4
votes
Accepted
Residue at infinity is the coefficient of $z$ of the Laurent expansion?
The main point is that we calculate the residue of a meromorphic differential ($1$-form) rather than of a meromorphic function. This is the notion that stays well-defined if you change coordinates. So ...
- 121k
2
votes
How to find the series expansion or asymptotic behavior of $Q(x)=\int_{0}^{+\infty}\frac{e^{-t}}{x-t}dt, x\in(0,+\infty)$?
We can compute every thing for
$$Q(x,y)=\int_{0}^{+\infty}\frac{(x-t)\,e^{-t}}{(x-t)^2+y^2}\,dt$$
First, the antiderivative
$$\int\frac{(x-t)\,e^{-t}}{(x-t)^2+y^2}\,dt=-e^{-x}\int \frac{e^{-u}\, u}{u^...
- 274k
1
vote
Accepted
Expand $f(z) = \frac{z-\sin z}{z}$ when $0<|z|<\infty$ as Laurent series?
As suggest comments use that :
$$ \dfrac{z-\sum_{n=0}^\infty (-1)^n \dfrac{z^{2n+1}}{(n+1)!}}{z}= \sum_{n=1}^\infty (-1)^n \dfrac{z^{2n}}{(n+1)!} $$
Which gives the Laurent Series development around $...
- 2,282
1
vote
How to find the series expansion or asymptotic behavior of $Q(x)=\int_{0}^{+\infty}\frac{e^{-t}}{x-t}dt, x\in(0,+\infty)$?
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...
- 92.3k
1
vote
Accepted
Question Regarding the Existence of a Laurent Series
By directly computing you get for $0 \neq |z| < 2$
$\begin{align*}
\frac{z+4}{z(z+2)} &= \frac{1}{z} + \frac{2}{z(z+2)} = \frac{1}{z} + \frac{1}{z} - \frac{1}{z+2} = \frac{2}{z} - \frac{1}{2}\...
- 4,988
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