# How do you solve the cauchy integral equation?

Show that $$\frac{1}{2πi} ∮\frac{e^{zt}}{z^2+ 1}dz = \sin t$$

if $t > 0$ and $C$ is the circle $|z|=3$.

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Hint: You have two poles inside the contour, namely $z=i$ and $z=-i$. Calculate the residue at each pole and add them. Or just use the Cauchy integral formula

$$\frac{1}{2πi} ∮\frac{e^{zt}}{z^2+ 1}dz = \frac{1}{2πi}\left(\frac{-i}{2}\right) ∮\frac{e^{zt}}{z-i}dz – \frac{1}{2πi} \left(\frac{-i}{2}\right)∮\frac{e^{zt}}{z+i}dz = \dots$$

Cauchy’s Residue Theorem says the following:

Let $f$ be a holomorphic function everywhere on a circle and its interior, except maybe at the points $z_1,\dots,z_N$ where $f$ has poles of order $n_1,\dots,n_N$ (meaning $f^{-1}$ has a zero of order $n_N$). Then:

$$\frac{1}{2πi} ∮f(z) dz = \sum_{k=1}^{N} {\textrm res}_k f$$

where:

$${\textrm res}_i f = \lim_{z\rightarrow z_k}\frac{1}{(n_k-1)!}(\frac{d}{dz})^{n_k-1}(z-z_k)^{n_k}f(z)$$.

Here, $f = \dfrac{e^{zt}}{z^2+ 1}$. Compute the rest and you will see.