Mathematics Asked by Gray on December 5, 2020

I have a question where I must evaluate the following integral over a circle where $lvert z rvert = 2$.

$$I = oint frac{z^3e^{frac{1}{z}}}{1+z}dz$$

I have tried the $z^3 = 8 cdot mathrm{e}^{3it}$ approach and ended up with:

$$I = iint_0^{2pi} frac{16e^{4it}e^{frac{1}{2e^{it}}}}{1+2e^{it}}dt$$

But I have no idea what to do from here! Or if this is even the correct approach. Thanks in advance to anyone who knows

This looks ready-made for the residue theorem. I will assume you know it and go from there. The pole at $z=-1$ is simple, and its residue is straightforward:

$$operatorname*{Res}_{z=-1} frac{z^3 e^{1/z}}{1+z} = -frac1{e}$$

The pole at $z=0$, however, is essential. Still, we seek to find the coefficient of $1/z$ in the Laurent expansion of the integrand about $z=0$:

$$begin{align}frac{z^3 e^{1/z}}{1+z} &= [z^{-1}] left [z^3 (1-z+z^2-z^3+cdots) left (1+frac1{z}+frac1{2! z^2} +frac1{3! z^3}+cdots right )right ]\ &= frac1{4!} - frac1{5!}+frac1{6!}-cdots\&= frac1{e}-frac1{2!}+frac1{3!}end{align}$$

The integral is, by the residue theorem,

$$i 2 pi left ( -frac1{e} + frac1{e}-frac1{2!}+frac1{3!}right ) = -i frac{2 pi}{3}$$

Correct answer by Ron Gordon on December 5, 2020

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},} newcommand{braces}[1]{leftlbrace,{#1},rightrbrace} newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack} newcommand{dd}{mathrm{d}} newcommand{ds}[1]{displaystyle{#1}} newcommand{expo}[1]{,mathrm{e}^{#1},} newcommand{ic}{mathrm{i}} newcommand{mc}[1]{mathcal{#1}} newcommand{mrm}[1]{mathrm{#1}} newcommand{on}[1]{operatorname{#1}} newcommand{pars}[1]{left(,{#1},right)} newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}} newcommand{root}[2][]{,sqrt[#1]{,{#2},},} newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}} newcommand{verts}[1]{leftvert,{#1},rightvert}$ begin{align} I & equiv bbox[5px,#ffd]{oint_{verts{z} = 2}, {z^{3}expo{1/z} over 1 + z},dd z} ,,,stackrel{z mapsto 1/z}{=},,, oint_{verts{z} = 1/2},,, {expo{z} over z^{4}pars{1 + z}},dd z \[5mm] = & 2piic,{1 over 3!},lim_{z to 0},totald[3]{}{z} pars{expo{z} over 1 + z} = bbx{-,{2pi over 3},ic} \ & end{align}

Answered by Felix Marin on December 5, 2020

Since $ displaystyle f(z) = frac{z^{3} e^{1/z}}{1+z} $ is meromorphic outside of the contour,

$$ displaystyleint_{|z|=2} frac{z^{3} e^{1/z}}{1+z} = -2 pi i text{Res} [f(z), infty]$$

Using the transformation to find the residue at infinity,

$$ text{Res}[f(z), infty] = -text{Res} Big[frac{1}{z^{2}} f left(frac{1}{z} right),0 Big] = - text{Res} Big[ frac{e^{z}}{z^{4}(z+1)} ,0 Big]$$

$$ = - frac{1}{3!} lim_{z to 0} frac{d}{dz^{3}}frac{e^{z}}{z+1} = - frac{1}{6}lim_{z to 0} frac{e^{z}(z^{3}+3z-2)}{(z+1)^{4}} = frac{1}{3}$$

So

$$ displaystyle int_{|z|=2} frac{z^{3} e^{1/z}}{1+z} = -2 pi i left(frac{1}{3} right) = -frac{2 pi i}{3}$$

Answered by Random Variable on December 5, 2020

The "best" way to compute this integral is certainly via the residue theorem, as Ron and Random Variable did.

On can also avoid the residue theorem, as follows.

Since $e^{1/z}=sum_0^infty frac{z^{-k}}{k!}$, one can write $$I=sum_{k=0}^infty frac1{k!},int_{vert zvert=2} frac{z^{3-k}}{1+z}, dz:=sum_{k=0}^infty frac1{k!}, J_k, . $$

Next, for each fixed $k$ we have begin{eqnarray}J_k&=&int_{vert zvert=2}frac{z^{2-k}}{1+frac1z}, dz\ &=&int_{vert zvert=2}z^{2-k}sum_{n=0}^infty frac{(-1)^n}{z^n}, dz\ &=&sum_{n=0}^infty (-1)^nint_{vert zvert=2}frac{dz}{z^{n+k-2}}cdot end{eqnarray} Now, the integral $int_{vert zvert=2}frac{dz}{z^{n+k-2}}$ is equal to $0$ if $n+k-2neq 1$, i.e. $nneq 3-k$, and it is equal to $2ipi$ if $n=3-k$. However, $n=3-k$ never happens if $k>3$ (many thanks to Ron for finding the mistake in a previous version of this post!!). So we have $$J_k=(-1)^{3-k}times 2ipi=-2ipitimes (-1)^{k}quad {rm if}quad kleq 3,$$ and $J_k=0$ for all $k>3$. It follows that $$I=-2ipisum_{k=0}^3frac{(-1)^k}{k!}=-{2ipi}left(1-1+frac12-frac16right)=-frac{2ipi}3cdot $$

Answered by Etienne on December 5, 2020

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