# An equality involving the mean value property

I am trying to establish that

$$\frac2{\pi}\int_0^{2\pi} f(e^{i\theta})\cos^2\left(\frac{\theta}{2}\right)\;\mathrm d\theta=2f(0)+f^\prime(0)$$

for $f$ analytic on $|z| < R,\; R > 1$

I think I am making a mistake somewhere, please point it out to me.Here's what I've done unsing the mean value property and Cauchy's integral formula:

$$2f(0)+f'(0)=2(\frac1{2\pi}\int_0^{2\pi} f(e^{i\theta})d\theta + \frac1{2\pi i}\int_0^{2\pi}\frac{f(e^{i\theta})ie^{i\theta}}{e^{i2\theta}}=\frac{1}{\pi}\int_0^{2\pi} f(e^{i\theta})d\theta\frac1{2\pi}\int_0^{2\pi}\frac{f(e^{i\theta})}{e^{i\theta}}\quad\text{(1)}$$

But when I develop the expression $\frac2{\pi}\int_0^{2\pi} f(e^{i\theta})\cos^2(\frac{\theta}{2}) d\theta$ I get:

$$\frac1{\pi}\int_0^{2\pi} f(e^{i\theta})d\theta+\frac1{2\pi}\int_0^{2\pi}f(e^{i\theta})[{e^{i\theta}}+e^{i-\theta}]d\theta \quad\text{(2)}$$

The second integral in (1) differs from the second integral in (2). Thanks.

Ooh, interesting formula! Let me try...

$$\frac2{\pi}\int_0^{2\pi} f(e^{i\theta})\cos^2\left(\frac{\theta}{2}\right)\;\mathrm d\theta=\frac2{\pi}\int_0^{2\pi} f(e^{i\theta})\left(\frac{1+\cos\,\theta}{2}\right)\;\mathrm d\theta$$

$$=\frac1{\pi}\int_0^{2\pi} f(e^{i\theta})(1+\cos\,\theta)\;\mathrm d\theta$$

$$=\frac1{\pi}\left(\int_0^{2\pi} f(e^{i\theta})\;\mathrm d\theta+\int_0^{2\pi} f(e^{i\theta})\cos\,\theta\;\mathrm d\theta\right)$$

Now, let's look at the Cauchy formula

$$f(0)=\frac1{2\pi i}\oint_\gamma \frac{f(u)}{u}\mathrm du=\frac1{2\pi i}\int_0^{2\pi} \frac{f(\exp(i\theta))}{\exp(i\theta)}\mathrm d(\exp(i\theta))=\frac1{2\pi}\int_0^{2\pi}f(\exp(i\theta))\mathrm d\theta$$

and going back to the original expression we have

$$2f(0)+\frac1{\pi}\int_0^{2\pi} f(\exp(i\theta))\cos\,\theta\;\mathrm d\theta$$

How to take care of that other piece... let's start at that Cauchy formula again:

$$f(z)=\frac1{2\pi i}\oint_\gamma \frac{f(u)}{u-z}\mathrm du$$

If we differentiate with respect to $z$, we have

$$f^{\prime}(z)=\frac1{2\pi i}\oint_\gamma \frac{f(u)}{(u-z)^2}\mathrm du$$

and now we can try substituting in an anticlockwise unit-circular contour yet again:

$$f^{\prime}(0)=\frac1{2\pi i}\int_0^{2\pi} \frac{f(\exp(i\theta))}{\exp(2i\theta)}\mathrm d(\exp(i\theta))$$

$$=\frac1{2\pi}\int_0^{2\pi} f(\exp(i\theta))\exp(-i\theta)\mathrm d\theta$$

$$=\frac1{2\pi}\int_0^{2\pi} f(\exp(i\theta))\cos\,\theta\;\mathrm d\theta-\frac{i}{2\pi}\int_0^{2\pi} f(\exp(i\theta))\sin\,\theta\;\mathrm d\theta$$

Luckily, the sine is antisymmetric about $\theta=\pi$, so that the imaginary part is easily removed:

$$f^{\prime}(0)=\frac1{2\pi}\int_0^{2\pi} f(\exp(i\theta))\cos\,\theta\;\mathrm d\theta$$

Long story short... the correct expression is actually $2f(0)+2f^\prime(0)$...