for $b > 0$, $k\in{\bf Z}_+$ and $\gamma$ a closed contour of the reader's choice which contains $b$.

My complex analysis is really rusty, but my understanding is that since $b$ is strictly positive, then we can choose the branch of the square root to be the negative real-axis and then make the contour small enough to avoid crossing this branch, and thus we don't have to worry about the branch cut when evaluating the integral, is that correct?

Also, does this integral have a closed form? Essentially my goal is to use it as a method to quickly evaluate the $k^{th}$ derivative of $e^{-\sqrt{z}}$.

I was thinking of writing the exponential function as a power series and then breaking the integral up and evaluating each of the resulting integrals individually,

$$\sum_{n=0}^\infty\frac{(-1)^n}{n!}\int_\gamma\frac{z^\frac{n}{2}}{(z-b)^{k+1}}dz$$

hoping that by the residue theorem most of these integrals will be zero, but like I said my complex analysis is really rusty, so I was hoping someone could help me do this.

Answer 1

As discussed in the OP, the function $f(z)=\frac{e^{-\sqrt{z}}}{(z-b)^{k+1}}$, where $b>0$ and $k$ is a positive integer is meromorphic on an open domain that excluded the origin.

Hence, if $\gamma$ is a circle centered at $b$ with radius less than $b$, then Cauchy's Integral Formula guarantees that

$$\bbox[5px,border:2px solid #C0A000]{\oint_\gamma \frac{e^{-\sqrt{z}}}{(z-b)^{k+1}}\,dz=\frac{2\pi i }{k!}\left.\left(\frac{d^k e^{-\sqrt{z}}}{dz^k}\right)\right|_{z=b} }\tag 1$$

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DERIVATIVES OF COMPOSITE FUNCTIONS: Faa di Bruno's Formula

The derivatives of a composite function $f(g(z))$ can be written using Faa di Bruno's Formula in terms of Bell polynomials as

$$\frac{d^k}{dx^k}f(g(z))=\sum_{n=1}^k f^{(n)}(g(z))B_{k,n}\left(g'(z),g''(z),\dots,g^{(k-n+1)(z)}\right) \tag 2$$

If $f(z)=e^{z}$, then $f^{(n)}(g(z))=e^{g(z)}$ and $(2)$ simplifies to

$$\frac{d^k}{dx^k}e^{g(z)}=e^{g(z)}B_k(g'(x),g''(x),\dots g^{(k)}(z)) \tag 3$$

where $B_k(x_1,\dots,x_k)$ is the $k$'th complete exponential Bell polynomial.

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PUTTING IT ALL TOGETHER:

Finally, using $(3)$ in $(1)$ yields

$$\bbox[5px,border:2px solid #C0A000]{\oint_\gamma \frac{e^{-\sqrt{z}}}{(z-b)^{k+1}}\,dz=\frac{2\pi i }{k!}e^{-\sqrt{b}}B_k\left(-\frac12b^{-1/2},\dots,\frac{(-1)^k (2k-3)!!}{2^k}b^{-(2k-1)/2}\right)} $$

Answer 2

If $k=0$ then the result is trivially $\Bbb e ^{-\sqrt b}$. Let us concentrate on $k \ge 1$.

Indeed, by choosing the contour $\gamma$ small enough as not to intersect $(-\infty, 0)$ and using the power series for the exponential, one arrives (as you have done) at

$$\sum _{n=0} ^\infty \frac {(-1)^n} {n!} \int \limits _\gamma \frac {z ^\frac n 2} {(z-b)^{k+1}} \ \Bbb dz = \sum _{n=0} ^\infty \frac {(-1)^n} {n!} (z^ \frac n 2) ^{(k)} (b) = \sum _{n=0} ^\infty \frac {(-1)^n} {n!} \ \frac n 2 \dots \left( \frac n 2 - (k-1) \right) b ^{\frac n 2 - k} = \\ \sum _{n=0} ^\infty \frac {(-1)^n} {n!} \ \frac {n (n-2) \dots (n - 2k + 2)} {2^k} b ^{\frac n 2 - k} = \frac 1 {(2b)^k} \sum _{n=0} ^\infty \frac {(-1)^n} {n!} \ \frac {n!!} {(n-2k)!!} b ^{\frac n 2} ,$$

where

$$n!! = \begin{cases} n (n-2) (n-4) \dots 2, & \text{if $n$ is even} \\ n (n-2) (n-4) \dots 1, & \text{if $n$ is odd} .\end{cases}$$