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(1) I would like to know how $\int_0^{2\pi} e^{imt}(ie^{it})\,dt = \int_0^{2\pi} e^{i(m+1)}dt$ holds for non-negative integer $m$. (2) I am also struggling to calculate the integral $\int_C \frac{sinz}{z^2}\,dz$ where $C(t)=2+e^{2it\pi}$ with $0\leq t\leq1$. I do have an intution that I need to use Cauchy's integral formula but it is very difficult to apply it as my book does not contain applied examples. Any help will be greatly appreciated.

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    (1): it is easy to calculate both integrals explicitly. As for the other sine integral, you can use Cauchy's theorem (not the integral formula). The integrand is holomorphic inside the contour of integration. – 2017-01-22
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    As for (1) it is difficult to see how $\int_0^{2\pi} e^{it(m+1)}idt = \int_0^{2\pi} e^{i(m+1)}dt$ holds. – 2017-01-22

1 Answers 1

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Question 1

\begin{align} \int_0^{2\pi}e^{imt}\left(ie^{it}\right)dt&=i\int_0^{2\pi}e^{i(m+1)t}dt\\ &=\frac{i}{i(m+1)}\left[e^{i(m+1)t}\right]_{t=0}^{2\pi}\\ &=\frac{1}{m+1}\left(\cos(2(m+1)\pi)-1+i\sin(2(m+1)\pi)\right)\\ &=0 \end{align}

The last line follows from the fact that $m+1$ is a whole number and $\cos(2n\pi)$ for any whole number $n$ is always equal to $1$, whereas the $\sin(2(m+1)\pi)$ is always $0$ for any whole number $m$.

\begin{align} \int_0^{2\pi}e^{i(m+1)t}dt&=\frac{1}{i(m+1)}\left[e^{i(m+1)t}\right]_{t=0}^{2\pi}\\ &=-\frac{i}{m+1}\left(\cos(2(m+1)\pi)-1+i\sin(2(m+1)\pi)\right)\\ &=0 \end{align}

by the same argument as for the first integral.

Question 2

$f(z):=\frac{\sin z}{z^2}$ is an entire function for any $z\neq0$. Your contour is a circle of radius $1$ around the point $(2,0)$ in the complex plane. Thus, no singularities of $f(z)$ are contained within $C$ and the integral evaluates to $0$.

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    Thank you very much, it makes sense. – 2017-01-22