Evaluating past exam problem: $\int_C \frac{\sin z}{(z+1)^7} \mathrm{d}z$

Question

I want to evaluate the following: $$\int_C \frac{\sin z}{(z+1)^7} \mathrm{d}z$$ Where $C$ is the circle of radius $5$, centre $0$, positively oriented.

Now this has one root at $z=-1$. Now I should probably use the Cauchy integral formula:

Then I get $$\int_C \frac{\sin z}{(z+1)^7} \mathrm{d}z=\int_C \frac{\sin z /(z+1)^6}{z-(-1)}\mathrm{d}z$$ But I have the same problem in regards to wanting to avoid $z=1$

How else should I go about this.

Answer 1

A second way of proceeding which is similar to @Icuttrees is to use the residue theorem. Here, the residue of an $n$'th order pole is given by

$$\text{Res} \left(f(z),z=z_0\right)=\frac{1}{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}\left((z-z_0)^nf(z)\right)$$

Here, we have $f(z)=\frac{\sin z}{(z+1)^7}$, $z_0=-1$, and $n=7$. Thus,

$$\text{Res} \left(\frac{\sin z}{(z+1)^7},z=-1\right)=\frac{1}{6!}\frac{d^{6}}{dx^{6}}\left(\sin z \right)=\frac{\sin 1}{6!}$$

whereupon the value of the integral of interest is

$$\oint_C \frac{\sin z}{(z+1)^7}dz=\frac{2\pi i}{6!}\sin 1$$

Answer 2

$$\int_C \frac{\sin z}{(z+1)^7} \mathrm{d}z$$ Where $C$ is the circle of radius $5$, centre $0$, positively oriented.

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By David's advice, we will use the generalized Cauchy integral formula:

$$\int_C \frac{f(z)}{(z-z_0)^{n+1}}\,dz =\frac{2\pi i}{n!}f^{(n)}(z_0)\ ,$$

Where we must note that $z_0=1$ is within our contour, that is otherwise analytic within and on the contour and is positively transversed.

$$\int_C \frac{\sin z}{(z-(-1))^{6+1}} \mathrm{d}z=\frac{2\pi i}{6!} \sin^{(6)}(-1)$$

Where $\sin^{(6)} =-\sin$, and hence we obtain $I=-\frac{2\pi i}{6!}\sin(-1)$

Answer 3

Since the coefficient of $z^6$ in $\sin(z)$ is $0$ and the coefficient of $z^6$ in $\cos(z)$ is $-\frac1{6!}$, we get $$ \newcommand{\Res}{\operatorname*{Res}} \begin{align} \Res_{z=-1}\left(\frac{\sin(z)}{(z+1)^7}\right) &=\Res_{z=0}\left(\frac{\sin(z-1)}{z^7}\right)\\ &=\Res_{z=0}\left(\frac{\sin(z)\cos(1)-\cos(z)\sin(1)}{z^7}\right)\\ &=\cos(1)\Res_{z=0}\left(\frac{\sin(z)}{z^7}\right)-\sin(1)\Res_{z=0}\left(\frac{\cos(z)}{z^7}\right)\\ &=\cos(1)\cdot0-\sin(1)\cdot\left(-\frac1{6!}\right)\\ &=\frac{\sin(1)}{720} \end{align} $$ Thus, the integral is $2\pi i$ times the sum of the residues (times the winding number of the curve around each residue): $$ \int_C\frac{\sin(z)}{(z+1)^7}\,\mathrm{d}z=\frac{\pi\sin(1)}{360}\,i $$