# Evaluate the given integral by using Cauchy's residue theorem

Evaluate the given integral by using Cauchy residue theorem

$$\int_{|z|=1} \frac{(1-\cos z)}{(e^z-1)\sin z}$$

My attempts : I take $$f(z) = \frac{(1-\cos z)}{\sin z}$$ as Residue $$z=0$$ now I got $$f(0)=0$$ $$\int_{|z|=1} \frac{(1-\cos z)}{(e^z-1)\sin z}= 2\pi if(z) _{\operatorname{res}z=0}=0$$

is its true ?

any hints/solution

• Your result is correct. Recall $1-\cos(z)=2\sin^2(z/2)$ and $\sin(z)=2\sin(z/2)\cos(z/2)$. – Mark Viola Feb 20 at 17:50

The integrand $$\frac{1-\cos(z)}{(e^z-1)\sin(z)}=\frac{\tan(z/2)}{e^z-1}$$ has a removable discontinuity at $$z=0$$. Hence, there is no singularity in or on $$|z|\le 1$$.

Cauchy's Integral Theorem guarantees that

$$\oint_{|z|=1}\frac{1-\cos(z)}{(e^z-1)\sin(z)}\,dz=0$$

No, that is not correct. The answer is indeed $$0$$, but in order to prove that the best way is to see that$$\frac{1-\cos(z)}{(e^z-1)\sin(z)}=\frac{\frac{z^2}{2}-\frac{z^4}{24}+\cdots}{z^2+\frac{z^3}{2}-\frac{z^5}{24}+\cdots}=\frac{\frac12-\frac{z^2}{24}+\cdots}{1+\frac z2-\frac{z^3}{24}+\cdots}$$and that therefore $$0$$ is a removable singularity.

Note that the singularity at $$z= 0$$ is removable, since

• $$\lim_{z\to 0}\frac{1-\cos z}{z^2} = \frac{1}{2}$$ and
• $$\lim_{z\to 0}\frac{e^z-1}{z} = 1$$ and
• $$\lim_{z\to 0}\frac{\sin z}{z} = 1$$

Now write

$$\frac{1-\cos z}{(e^z-1)\sin z}=\frac{1-\cos z}{z^2}\frac{z^2}{(e^z-1)\sin z}$$

So, the residue is equal to $$0$$ as the only singularity within the region bounded by $$|z|=1$$ is removable.

• ..@trancelocation i thinks $\lim_{z\to 0}\frac{1-e^z}{z} = 0$ – jasmine Feb 20 at 17:59
• @jasmine : No. It is $\frac{d}{dz}\left(e^z\right)$ at $z = 0$. Btw.: I edited my typo from $1-e^z$ to $e^z-1$. – trancelocation Feb 20 at 18:01
• @jasmine Using L'Hospital's Rule, we see that $\lim_{z\to0}\frac{e^z-1}{z}=\lim_{z\to0}e^z=1$. Alternatively, $e^z=1+z+O(z^2)$ so that $\frac{e^z-1}{z}=1+O(z)\to 1$ as $z\to0$. – Mark Viola Feb 20 at 18:02
• @Tancelocation ya i missed that – jasmine Feb 20 at 18:02