Trying to evaluate a complex integral?

Question

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

Ron Gordon · Accepted Answer

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,

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

Felix Marin · Answer

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
 newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
 newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
 newcommand{dd}{mathrm{d}}
 newcommand{ds}[1]{displaystyle{#1}}
 newcommand{expo}[1]{,mathrm{e}^{#1},}
 newcommand{ic}{mathrm{i}}
 newcommand{mc}[1]{mathcal{#1}}
 newcommand{mrm}[1]{mathrm{#1}}
 newcommand{on}[1]{operatorname{#1}}
 newcommand{pars}[1]{left(,{#1},right)}
 newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
 newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
 newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
 newcommand{verts}[1]{leftvert,{#1},rightvert}$
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}

Random Variable · Answer

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

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

Etienne · Answer

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 
$$I=-2ipisum_{k=0}^3frac{(-1)^k}{k!}=-{2ipi}left(1-1+frac12-frac16right)=-frac{2ipi}3cdot $$

Trying to evaluate a complex integral?

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