# Compute the following integral over a closed curve

Let $$\gamma$$ denote the unit circle center at origin. compute the following integral. $$\int_\gamma \frac{e^z-e^{-z}}{z^4}dz$$

I guess I can solve this by integral by parts, $$\int_\gamma \frac{e^z-e^{-z}}{z^4}dz=\int^{2\pi}_0 \frac{e^{e^{it}}-e^{-e^{it}}}{e^{3it}}dt=\int^{2\pi}_0 \frac{e^{e^{it}}}{e^{3it}}dt-\int^{2\pi}_0 \frac{e^{-e^{it}}}{e^{3it}}dt$$

we first compute $$\int^{2\pi}_0 \frac{e^{e^{it}}}{e^{4it}}dt$$. Let $$u=e^{it}\implies du=ie^{it}dt$$ and $$dv=e^{3it}\implies v=e^{3it}/3i$$, hence we have $$[\frac{e^{it}e^{3it}}{3i}]^{2\pi}_0-\frac{1}{3}\int^{2\pi}_0e^{4it}dt=0$$ Similarily, $$\int^{2\pi}_0 \frac{e^{-e^{it}}}{e^{3it}}dt=0$$ Hence, we have the integral to be zero.

But I was wondering if I can use any propositions relating to holomorphic functions on a disc to conclude that the above integral is zero since $$\gamma$$ is a closed curve. If I want to compute the integral this way, how should I begin?

Thanks!

• By Residue Theorem the value of the integral is $2\pi i /3$. Oct 30, 2022 at 4:47

we first compute $$\int^{2\pi}_0 \frac{e^{e^{it}}}{e^{4it}}dt$$. Let $$u=e^{it}\implies du=ie^{it}dt$$ and $$dv=e^{3it}\implies v=e^{3it}/3i$$, hence we have $$[\frac{e^{it}e^{3it}}{3i}]^{2\pi}_0-\frac{1}{3}\int^{2\pi}_0e^{4it}dt=0$$

This is incorrect. The substitution $$u$$ should not be an exponent of the exponential function $$\exp$$ if you are doing integration by parts. (And did you mean to write $$e^{4it}$$?)

But I was wondering if I can use any propositions relating to holomorphic functions on a disc to conclude that the above integral is zero since $$\gamma$$ is a closed curve. If I want to compute the integral this way, how should I begin?

That depends on what you define your disc as. If $$f(z)$$ is holomorphic at all points interior to and on a simple closed contour $$\gamma$$, then $$\int_{\gamma}f(z)dz = 0$$. But in order to use this statement, the function to be integrated has to be holomorphic inside and on $$\gamma$$. This does not work for the unit circle centered at the origin because the function $$\dfrac{e^z-e^{-z}}{z^4}$$ has a singularity at $$z=0$$.

The best way to calculate this is through the Residue Theorem. Note that $$z=0$$ is a pole of order $$4$$. Then

$$\int_{\gamma}\frac{e^z-e^{-z}}{z^4}dz = 2\pi i \operatorname{Res}\left(\frac{e^z-e^{-z}}{(z-0)^4}, z=0\right) = 2\pi i \cdot \frac{\dfrac{d^{4-1}}{dz^{4-1}}\left[e^z-e^{-z}\right]_{z=0}}{(4-1)!} = \frac{2\pi i}{3}.$$

Using the residue theorem, we get $$2\pi i/3$$, because the residue is $$1/3$$, the coefficient of the $$z^3$$ term of $$e^z-e^{-z}$$.

(I tried to correct your integration by parts, without success.)

You can directly use Cauchy's differentiation formula: $$f^{(n)}(a) = \frac{n!}{2\pi i}\oint_{\gamma}\frac{f(z)}{(z-a)^{n+1}}$$

where

$$a=0, f(z) = e^z-e^{-z} = 2\sinh z \text{ and } \gamma: |z|=1$$

Just noting that $$\sinh^{(3)} z = \cosh z$$ you get

$$\int_\gamma \frac{e^z-e^{-z}}{z^4}dz = \frac{2\pi i}{3!}\cdot 2\cosh 0 = \frac 23 \pi i$$