how to determine the complex integral ∫f(e^it) cost dt

**sorv1986** · December 5th 2011, 09:10 PM

If f(z) is an analytic function, then what will be the value of ∫f(e^it) cost dt,

,where (limit of integration) lower limit=0, upper limit =2π.

Thanks for your help in advance. regards.

**Prove It** · December 5th 2011, 09:36 PM

Originally Posted by sorv1986

If f(z) is an analytic function, then what will be the value of ∫f(e^it) cost dt,

,where (limit of integration) lower limit=0, upper limit =2π.

Thanks for your help in advance. regards.

$\displaystyle \begin{align*} \int_0^{2\pi}{f\left(e^{it}\right)\cos{t}\,dt} &= \int_0^{2\pi}{f\left(e^{it}\right)\left(\frac{e^{i t} + e^{-it}}{2}\right)dt} \\ &= \int_0^{2\pi}{f\left(e^{it}\right)\left(\frac{1 + e^{-2it}}{2}\right)e^{it}\,dt} \\ &= \frac{1}{2i}\int_0^{2\pi}{f\left( e^{it} \right)\left[1 + \left(e^{it}\right)^{-2} \right] i\,e^{it}\,dt} \end{align*}$

Now make the substitution $\displaystyle \begin{align*} u = e^{it} \implies du = i\,e^{it}\,dt \end{align*}$ and note that $\displaystyle \begin{align*} u(0) = 1\end{align*}$ and $\displaystyle \begin{align*} u(2\pi) = 1 \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \frac{1}{2i}\int_1^1{f(u)\left(1 + u^{-2}\right)du} = 0 \end{align*}$

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