Cauchy integral formula
Can someone please help me answer this question as I cannot seem to get to the answer. Please note that the Cauchy integral formula must be used in order to solve it.
Many thanks in advance! \begin{equation*} \int_{|z|=3}\frac{e^{zt}}{z^2+4}=\pi i\sin(2t). \end{equation*}
Also $|z| = 3$ is given the counterclockwise direction.
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$\begingroup$You should know that using the residue theorem would be easier, but if we are to restrict ourselves to Cauchy's integral formula, then here's one way of attacking it:
First, note that the integrand is a quotient of two entire functions. As such, the integrand is analytic everywhere except the points at which the denominator is zero. Since $z^2+4$ can be factored as $(z-2i)(z+2i)$, then the only points at which the integrand is not analytic are $\pm 2i$. Unfortunately, both of these points are inside the circle $|z| = 3$, so in order to apply Cauchy's integral formula, we will have to be clever.
Let $C$ be the circle $|z| = 3$ oriented counterclockwise. Let $C_1$ be the upper half of the circle $|z| = 3$ together with the line segment $[-3, 3]$ oriented counterclockwise. Let $C_2$ be the lower half together with the line segment $[-3, 3]$ oriented counterclockwise. Notice we have:
$$\int_C \frac{e^{zt}}{z^2+4} \ dz = \int_{C_1} \frac{e^{zt}}{z^2+4} \ dz + \int_{C_2} \frac{e^{zt}}{z^2+4} \ dz$$
Now we can attack the two integrals on the right hand side separately using Cauchy's integral formula. For the first, e.g., you can let $\displaystyle f(z) = \frac{e^{zt}}{z+2i}$, which is analytic everywhere inside $C_1$, and your integrand becomes $\displaystyle \frac{f(z)}{z-2i}$.
Alternative method:
Using partial fraction decomposition, we have $\displaystyle \frac{1}{z^2+4} = \frac{i}{4(z+2i)} - \frac{i}{4(z-2i)}$. Hence:
$$\int_C \frac{1}{z^2+4} \ dz = \int_C \frac{i}{4(z+2i)} \ dz - \int_C \frac{i}{4(z-2i)} \ dz$$
And then one can apply Cauchy's integral formula on the two separate pieces without having to split the contour.
$\endgroup$ 5 $\begingroup$Write integral in the form the function in the form $\frac{e^{zt}}{(z-2i)(z+2i)}$. Then you should split the contour in two parts such that interior of each part contain only one point $2i$ or $-2i$.
Then apply Cauchy integral formula for each contour.
$\endgroup$ $\begingroup$Since all the poles of the integrand are enclosed by the contour, we have \begin{eqnarray} \int_{|z|=3}\frac{e^{zt}}{z^2+4}dz=2\pi i\text{Res}_{z=0}\frac{1}{z^2}\frac{e^{\frac{t}{z}}}{\frac{1}{z^2}+4}=2\pi i\text{Res}_{z=0}\frac{e^{\frac{t}{z}}}{1+4z^2} \end{eqnarray} Note that $\displaystyle \frac{e^{\frac{t}{z}}}{1+4z^2}=\left(\sum_{n=0}^\infty (-1)^n4^nz^{2n}\right)\left(\sum_{n=0}^\infty \frac{t^n}{n! z^n}\right)$. So $\displaystyle\text{Res}_{z=0}=\sum_{n=0}^\infty \frac{(-1)^n4^nt^{2n+1}}{(2n+1)!}=\frac{1}{2}\sin 2t$
$\endgroup$ $\begingroup$Hint. The denominator $\displaystyle z^2+4=(z-2i)(z+2i)$ of the function $$ z \longmapsto \frac{e^{zt}}{z^2+4} $$ gives two poles inside $|z|=3$ and by the Cauchy integral formula we have $$ \int_{|z|=3}\frac{e^{zt}}{z^2+4}dz=2i\pi\left({\rm{Res}}_{z=-2i}f(z)+{\rm{Res}}_{z=2i}f(z)\right). $$ You conclude using for example $$ {\rm{Res}}_{z=-2i}f(z)=\lim_{z\to -2i}\left((z+2i)\times\frac{e^{zt}}{z^2+4}\right)=\ldots $$
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