# Integral

• September 6th 2010, 10:10 PM
chiph588@
Integral
Given $\displaystyle a,b\in\mathbb{R}$, compute $\displaystyle \int_0^\infty e^{-ax^2}\cos(bx)dx$.
• September 6th 2010, 11:27 PM
NonCommAlg
Quote:

Originally Posted by chiph588@
Given $\displaystyle a,b\in\mathbb{R}$, compute $\displaystyle \int_0^\infty e^{-ax^2}\cos(bx)dx$.

i think we also need $a > 0$ for the convergence. anyway, i think the answer is:

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

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

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

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

solving this differential equation with the fact that $f(0)=\frac{1}{2} \sqrt{\frac{\pi}{a}}$ will give us the result.
• September 7th 2010, 07:22 AM
chisigma
Is...

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

Kind regards

$\chi$ $\sigma$
• September 7th 2010, 09:28 PM
TheCoffeeMachine
Quote:

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

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

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

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

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

Dear me, that was very clever!