1. Integral

Given $\displaystyle \displaystyle a,b\in\mathbb{R}$, compute $\displaystyle \displaystyle \int_0^\infty e^{-ax^2}\cos(bx)dx$.

2. Originally Posted by chiph588@
Given $\displaystyle \displaystyle a,b\in\mathbb{R}$, compute $\displaystyle \displaystyle \int_0^\infty e^{-ax^2}\cos(bx)dx$.
i think we also need $\displaystyle a > 0$ for the convergence. anyway, i think the answer is:

Spoiler:
$\displaystyle \displaystyle \frac{1}{2}\sqrt{\frac{\pi}{a}} \exp \left( \frac{-b^2}{4a} \right)$.

why? first do the substitution $\displaystyle x \to \sqrt{a}x$ and put $\displaystyle \frac{b}{\sqrt{a}}=c.$ then your integral becomes $\displaystyle f(c)=\frac{1}{\sqrt{a}} \int_0^{\infty} e^{-x^2} \cos (cx) \ dx.$

differentiating with respect to $\displaystyle c$ (check the conditions that are needed) will give us: $\displaystyle f'(c)=\frac{1}{\sqrt{a}} \int_0^{\infty} -xe^{-x^2} \sin(cx) \ dx.$

now let $\displaystyle u=\sin(cx)$ and $\displaystyle dv=-xe^{-x^2} dx$ and apply by parts to get: $\displaystyle f'(c)=\frac{-c}{2\sqrt{a}} f(c).$

solving this differential equation with the fact that $\displaystyle f(0)=\frac{1}{2} \sqrt{\frac{\pi}{a}}$ will give us the result.

3. Is...

$\displaystyle \displaystyle \int_{0}^{\infty} e^{-a\ x^{2}}\ \cos b x\ dx = \frac{1}{2} \ \mathcal{F} \{e^{-a\ x^{2}}\} = \sqrt{\frac{\pi}{4\ a}}\ e^{-\frac{b^{2}}{4a}}$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

4. Originally Posted by NonCommAlg
Spoiler:
$\displaystyle \displaystyle \frac{1}{2}\sqrt{\frac{\pi}{a}} \exp \left( \frac{-b^2}{4a} \right)$.

why? first do the substitution $\displaystyle x \to \sqrt{a}x$ and put $\displaystyle \frac{b}{\sqrt{a}}=c.$ then your integral becomes $\displaystyle f(c)=\frac{1}{\sqrt{a}} \int_0^{\infty} e^{-x^2} \cos (cx) \ dx.$

differentiating with respect to $\displaystyle c$ (check the conditions that are needed) will give us: $\displaystyle f'(c)=\frac{1}{\sqrt{a}} \int_0^{\infty} -xe^{-x^2} \sin(cx) \ dx.$

now let $\displaystyle u=\sin(cx)$ and $\displaystyle dv=-xe^{-x^2} dx$ and apply by parts to get: $\displaystyle f'(c)=\frac{-c}{2\sqrt{a}} f(c).$

solving this differential equation with the fact that $\displaystyle f(0)=\frac{1}{2} \sqrt{\frac{\pi}{a}}$ will give us the result.
Dear me, that was very clever!