For $f$ an analytic function, what is

$f$ be analytic function, could any one tell me how to find the value of $$\int_{0}^{2\pi} f(e^{it})\cos t \,\mathrm dt$$

I am not able to apply any complex analysis result here, could any one give me hint?

Solution 1:

Make a change of variables, $z = e^{it}$, which maps the interval $[0,2\pi]$ to the unit circle. You get

\begin{align} \int_0^{2\pi} f(e^{it})\cos t\,dt &= \int_0^{2\pi} f(e^{it}) \frac{e^{it}+e^{-it}}{2} \,dt \\ &=\int_{|z|=1} f(z) \frac{z+\frac1z}{2} \frac{dz}{iz} \\ &=\frac{1}{2i} \int_{|z|=1} f(z) \frac{z^2+1}{z^2}\,dz \\ &=\frac{1}{2i} \int_{|z|=1} \frac{f(z)}{z^2}\,dz = \pi f'(0) \end{align} assuming $f$ is analytic on a neighborhood of the closed unit disc (using Cauchy's integral theorem and Cauchy's integral formula for $f'$).