Contour integration around pole and essential singularity

Question

I've got the following integral:

$$ \frac{1}{2\pi i}\oint_\limits{|z|\,=\,R} \frac{z^5}{z-1} e^\frac{2}{z} dz, $$

where $R$ - sufficiently large number.

I've tried to evaluate it with residues. The integrand has a simple pole at $z = 1$ and an essential singularity at $z=0$.

Evaluation of the residue about the pole is trivial: it equals to $e^2$. I tried to deal with the singularity:

$$ \frac{z^5}{z-1} e^\frac{2}{z} = z^4 \frac{1}{1-\frac{1}{z}} e^\frac{2}{z} = \left[z^4 + z^3 + z^2 + z + 1 + \frac{1}{z} + ... \right]\left[1 + \frac{2}{z} + \frac{2}{z^2} + \frac{4}{3z^3} + \frac{2}{3z^4} + \frac{4}{15z^5} + ... \right]. $$

After term-by-term multiplication and collecting terms with $\frac{1}{z}$ I got the residue $\frac{109}{15}$. Therefore, the integral equals to $e^2 + \frac{109}{15}$.

But I am told it's wrong, and I can't find a problem for a long time. Can someone tell, is that solution wrong? Very grateful in advance for your help and time. Thanks!

Answer 1

There are two errors. A minor one is that you forgot to divide the residue by $2\pi i$. And a major one is that, near $0$, you have$$\frac{z^5}{z-1}=-z^5 - z^6 - z^7 -\cdots$$instead of$$\frac{z^5}{z-1}=z^4+z^3+z^2+\cdots$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{R\ \mbox{is a}\ \underline{\it sufficiently\ large\ number}}$: \begin{align} &\bbox[5px,#ffd]{{1 \over 2\pi\ic} \oint_{\verts{z}\ =\ R}\,\,\, {z^{5} \over z - 1}\expo{2/z}\,\dd z} \\[5mm] \stackrel{z\ \mapsto\ 1/z}{=}\,\,\,& {1 \over 2\pi\ic}\bracks{% -\oint_{\verts{z}\ =\ 1/R}\,\,\, {1/z^{5} \over 1/z - 1}\expo{2z} \,\pars{-\,{\dd z \over z^{2}}}} \\[5mm] = & {1 \over 2\pi\ic} \oint_{\verts{z}\ =\ 1/R}\,\,\, {\expo{2z} \over z^{6}\pars{1 - z}}\,\dd z \\[5mm] = & {1 \over 2\pi\ic}\bracks{2\pi\ic\,{1 \over \pars{6 - 1}!} \,\lim_{z \to 0}\,\,\totald[5]{}{z} \bracks{z^{6}\,{\expo{2z} \over z^{6}\pars{1 - z}}}} \\[5mm] = & {1 \over 120} \,\ \underbrace{\lim_{z \to 0}\,\,\totald[5]{}{z} \pars{{\expo{2z} \over 1 - z}}}_{\ds{872}}\ =\ \bbx{109 \over 15} \approx 7.2667 \end{align}