Integral of $\int_{-\infty}^{\infty} \, \cos^{2}(\theta) f_{\Theta}(\theta) \, d\theta$

Question

What is the value of the integral

$$I=\int_{-\infty}^{\infty} \, \cos^{2}(\theta) f_{\Theta}(\theta) \, d\theta$$

where $f_{\Theta}(\theta)$ is the gaussian pdf with mean $0$ and variance $1$

Answer 1

By substituting $$\cos^2\theta=0.5+0.5\cos 2\theta=0.5+0.25(e^{2i\theta}+e^{-2i\theta})$$we have $$I=\int_{-\infty}^{\infty} \cos^{2}(\theta) f_{\Theta}(\theta) d\theta=0.5+0.25\int_{-\infty}^\infty {1\over \sqrt{2\pi}} e^{-\theta^2\over 2}(e^{2i\theta}+e^{-2i\theta})d\theta$$From the other side $$ \int_{-\infty}^\infty e^{-\theta^2\over 2}e^{ki\theta}d\theta {= \int_{-\infty}^\infty e^{-\theta^2\over 2}e^{ki\theta}e^{-{(ki)^2\over 2}}e^{{(ki)^2\over 2}}d\theta \\= \int_{-\infty}^\infty e^{-(\theta-ki)^2\over 2}e^{-{k^2\over 2}}d\theta \\= \int_{-\infty}^\infty e^{-\theta^2\over 2}e^{-{k^2\over 2}}d\theta \\= \sqrt{2\pi}e^{-{k^2\over 2}} } $$hence $$ I=0.5+0.5e^{-2} $$