I'm trying to solve the following integral
int_0^{+infintity} 1/(1+a^2*x^2)*exp(-b^2*x^2)*cos(c*x) dx
where a,b and c are real.
thanks a lot
Marco
I though it was not necessary to have $\displaystyle b\in N$
It is not sufficient that $\displaystyle b\in R $
why do you think it will diverge? Does not the following integral converge?
$\displaystyle \int_0^{+\infty} e^{-b^2 x^2} dx $
I'm maybe missing something...
$\displaystyle \int_0^{\infty}e^{-b^2x^2}~dx=\int_0^{\infty}e^{-(bx)^2}$
Now let $\displaystyle u=bx\Rightarrow\frac{u}{b}=x$
So $\displaystyle dx=\frac{1}{b}$
so we have that when $\displaystyle x\to\infty\Rightarrow{u\to\infty}$
and when
$\displaystyle x\to{0}\Rightarrow{u\to{0}}$
So we have
$\displaystyle \frac{1}{b}\int_0^{\infty}e^{-u^2}~du=\frac{\sqrt{\pi}}{2b}$
Yeah you are right, I misread it at first as $\displaystyle b$ not $\displaystyle b^2$
I can do little more than show it converges, are you sure it is integralbe? Have you tried differentiation under the integral sign after a sub of $\displaystyle bx=u$?
$\displaystyle \forall{x}\in\mathbb{R}\quad{-1\leq\cos(cx)\leq{1}}$
So
$\displaystyle -\int_0^{\infty}\frac{e^{-b^2x^2}}{1+a^2x^2}~dx\leq\int_0^{\infty}\frac{\cos (cx)\cdot{e^{-b^2x^2}}}{1+a^2x^2}~dx\leq\int_0^{\infty}\frac{e^{-b^2x^2}}{1+a^2x^2}~dx$
Or in other words
$\displaystyle \bigg|\int_0^{\infty}\frac{\cos(cx)e^{-b^2x^2}}{1+a^2x^2}~dx\bigg|\leq\int_0^{\infty}\fra c{e^{-b^2x^2}}{1+a^2x^2}~dx$
and obviously since
$\displaystyle 0\leq\int_0^{\infty}\frac{e^{-b^2x^2}}{1+a^2x^2}~dx\ll\int_0^{\infty}\frac{dx}{1 +a^2x^2}$
Now
$\displaystyle \int_0^{\infty}\frac{dx}{1+(ax)^2}$
Let $\displaystyle ax=\varphi\Rightarrow\frac{\varphi}{a}=x$
So $\displaystyle dx=\frac{1}{a}$
And $\displaystyle \text{As }x\to\infty\Rightarrow\varphi\to\infty$
and $\displaystyle \text{As }x\to{0}\Rightarrow\varphi\to{0}$
So we have
$\displaystyle \frac{1}{a}\int_0^{\infty}\frac{d\varphi}{1+\varph i^2}=\frac{1}{a}\arctan\left(\varphi\right)\bigg|_ 0^{\infty}=\frac{\pi}{2a}$
Therefore $\displaystyle \int_0^{\infty}\frac{e^{-b^2x^2}}{1+a^2x^2}~dx$
So we can finally conclude that
$\displaystyle \int_0^{\infty}\frac{e^{-b^2x^2}\cos(cx)}{1+a^2x^2}~dx\quad\text{converges} ~\forall~(a,b,c)\in\mathbb{R}$
Furthermore we see that
$\displaystyle \bigg|\int_0^{\infty}\frac{e^{-b^2x^2}\cos(cx)}{1+a^2x^2}~dx\bigg|\leq\int_0^{\in fty}\frac{e^{-b^2x^2}}{1+a^2x^2}~dx\leq\frac{\pi}{2a}$
Hope this has been some help.