Calculate $\int\limits_{\left | z \right |=3} e^{\frac{1}{z+1}}\sin\left ( \frac{1}{z-1} \right ) dz$

Question

Calculate $\int\limits_{\left | z \right |=3} e^{\frac{1}{z+1}}\sin\left ( \frac{1}{z-1} \right ) dz$ over the positively oriented circle with radius 3.

I'm struggle with this problem. Any help please. Thank you.

Answer 1

By residue theorem (using a change of variable)

$$ \int_{\left | z \right |=3} e^{\frac{1}{z+1}}\sin\left ( \frac{1}{z-1} \right ) dz = \int_{\left | w\right |=1/3} e^{\frac{w}{w+1}}\sin\left ( \frac{w}{1-w} \right ) \frac{dw}{w^2} $$

It has a double pole at $w=0$ so the residue is $$ \lim_{w\to 0} \frac{\partial}{\partial w}(e^{\frac{w}{w+1}}sin\left ( \frac{w}{1-w} \right ) ) = 1 $$ So the integral is $2\pi i$.

Answer 2

Slightly different way although it is not as elegant as Rafael's solution:

As integrand is not holomorphic at $z =1, -1$, we just need to write the function as Laurentz series about $z=1$ and $z=-1$. We can't directly use Residues Theorem, but as we know the series expansion, we can still calculate the integral by obtaining the coefficient of $(z+1)^{-1}$ and $(z-1)^{-1}$

Write $\sin \left(\frac{1}{z-1}\right) = \frac1{2i}(e^{\frac i{z-1}} - e^{\frac {-i}{z-1}})$

Note that
$e^{{\frac i{z-1}}}=e^{{\frac i{(z+1)-2}}} = e^{{\frac i{(z+1)(1-\frac2{z+1})}}} = e^{\frac i {z+1}[ \sum_{k=0}^{\infty}\left(\frac{-2}{z+1}\right)^{k}]} = e^{\frac i {z+1}} \prod_{k=1}^{\infty}\left(e^{\frac i{z+1}}\right)^{(\frac{-2}{z+1})^k} $

From here we can obtain coefficient of $(z+1)^{-1}$ in $e^{\frac1{z+1}}e^{\frac i{z-1}}$ which is coming out to be $1+i$

Similarly, coefficient of $(z+1)^{-1}$ in $e^{\frac1{z+1}}e^{\frac {-i}{z-1}}$ is $1-i$

Going exactly like above, we can get coefficient of $(z-1)^{-1}$ in expression $e^{\frac1{z+1}}e^{\frac i{z-1}}$ and $e^{\frac1{z+1}}e^{\frac {-i}{z-1}}$ by writing $e^{\frac 1{z+1}} = e^{\frac{1}{(z-1) + 2}} = e^{\frac{1}{(z-1)(1 + \frac2{z-1})}} $

Coefficients again are coming out to be $1+i$ and $1-i$ respectively.

$\implies \int\limits_{\left | z \right |=3} e^{\frac{1}{z+1}}\sin\left ( \frac{1}{z-1} \right ) dz = 2 \pi i[\frac1{2i}\left( (1+i) - (1-i) +(1+i) - (1-i)\right)] = 4 \pi i$