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

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.

• Your understanding of the choice of $\gamma$ around $b$ is correct. I doubt, though, that you will be able to obtain a closed form for that integral - simply because the derivative of $\Bbb e ^{\sqrt z}$ when computed naively has no closed form. For instance, if you choose $\gamma$ to be a circle, you'll get $\Bbb e ^{-\sqrt {b + \Bbb e ^{\Bbb i t}}}$ in the numerator - clearly a dead end. Commented Oct 13, 2016 at 15:12
• @AlexM. It's good to know my intuition was correct. My hope was that if I couldn't get a true closed form, that maybe in my second expression (where I've broken it up into an infinite number of integrals), that the number of these integrals which don't evaluate to zero grows linearly with $k$, this seems to be suggested by looking at the $k^{th}$ derivatives of $e^{-\sqrt{z}}$ on WolframAlpha.
– Set
Commented Oct 13, 2016 at 15:16
• @AlexM. Ah yes I see what you're saying. So I guess the question is, is there some other clever closed contour we could use besides just a circle, which would make this integral tractable.
– Set
Commented Oct 13, 2016 at 15:31
• No, you've got it all wrong. :) I'm saying that I highly doubt that there exist any contour making this integral tractable... I believe that we are forced to use power series in this problem. Commented Oct 13, 2016 at 15:34
• @AlexM. Faa di Bruno's Formula provides a closed-form expression for the $n$'th derivative of a composite function. One of its forms can be expressed in terms of Bell polynomials. While the result might not be appealing, it is nevertheless a closed form. Commented Oct 13, 2016 at 15:44

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

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

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.

PUTTING IT ALL TOGETHER:

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

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

• God damn that's dirty, thanks B, looks like at the end of it all this method is more complicated than just evaluating the derivative symbolically $k$ times.
– Set
Commented Oct 13, 2016 at 17:27
• Well, it's not more complicated if you have the complete exponential Bell polynomial, especially for "large" $k$. ;-)) Commented Oct 13, 2016 at 17:50
• And you're quite welcome. My pleasure. Just curious ... to what does "B" refer in your comment? -Mark Commented Oct 13, 2016 at 17:51
• urbandictionary.com/define.php?term=b
– Set
Commented Oct 13, 2016 at 17:57
• @FelixMarin Thank you!! Commented Oct 13, 2016 at 19:50

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

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