2
$\begingroup$

I have to evaluate $<cos(x)>$ with $x$ distributed according to $Q^{-1}e^-{\frac{x^2}{2\sigma^2}}$.

I have gone this far:

$$<cos(x)>=\frac{\int\limits_{-\infty}^\infty \cos (x) Q^{-1}e^{-\frac{x^2}{2\sigma^2}} \mathrm{d}x}{\int\limits_{-\infty}^\infty Q^{-1}e^{-\frac{x^2}{2\sigma^2}}\mathrm{d}x}=\sigma\sqrt{2\pi}\int\limits_{-\infty}^\infty \cos (x) e^{-\frac{x^2}{2\sigma^2}}\mathrm{d}x$$

How to deal with this integral? I considered partial integration, but that seems impossible because I would have to evaluate $\cos(x)$ at infinity.

$\endgroup$
2
$\begingroup$

Since $\sin(x)$ is an odd function, we have $\int_{-\infty}^{\infty}\sin(x)\exp\{-\frac{x^2}{2\sigma^2}\}dx = 0$. Therefore, after simple algebra

$$\int_{-\infty}^{\infty}\cos(x)\exp\{-\frac{x^2}{2\sigma^2}\}dx = \int_{-\infty}^{\infty}\exp\{i x\}\exp\{-\frac{x^2}{2\sigma^2}\}dx \\ = e^{-\frac{\sigma^2}{2}} \int_{-\infty}^{\infty}\exp\{-\frac{1}{2\sigma^2}(x-i\sigma)^2\}dx\\ = \sqrt{2\pi}\sigma e^{-\frac{\sigma^2}{2}} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.