Integrate $\Sigma \frac{1}{n!}\int z^{n}e^{1/z}dz$

[HACR]

Homework Statement

Integrate $\Sigma \frac{1}{n!}\int z^{n}e^{1/z}dz$

Homework Equations

The Attempt at a Solution

Wrote out the first couple of terms, with $\frac{1}{z}=w$, making the integral $\Sigma \frac{1}{n!} (-w^{2-n}e^{w}+(2-n)(w^{1-n}e^{w}+(1-n)(w^{1-n}e^w)-(1-n)^2(w^(-n)e^w)...)$

But wasn't sure how to go from here.

 

[jackmell]

Homework Statement

Integrate \Sigma \frac{1}{n!}\int z^{n}e^{1/z}dz

Homework Equations

The Attempt at a Solution

Wrote out the first couple of terms, with \frac{1}{z}=w, making the integral \Sigma \frac{1}{n!} (-w^{2-n}e^{w}+(2-n)(w^{1-n}e^{w}+(1-n)(w^{1-n}e^w)-(1-n)^2(w^(-n)e^w)...)

But wasn't sure how to go from here.

Need to use tex delimiters and not $ as in:

<br /> \int<br />

Do a quote of my post to see that. Also, what happen to the integral signs. Would be easy if those were just integrations around closed contours about the origin. Otherwise, the antiderivatives are not elementary.

 

[HACR]

<br /> \Sigma \frac{1}{n!}\int z^{n}e^{1/z}dz <br /><br /> \Sigma \frac{1}{n!} (-w^{2-n}e^{w}+(2-n)(w^{1-n}e^{w}+(1-n)(w^{1-n}e^w)-(1-n)^2(w^{-n}e^w)...) <br />

Sorry I was used to a different forum. Depending on the forum, a different style is used, I guess. Thanks for pointing this out.

Yes, the contour integration is done on a unit circle, |z|=1, but it's multiply connected because of the isolated singularity at the origin.

 

[jackmell]

<br /> \Sigma \frac{1}{n!}\int z^{n}e^{1/z}dz <br />


<br /> \Sigma \frac{1}{n!} (-w^{2-n}e^{w}+(2-n)(w^{1-n}e^{w}+(1-n)(w^{1-n}e^w)-(1-n)^2(w^{-n}e^w)...) <br />

Sorry I was used to a different forum. Depending on the forum, a different style is used, I guess. Thanks for pointing this out.

Yes, the contour integration is done on a unit circle, |z|=1, but it's multiply connected because of the isolated singularity at the origin.

So you want to find:

\sum_{n=1}^{\infty}\frac{1}{n!}\mathop\oint\limits_{|z|=1} z^n e^{1/z}dz

How about using the Residue theorem? You can do that easily (I think, haven't gone over all of it yet) by taking the residue at infinity.

 

[HACR]

Right, the residue is
<br /> \frac{1}{(n+1)!}<br />

So then the expression becomes
\Sigma \frac{1}{n!(n+1)!}